Explanation:
Okay, so I need to find the maximum value of the expression |1 + z| + |1 - z + z²| where z is a complex number with |z| = 1. Hmm, let me start by recalling that if |z| = 1, then z lies on the unit circle in the complex plane. That means I can represent z as e^(iθ) where θ is the argument of z. Maybe that will help me convert the problem into trigonometric terms?

Let me write z = e^(iθ). Then, I can express each modulus in terms of θ. Let me compute each part step by step.

First, compute |1 + z|. Since z = e^(iθ), then 1 + z = 1 + e^(iθ). The modulus of this is |1 + e^(iθ)|. I remember that |1 + e^(iθ)| can be simplified using the formula for the modulus of a complex number. Let me recall that |a + b|² = |a|² + |b|² + 2Re(a\overline{b}). But maybe there's a simpler way. Alternatively, since 1 + e^(iθ) is 2cos(θ/2) e^(iθ/2), right? Because e^(iθ) + 1 = e^(iθ/2)(e^(-iθ/2) + e^(iθ/2)) = 2cos(θ/2) e^(iθ/2). Therefore, the modulus is 2|cos(θ/2)|. So |1 + z| = 2|cos(θ/2)|. Wait, but since modulus is always non-negative, maybe I can drop the absolute value? Wait, no, because cos(θ/2) can be positive or negative. But since modulus is non-negative, it's 2|cos(θ/2)|. Okay, so that's the first term.

Now, the second term is |1 - z + z²|. Let me try to simplify this expression. Let me write it as |z² - z + 1|. Maybe I can factor this or find a way to express it in terms of θ. Let me try substituting z = e^(iθ). Then z² = e^(i2θ), so the expression becomes |e^(i2θ) - e^(iθ) + 1|. Let me compute this modulus. Let me denote this as |1 - e^(iθ) + e^(i2θ)|. Maybe I can write this as |e^(i2θ) - e^(iθ) + 1|. Let me see if there's a way to combine terms or use trigonometric identities.

Alternatively, maybe I can factor this expression. Let me think. Let me consider the polynomial z² - z + 1. The roots of z² - z + 1 = 0 are [1 ± sqrt(1 - 4)] / 2 = [1 ± i√3]/2. So they are complex numbers. But I don't know if that helps here. Alternatively, maybe I can write the expression as z² - z + 1. Let me try to write it in terms of z + 1/z or something? Wait, since |z| = 1, then z̄ = 1/z. Maybe that can help. Let me try multiplying numerator and denominator by something? Wait, but here we have modulus, so maybe I can compute the modulus squared?

Let me compute |1 - z + z²|². Let me expand that. Let me recall that |a + b + c|² = |a|² + |b|² + |c|² + 2Re(ab̄ + ac̄ + bc̄). But maybe it's easier to compute directly. Let me write 1 - z + z² as z² - z + 1. Let me compute its modulus squared:

|z² - z + 1|² = (z² - z + 1)(\overline{z²} - \overline{z} + 1). Since |z| = 1, \overline{z} = 1/z, so \overline{z²} = 1/z². Therefore, substituting:

(z² - z + 1)(1/z² - 1/z + 1). Let me compute this product.

First, multiply term by term:

First term: z² * 1/z² = 1

Second term: z² * (-1/z) = -z

Third term: z² * 1 = z²

Fourth term: (-z) * 1/z² = -1/z

Fifth term: (-z) * (-1/z) = 1

Sixth term: (-z) * 1 = -z

Seventh term: 1 * 1/z² = 1/z²

Eighth term: 1 * (-1/z) = -1/z

Ninth term: 1 * 1 = 1

Wait, that seems complicated. Maybe there's a better way. Alternatively, maybe I can write the expression as z² - z + 1 and then compute its modulus squared. Let me note that for any complex number w, |w|² = w * \overline{w}. So here, w = z² - z + 1, so |w|² = (z² - z + 1)(\overline{z}² - \overline{z} + 1). Since |z| = 1, \overline{z} = 1/z, so \overline{z}² = 1/z². Therefore:

|w|² = (z² - z + 1)(1/z² - 1/z + 1). Let me compute this:

Multiply (z² - z + 1) by (1/z² - 1/z + 1). Let me denote A = z² - z + 1 and B = 1/z² - 1/z + 1. Let me compute A*B.

First, expand A*B:

= z²*(1/z²) + z²*(-1/z) + z²*1 + (-z)*(1/z²) + (-z)*(-1/z) + (-z)*1 + 1*(1/z²) + 1*(-1/z) + 1*1

Wait, that's the same as before. Let me compute term by term:

First term: z² * 1/z² = 1

Second term: z² * (-1/z) = -z

Third term: z² * 1 = z²

Fourth term: (-z) * 1/z² = -1/z

Fifth term: (-z) * (-1/z) = 1

Sixth term: (-z) * 1 = -z

Seventh term: 1 * 1/z² = 1/z²

Eighth term: 1 * (-1/z) = -1/z

Ninth term: 1 * 1 = 1

Now, let me combine all these terms:

1 - z + z² - 1/z + 1 - z + 1/z² - 1/z + 1

Combine like terms:

Constants: 1 + 1 + 1 = 3

Terms with z²: z²

Terms with -z: -z - z = -2z

Terms with 1/z: -1/z -1/z = -2/z

Terms with 1/z²: 1/z²

So altogether:

z² + 1/z² - 2z - 2/z + 3

Hmm, okay. Now, let me note that z² + 1/z² can be written as (z + 1/z)^2 - 2. Similarly, -2z - 2/z = -2(z + 1/z). Let me check:

(z + 1/z)^2 = z² + 2*(z)*(1/z) + 1/z² = z² + 2 + 1/z². Therefore, z² + 1/z² = (z + 1/z)^2 - 2. So substituting:

(z + 1/z)^2 - 2 - 2(z + 1/z) + 3 = (z + 1/z)^2 - 2(z + 1/z) + 1.

So |w|² = (z + 1/z)^2 - 2(z + 1/z) + 1. Let me denote y = z + 1/z. Then |w|² = y² - 2y + 1 = (y - 1)^2. Therefore, |w| = |y - 1|. But y = z + 1/z. Since |z| = 1, z + 1/z = z + \overline{z} = 2 Re(z) = 2 cos θ. Therefore, y = 2 cos θ. Therefore, |w| = |2 cos θ - 1|. Wait, so |1 - z + z²| = |2 cos θ - 1|? Wait, but modulus is non-negative, so |w| = |2 cos θ - 1|. Wait, but let me check that again. Because |w|² = (y - 1)^2, so |w| = |y - 1|. Since y = 2 cos θ, then |w| = |2 cos θ - 1|. Therefore, |1 - z + z²| = |2 cos θ - 1|. Wait, is that correct? Let me check with a specific value. Let me take θ = 0. Then z = 1. Then 1 - z + z² = 1 - 1 + 1 = 1, so |1| = 1. On the other hand, 2 cos 0 - 1 = 2*1 - 1 = 1. So that works. Let me take θ = π. Then z = -1. Then 1 - z + z² = 1 - (-1) + 1 = 1 + 1 + 1 = 3. On the other hand, 2 cos π - 1 = 2*(-1) - 1 = -3, absolute value is 3. That works. Another test: θ = π/2. Then z = i. Then 1 - z + z² = 1 - i + (-1) = 0 - i, modulus is 1. On the other hand, 2 cos(π/2) - 1 = 0 - 1 = -1, absolute value is 1. That works too. Okay, so that seems correct. Therefore, |1 - z + z²| = |2 cos θ - 1|.

So putting it all together, the original expression is |1 + z| + |1 - z + z²| = 2|cos(θ/2)| + |2 cos θ - 1|. Therefore, the problem reduces to finding the maximum value of 2|cos(θ/2)| + |2 cos θ - 1| where θ is a real number (since z is on the unit circle).

Now, I need to find the maximum of this function. Let me denote f(θ) = 2|cos(θ/2)| + |2 cos θ - 1|. To find its maximum, perhaps I can analyze it over θ in [0, 2π), since cosine is periodic with period 2π. Also, since cos(θ/2) has period 4π, but since θ is in [0, 2π), θ/2 is in [0, π), so maybe I can consider θ in [0, 2π) and analyze the function accordingly.

But maybe it's easier to consider θ in [0, π] and [π, 2π] separately, since cos(θ/2) is non-negative in [0, π] and non-positive in [π, 2π], but since it's inside absolute value, maybe we can write 2|cos(θ/2)| as 2 cos(θ/2) when θ ∈ [0, π], and 2 cos(θ/2) when θ ∈ [π, 2π] but cos(θ/2) is negative there, so absolute value would make it positive. Wait, θ ∈ [π, 2π], θ/2 ∈ [π/2, π], so cos(θ/2) is negative there. Therefore, |cos(θ/2)| = -cos(θ/2). Therefore, 2|cos(θ/2)| = -2 cos(θ/2) for θ ∈ [π, 2π]. Therefore, maybe I can split the function into two cases:

Case 1: θ ∈ [0, π], so f(θ) = 2 cos(θ/2) + |2 cos θ - 1|

Case 2: θ ∈ [π, 2π], so f(θ) = -2 cos(θ/2) + |2 cos θ - 1|

But maybe even better to consider θ ∈ [0, π] and θ ∈ [π, 2π], but also analyze the absolute value term |2 cos θ - 1|. Let me check when 2 cos θ - 1 is positive or negative. Let me solve 2 cos θ - 1 ≥ 0 => cos θ ≥ 1/2. Which occurs when θ ∈ [-π/3, π/3] modulo 2π. Therefore, in [0, 2π), cos θ ≥ 1/2 when θ ∈ [0, π/3] ∪ [5π/3, 2π). Therefore, in these intervals, |2 cos θ - 1| = 2 cos θ - 1, and in [π/3, 5π/3], it's 1 - 2 cos θ.

Therefore, combining this with the previous cases, maybe I can split the domain into more intervals. Let me try to structure this.

First, let me consider θ ∈ [0, π/3]. In this interval, cos θ ≥ 1/2, so |2 cos θ - 1| = 2 cos θ - 1. Also, θ ∈ [0, π], so 2|cos(θ/2)| = 2 cos(θ/2). Therefore, f(θ) = 2 cos(θ/2) + 2 cos θ - 1.

Next, θ ∈ [π/3, π]. Here, cos θ ≤ 1/2, so |2 cos θ - 1| = 1 - 2 cos θ. Also, θ ∈ [0, π], so 2|cos(θ/2)| = 2 cos(θ/2). Therefore, f(θ) = 2 cos(θ/2) + 1 - 2 cos θ.

Then, θ ∈ [π, 5π/3]. Here, cos θ ≤ 1/2, so |2 cos θ - 1| = 1 - 2 cos θ. Also, θ ∈ [π, 2π], so 2|cos(θ/2)| = -2 cos(θ/2). Therefore, f(θ) = -2 cos(θ/2) + 1 - 2 cos θ.

Finally, θ ∈ [5π/3, 2π]. Here, cos θ ≥ 1/2, so |2 cos θ - 1| = 2 cos θ - 1. Also, θ ∈ [π, 2π], so 2|cos(θ/2)| = -2 cos(θ/2). Therefore, f(θ) = -2 cos(θ/2) + 2 cos θ - 1.

Therefore, the function f(θ) can be broken down into four intervals:

1. [0, π/3]: f(θ) = 2 cos(θ/2) + 2 cos θ - 1

2. [π/3, π]: f(θ) = 2 cos(θ/2) + 1 - 2 cos θ

3. [π, 5π/3]: f(θ) = -2 cos(θ/2) + 1 - 2 cos θ

4. [5π/3, 2π]: f(θ) = -2 cos(θ/2) + 2 cos θ - 1

Now, to find the maximum of f(θ), I need to analyze each interval separately and find the maximum in each, then compare them.

Let me start with interval 1: [0, π/3]. Let me denote f1(θ) = 2 cos(θ/2) + 2 cos θ - 1. Let me compute its derivative to find critical points.

First, compute f1’(θ):

d/dθ [2 cos(θ/2)] = 2 * (-sin(θ/2)) * (1/2) = -sin(θ/2)

d/dθ [2 cos θ] = -2 sin θ

d/dθ [-1] = 0

Therefore, f1’(θ) = -sin(θ/2) - 2 sin θ

Set derivative to zero:

-sin(θ/2) - 2 sin θ = 0 => sin(θ/2) + 2 sin θ = 0

Let me solve this equation for θ in [0, π/3].

Let me express sin θ in terms of sin(θ/2). Recall that sin θ = 2 sin(θ/2) cos(θ/2). Therefore:

sin(θ/2) + 2*(2 sin(θ/2) cos(θ/2)) = 0 => sin(θ/2) + 4 sin(θ/2) cos(θ/2) = 0

Factor out sin(θ/2):

sin(θ/2) [1 + 4 cos(θ/2)] = 0

Therefore, either sin(θ/2) = 0 or 1 + 4 cos(θ/2) = 0.

In the interval θ ∈ [0, π/3], θ/2 ∈ [0, π/6]. Therefore, sin(θ/2) = 0 only when θ = 0. The other equation: 1 + 4 cos(θ/2) = 0 => cos(θ/2) = -1/4. However, θ/2 ∈ [0, π/6], so cos(θ/2) is in [cos(π/6), 1] = [√3/2 ≈ 0.866, 1]. Therefore, cos(θ/2) cannot be -1/4 here. Therefore, the only critical point in this interval is θ = 0.

Therefore, in interval [0, π/3], the critical points are θ = 0 and endpoints θ = π/3. Let me compute f1(θ) at these points.

At θ = 0:

f1(0) = 2 cos(0) + 2 cos(0) - 1 = 2*1 + 2*1 - 1 = 2 + 2 - 1 = 3.

At θ = π/3:

Compute cos(π/6) = √3/2 ≈ 0.866, cos(π/3) = 1/2.

So f1(π/3) = 2*(√3/2) + 2*(1/2) - 1 = √3 + 1 - 1 = √3 ≈ 1.732.

Therefore, in this interval, the maximum is 3 at θ = 0.

Now, moving to interval 2: [π/3, π]. Here, f2(θ) = 2 cos(θ/2) + 1 - 2 cos θ. Let me compute its derivative.

f2’(θ) = derivative of 2 cos(θ/2) is -sin(θ/2), derivative of 1 is 0, derivative of -2 cos θ is 2 sin θ. Therefore:

f2’(θ) = -sin(θ/2) + 2 sin θ

Set derivative to zero:

-sin(θ/2) + 2 sin θ = 0 => 2 sin θ = sin(θ/2)

Again, express sin θ in terms of sin(θ/2):

sin θ = 2 sin(θ/2) cos(θ/2). Therefore:

2*(2 sin(θ/2) cos(θ/2)) = sin(θ/2) => 4 sin(θ/2) cos(θ/2) = sin(θ/2)

Assuming sin(θ/2) ≠ 0, we can divide both sides by sin(θ/2):

4 cos(θ/2) = 1 => cos(θ/2) = 1/4

Therefore, θ/2 = arccos(1/4) => θ = 2 arccos(1/4). Let me compute this value. Let me note that arccos(1/4) is some angle between 0 and π/2, since 1/4 is between 0 and 1. Therefore, θ = 2 arccos(1/4) is between 0 and π. Let me check if this θ lies in [π/3, π]. Since arccos(1/4) ≈ 1.318 radians (since cos(π/3) = 0.5, cos(π/2) = 0, so arccos(1/4) is between π/3 and π/2). Therefore, θ ≈ 2 * 1.318 ≈ 2.636 radians. π is approximately 3.1416, so 2.636 is between π/3 ≈ 1.047 and π. Therefore, this critical point is within the interval [π/3, π].

Therefore, critical points in this interval are θ = 2 arccos(1/4) and endpoints θ = π/3 and θ = π.

Let me compute f2(θ) at these points.

First, compute θ = 2 arccos(1/4). Let me denote α = arccos(1/4), so θ = 2α. Then cos(α) = 1/4, so cos(θ/2) = cos α = 1/4. Also, sin α = sqrt(1 - (1/4)^2) = sqrt(15)/4. Then, sin θ = sin(2α) = 2 sin α cos α = 2*(sqrt(15)/4)*(1/4) = sqrt(15)/8. Wait, but maybe I need to compute f2(θ) = 2 cos(θ/2) + 1 - 2 cos θ. Let me compute each term:

First, 2 cos(θ/2) = 2*(1/4) = 1/2.

Second, 1 is just 1.

Third, -2 cos θ. Let me compute cos θ. Since θ = 2α, cos θ = cos(2α) = 2 cos² α - 1 = 2*(1/4)^2 - 1 = 2*(1/16) - 1 = 1/8 - 1 = -7/8. Therefore, -2 cos θ = -2*(-7/8) = 14/8 = 7/4.

Therefore, f2(θ) = 1/2 + 1 + 7/4 = (2/4 + 4/4 + 7/4) = 13/4 = 3.25. So 3.25 is 13/4.

Now, check endpoints:

At θ = π/3:

f2(π/3) = 2 cos(π/6) + 1 - 2 cos(π/3) = 2*(√3/2) + 1 - 2*(1/2) = √3 + 1 - 1 = √3 ≈ 1.732.

At θ = π:

f2(π) = 2 cos(π/2) + 1 - 2 cos π = 2*0 + 1 - 2*(-1) = 0 + 1 + 2 = 3.

Therefore, in interval [π/3, π], the maximum is 13/4 = 3.25 at θ = 2 arccos(1/4).

Moving on to interval 3: [π, 5π/3]. Here, f3(θ) = -2 cos(θ/2) + 1 - 2 cos θ. Let me compute its derivative.

f3’(θ) = derivative of -2 cos(θ/2) is -2*(-sin(θ/2))*(1/2) = sin(θ/2)

Derivative of 1 is 0, derivative of -2 cos θ is 2 sin θ. Therefore:

f3’(θ) = sin(θ/2) + 2 sin θ

Set derivative to zero:

sin(θ/2) + 2 sin θ = 0

Again, express sin θ as 2 sin(θ/2) cos(θ/2):

sin(θ/2) + 2*(2 sin(θ/2) cos(θ/2)) = 0 => sin(θ/2) + 4 sin(θ/2) cos(θ/2) = 0

Factor out sin(θ/2):

sin(θ/2) [1 + 4 cos(θ/2)] = 0

Therefore, either sin(θ/2) = 0 or 1 + 4 cos(θ/2) = 0.

In interval θ ∈ [π, 5π/3], θ/2 ∈ [π/2, 5π/6]. Therefore, sin(θ/2) is non-negative here (since θ/2 is between π/2 and 5π/6, where sine is positive). Therefore, sin(θ/2) = 0 only when θ/2 = π, i.e., θ = 2π, which is the endpoint of the interval [π, 5π/3]. The other equation: 1 + 4 cos(θ/2) = 0 => cos(θ/2) = -1/4. Let me check if this is possible. θ/2 ∈ [π/2, 5π/6], so cos(θ/2) ∈ [cos(5π/6), cos(π/2)] = [-√3/2, 0]. Since -1/4 is within [-√3/2 ≈ -0.866, 0], so yes, there is a solution here. Let me find θ such that cos(θ/2) = -1/4. Let me denote β = θ/2, so β ∈ [π/2, 5π/6], and cos β = -1/4. Therefore, β = arccos(-1/4). Let me compute arccos(-1/4). Since cos(π - arccos(1/4)) = -cos(arccos(1/4)) = -1/4. Therefore, β = π - arccos(1/4). Therefore, θ = 2β = 2π - 2 arccos(1/4). Let me check if this θ is in [π, 5π/3]. Let me compute 2π - 2 arccos(1/4). Since arccos(1/4) ≈ 1.318 radians, so 2π - 2*1.318 ≈ 6.283 - 2.636 ≈ 3.647 radians. Now, π ≈ 3.1416, 5π/3 ≈ 5.236. Therefore, 3.647 is between π and 5π/3. Therefore, this critical point is within the interval.

Therefore, critical points in this interval are θ = 2π - 2 arccos(1/4) and endpoints θ = π and θ = 5π/3.

Let me compute f3(θ) at these points.

First, at θ = 2π - 2 arccos(1/4). Let me denote β = arccos(1/4), so θ = 2π - 2β. Then θ/2 = π - β. Therefore, cos(θ/2) = cos(π - β) = -cos β = -1/4. Also, sin(θ/2) = sin(π - β) = sin β = sqrt(1 - (1/4)^2) = sqrt(15)/4. Also, cos θ = cos(2π - 2β) = cos(2β) = 2 cos² β - 1 = 2*(1/4)^2 - 1 = 2*(1/16) - 1 = 1/8 - 1 = -7/8. Therefore, compute f3(θ):

-2 cos(θ/2) + 1 - 2 cos θ = -2*(-1/4) + 1 - 2*(-7/8) = 1/2 + 1 + 14/8 = 1/2 + 1 + 7/4 = (2/4 + 4/4 + 7/4) = 13/4 = 3.25. Same as before.

Now, check endpoints:

At θ = π:

f3(π) = -2 cos(π/2) + 1 - 2 cos π = -2*0 + 1 - 2*(-1) = 0 + 1 + 2 = 3.

At θ = 5π/3:

Compute cos(θ/2) = cos(5π/6) = -√3/2. Therefore, -2 cos(θ/2) = -2*(-√3/2) = √3.

Compute cos θ = cos(5π/3) = cos(π/3) = 1/2. Therefore, -2 cos θ = -2*(1/2) = -1. Therefore, f3(5π/3) = √3 + 1 - 1 = √3 ≈ 1.732.

Therefore, in interval [π, 5π/3], the maximum is 13/4 = 3.25 at θ = 2π - 2 arccos(1/4).

Now, moving to interval 4: [5π/3, 2π]. Here, f4(θ) = -2 cos(θ/2) + 2 cos θ - 1. Let me compute its derivative.

f4’(θ) = derivative of -2 cos(θ/2) is -2*(-sin(θ/2))*(1/2) = sin(θ/2)

Derivative of 2 cos θ is -2 sin θ

Derivative of -1 is 0. Therefore:

f4’(θ) = sin(θ/2) - 2 sin θ

Set derivative to zero:

sin(θ/2) - 2 sin θ = 0 => sin(θ/2) = 2 sin θ

Again, express sin θ as 2 sin(θ/2) cos(θ/2):

sin(θ/2) = 2*(2 sin(θ/2) cos(θ/2)) => sin(θ/2) = 4 sin(θ/2) cos(θ/2)

Assuming sin(θ/2) ≠ 0, divide both sides by sin(θ/2):

1 = 4 cos(θ/2) => cos(θ/2) = 1/4

Therefore, θ/2 = arccos(1/4) or θ/2 = -arccos(1/4). But θ ∈ [5π/3, 2π], so θ/2 ∈ [5π/6, π]. Therefore, arccos(1/4) is in [0, π/2], so θ/2 = arccos(1/4) would be in [0, π/2], which is not in [5π/6, π]. Therefore, no solution here. Alternatively, maybe θ/2 = 2π - arccos(1/4)? Wait, cos(θ/2) = 1/4. Since θ/2 ∈ [5π/6, π], cos(θ/2) is negative there (since cos(π) = -1, cos(5π/6) = -√3/2 ≈ -0.866). But 1/4 is positive, so there is no solution here. Therefore, the only critical points are when sin(θ/2) = 0. In θ ∈ [5π/3, 2π], θ/2 ∈ [5π/6, π]. sin(θ/2) is non-negative here (since θ/2 is between 5π/6 and π, where sine is positive). Therefore, sin(θ/2) = 0 only when θ/2 = π, i.e., θ = 2π, which is the endpoint. Therefore, the critical points in this interval are θ = 2π and endpoints θ = 5π/3 and θ = 2π. Wait, but θ = 2π is the same as θ = 0, but since we are considering θ ∈ [5π/3, 2π], the endpoints are θ = 5π/3 and θ = 2π.

Compute f4(θ) at these points.

At θ = 5π/3:

f4(5π/3) = -2 cos(5π/6) + 2 cos(5π/3) - 1 = -2*(-√3/2) + 2*(1/2) - 1 = √3 + 1 - 1 = √3 ≈ 1.732.

At θ = 2π:

f4(2π) = -2 cos(π) + 2 cos(2π) - 1 = -2*(-1) + 2*1 - 1 = 2 + 2 - 1 = 3.

Therefore, in interval [5π/3, 2π], the maximum is 3 at θ = 2π.

So summarizing the maximum values in each interval:

- Interval 1: 3

- Interval 2: 13/4 = 3.25

- Interval 3: 13/4 = 3.25

- Interval 4: 3

Therefore, the overall maximum is 13/4, which is 3.25. However, let me check if there are any other critical points or if I missed something. Wait, but in intervals 2 and 3, we found critical points with value 13/4, which is higher than 3. So 13/4 is 3.25, which is higher than 3. Therefore, the maximum value is 13/4. However, let me check if there are any other critical points or if I made a mistake in calculations.

Wait, but 13/4 is 3.25, which is 3 + 1/4. Let me check if that's correct. Let me recheck the calculation for the critical point in interval 2. When θ = 2 arccos(1/4), we had:

f2(θ) = 2 cos(θ/2) + 1 - 2 cos θ. We found that cos(θ/2) = 1/4, so 2*(1/4) = 1/2. Then cos θ = -7/8, so -2 cos θ = 14/8 = 7/4. Then total is 1/2 + 1 + 7/4 = 1/2 + 4/4 + 7/4 = (2/4 + 4/4 + 7/4) = 13/4. Yes, that's correct. Similarly for interval 3, same value. So 13/4 is indeed the maximum.

But let me check if there are other points where the maximum could be higher. For example, maybe at θ = π/2? Let me compute f(π/2). Let me compute |1 + z| + |1 - z + z²| when θ = π/2. z = i.

Compute |1 + i| = sqrt(1² + 1²) = √2 ≈ 1.414.

Compute |1 - i + i²| = |1 - i -1| = |-i| = 1. Therefore, total is √2 + 1 ≈ 2.414, which is less than 3.25. So no.

Another test: θ = π/2. Wait, already did that. How about θ = π/3? As before, gives √3 ≈ 1.732. How about θ = 0? Gives 3. How about θ = π? Gives 3. So 3.25 is higher. So seems like 13/4 is the maximum.

But let me check another angle. Let me take θ where cos θ = 1/2, which is θ = π/3 or 5π/3. Wait, already checked those. How about θ where cos θ = something else?

Alternatively, maybe I can think geometrically. Let me think about the original problem: |1 + z| + |1 - z + z²|. Since |z| = 1, maybe there's a geometric interpretation?

Alternatively, maybe I can parametrize z as e^(iθ) and write the expression in terms of θ, then find maximum. But I think the previous analysis is thorough. However, let me check if 13/4 is indeed the maximum. Let me compute 13/4 = 3.25. Let me check if there's a way to get higher.

Alternatively, maybe I can consider specific values. For example, suppose θ is such that z is a root of some equation? Not sure. Alternatively, think of the expression as |1 + z| + |z² - z + 1|. Maybe there's a way to relate these terms?

Alternatively, note that z² - z + 1 = (z³ + 1)/(z + 1) when z ≠ -1. Wait, let me check:

(z³ - 1) = (z - 1)(z² + z + 1). But here we have z² - z + 1. Not sure. Alternatively, maybe not helpful.

Alternatively, think of z² - z + 1. Let me write it as z² + 1 - z. Since |z| = 1, z² = \overline{z}^{-2}? Wait, no. Wait, z² = z * z, and since |z| = 1, z̄ = 1/z. Therefore, z² = z * z, but not sure.

Alternatively, maybe use triangle inequality? But since we have sum of two moduli, not sure.

Alternatively, think of |1 + z| + |z² - z + 1|. Maybe express z² - z + 1 in terms of z + 1/z? Wait, since z² + 1/z² = (z + 1/z)^2 - 2. Wait, earlier we had |z² - z + 1| = |2 cos θ - 1|. But maybe not helpful here.

Alternatively, since we already did the calculus approach and found 13/4 as maximum, which is 3.25, and verified with specific points, maybe that's the answer. However, let me check if 13/4 is indeed the maximum. Let me compute 13/4 = 3.25. Let me check if there's a way to get higher.

Alternatively, think of θ where the two terms are maximized. For example, |1 + z| is maximized when z = 1, giving 2, and |1 - z + z²| when z = 1 is |1 - 1 + 1| = 1, so total 3. When z = -1, |1 + (-1)| = 0, |1 - (-1) + 1| = |1 + 1 + 1| = 3, total 3. So maximum of 3. But we found 3.25, which is higher. Therefore, that seems possible.

Alternatively, maybe check θ where the critical point occurs. Let me compute θ = 2 arccos(1/4). Let me compute arccos(1/4) ≈ 1.318 radians, so θ ≈ 2.636 radians. Let me compute z = e^(iθ). Let me compute |1 + z| and |1 - z + z²| numerically.

First, compute θ ≈ 2.636 radians. Let me compute cos θ and sin θ. Since θ ≈ 2.636 radians, which is approximately 151 degrees (since π radians is 180 degrees, so 2.636 radians ≈ 151 degrees). Let me compute cos θ ≈ cos(2.636). Let me use calculator:

cos(2.636) ≈ cos(π - 0.505) ≈ -cos(0.505) ≈ -0.876. Wait, but earlier we had cos θ = -7/8 = -0.875. So that's consistent. Therefore, cos θ ≈ -0.875, sin θ ≈ sin(2.636) ≈ sin(π - 0.505) ≈ sin(0.505) ≈ 0.486. Therefore, z ≈ cos θ + i sin θ ≈ -0.875 + i*0.486.

Compute |1 + z|: 1 + z ≈ 1 - 0.875 + i*0.486 = 0.125 + i*0.486. The modulus is sqrt(0.125² + 0.486²) ≈ sqrt(0.0156 + 0.236) ≈ sqrt(0.2516) ≈ 0.5016. Wait, but according to our previous calculation, |1 + z| = 2|cos(θ/2)|. Let me compute θ/2 ≈ 1.318 radians, cos(θ/2) = 1/4 = 0.25, so 2*0.25 = 0.5. Which matches the approximate value. So |1 + z| ≈ 0.5.

Then |1 - z + z²|. Let me compute z². Since z ≈ -0.875 + i*0.486, z² = (-0.875)^2 - (0.486)^2 + 2*(-0.875)*(0.486)i ≈ 0.7656 - 0.236 - 0.848i ≈ 0.5296 - 0.848i. Then 1 - z + z² ≈ 1 - (-0.875 + i*0.486) + (0.5296 - 0.848i) ≈ 1 + 0.875 - i*0.486 + 0.5296 - 0.848i ≈ (1 + 0.875 + 0.5296) + (-0.486 - 0.848)i ≈ 2.4046 - 1.334i. The modulus is sqrt(2.4046² + 1.334²) ≈ sqrt(5.78 + 1.78) ≈ sqrt(7.56) ≈ 2.75. Which is approximately 13/4 = 3.25? Wait, no, 2.75 is 11/4. Wait, but according to our previous calculation, |1 - z + z²| = |2 cos θ - 1|. Let me compute 2 cos θ - 1 ≈ 2*(-0.875) - 1 = -1.75 - 1 = -2.75, so absolute value is 2.75. Therefore, total expression is |1 + z| + |1 - z + z²| ≈ 0.5 + 2.75 = 3.25, which is 13/4. So that checks out. Therefore, the calculation is correct.

Therefore, the maximum value is 13/4. However, let me check if 13/4 is indeed the maximum. Let me think if there's a way to get higher. For example, suppose θ is such that both terms are maximized? But since they are dependent on θ, it's not straightforward. However, since we did the calculus and found critical points, and verified with specific examples, I think 13/4 is correct.

But let me check another critical point. For example, suppose θ where cos θ = something else. Wait, but according to our analysis, the maximum occurs at those critical points. Therefore, I think 13/4 is the correct answer.

But let me check if 13/4 is the maximum. Let me convert 13/4 to decimal, which is 3.25. Let me check if there's a case where the sum is higher. For example, suppose θ where |1 + z| is large and |1 - z + z²| is also large. For example, suppose z is such that both terms are large. However, since |1 + z| is maximized at 2 (when z = 1), and |1 - z + z²| is 1 at that point. If z is somewhere else, maybe both terms can be larger? For example, if z is such that 1 + z is large in magnitude and 1 - z + z² is also large. But since |z| = 1, maybe not. Let me think.

Alternatively, think of the expression as |1 + z| + |z² - z + 1|. Let me try to write z² - z + 1 in terms of z + 1/z. Wait, since z² + 1/z² = (z + 1/z)^2 - 2. But z² - z + 1 = z² + 1 - z. Not sure. Alternatively, maybe write z² - z + 1 = z² + 1 - z. Since |z| = 1, z² = \overline{z}^{-2}? Wait, no. Wait, z² = z * z, and since |z| = 1, z̄ = 1/z. Therefore, z² = z * z, and z̄² = 1/z². Not sure.

Alternatively, maybe consider writing the entire expression in terms of real and imaginary parts. Let me try that. Let me let z = x + iy, with x² + y² = 1. Then compute |1 + z| + |1 - z + z²|.

First, |1 + z| = sqrt((1 + x)^2 + y^2). Since x² + y² = 1, this becomes sqrt(1 + 2x + x² + y²) = sqrt(1 + 2x + 1) = sqrt(2 + 2x) = sqrt(2(1 + x)) = √2 * sqrt(1 + x). Wait, that's interesting. Wait, let me check:

(1 + x)^2 + y^2 = 1 + 2x + x² + y² = 1 + 2x + (x² + y²) = 1 + 2x + 1 = 2 + 2x. Yes, so |1 + z| = sqrt(2 + 2x) = √(2(1 + x)) = √2 * sqrt(1 + x). Since x = Re(z), which is cos θ, so this matches our previous result of 2|cos(θ/2)|. Because sqrt(2(1 + cos θ)) = sqrt(2 * 2 cos²(θ/2)) = sqrt(4 cos²(θ/2)) = 2|cos(θ/2)|. Yes, that's correct.

Now, compute |1 - z + z²|. Let me compute 1 - z + z². Let me write z = x + iy. Then z² = (x + iy)^2 = x² - y² + 2ixy. Therefore, 1 - z + z² = 1 - x - iy + x² - y² + 2ixy. Combine real and imaginary parts:

Real part: 1 - x + x² - y²

Imaginary part: -y + 2xy

Since x² + y² = 1, we can replace y² with 1 - x². Therefore, real part becomes 1 - x + x² - (1 - x²) = 1 - x + x² - 1 + x² = 2x² - x.

Imaginary part: -y + 2xy = y(-1 + 2x).

Therefore, |1 - z + z²| = sqrt[(2x² - x)^2 + (y(-1 + 2x))^2]. Let me compute this:

= sqrt{(2x² - x)^2 + y²(2x - 1)^2}

But since y² = 1 - x², substitute:

= sqrt{(2x² - x)^2 + (1 - x²)(2x - 1)^2}

Let me expand this expression:

First, compute (2x² - x)^2:

= 4x^4 - 4x^3 + x²

Then compute (1 - x²)(2x - 1)^2:

First compute (2x - 1)^2 = 4x² - 4x + 1

Multiply by (1 - x²):

= (1 - x²)(4x² - 4x + 1) = 4x²(1 - x²) - 4x(1 - x²) + 1*(1 - x²)

= 4x² - 4x^4 - 4x + 4x^3 + 1 - x²

Combine like terms:

-4x^4 + 4x^3 + (4x² - x²) + (-4x) + 1

= -4x^4 + 4x^3 + 3x² - 4x + 1

Therefore, total expression under sqrt is:

(4x^4 - 4x^3 + x²) + (-4x^4 + 4x^3 + 3x² - 4x + 1) =

Let me add term by term:

4x^4 - 4x^4 = 0

-4x^3 + 4x^3 = 0

x² + 3x² = 4x²

-4x remains

+1 remains

Therefore, total expression under sqrt is 4x² - 4x + 1. Therefore, |1 - z + z²| = sqrt(4x² - 4x + 1). Let me note that 4x² - 4x + 1 = (2x - 1)^2. Therefore, |1 - z + z²| = |2x - 1|. Since modulus is non-negative, it's |2x - 1|. Therefore, this matches our previous result where we had |2 cos θ - 1|, since x = cos θ. Therefore, this confirms that |1 - z + z²| = |2x - 1| = |2 cos θ - 1|.

Therefore, the original expression is √2 * sqrt(1 + x) + |2x - 1|, where x = cos θ and x ∈ [-1, 1]. Therefore, we can rephrase the problem as finding the maximum of f(x) = √2 * sqrt(1 + x) + |2x - 1| for x ∈ [-1, 1].

This seems like a simpler one-variable calculus problem. Let me try this approach.

So f(x) = √2 * sqrt(1 + x) + |2x - 1|. Let me analyze this function over x ∈ [-1, 1].

First, note that 2x - 1 changes sign at x = 1/2. Therefore, we can split the domain into two intervals:

1. x ∈ [-1, 1/2], where 2x - 1 ≤ 0, so |2x - 1| = 1 - 2x.

2. x ∈ [1/2, 1], where 2x - 1 ≥ 0, so |2x - 1| = 2x - 1.

Therefore, f(x) can be written as:

For x ∈ [-1, 1/2]: f(x) = √2 * sqrt(1 + x) + 1 - 2x

For x ∈ [1/2, 1]: f(x) = √2 * sqrt(1 + x) + 2x - 1

Now, let's find the maximum of f(x) in each interval.

First, interval [-1, 1/2]:

Let me denote f1(x) = √2 * sqrt(1 + x) + 1 - 2x. Find its maximum on [-1, 1/2].

Compute derivative f1’(x):

d/dx [√2 * sqrt(1 + x)] = √2 * (1/(2 sqrt(1 + x))) = √2 / (2 sqrt(1 + x)) = 1 / sqrt(2(1 + x))

Derivative of 1 - 2x is -2.

Therefore, f1’(x) = 1 / sqrt(2(1 + x)) - 2.

Set derivative to zero:

1 / sqrt(2(1 + x)) - 2 = 0 => 1 / sqrt(2(1 + x)) = 2 => sqrt(2(1 + x)) = 1/2 => 2(1 + x) = 1/4 => 1 + x = 1/8 => x = 1/8 - 1 = -7/8.

Check if x = -7/8 is in [-1, 1/2]. Yes, since -7/8 ≈ -0.875 is between -1 and 0.5.

Therefore, critical points at x = -7/8 and endpoints x = -1 and x = 1/2.

Compute f1(-7/8):

First, compute sqrt(1 + x) = sqrt(1 - 7/8) = sqrt(1/8) = 1/(2√2). Therefore, √2 * sqrt(1 + x) = √2 * 1/(2√2) = 1/2.

Then, 1 - 2x = 1 - 2*(-7/8) = 1 + 14/8 = 1 + 7/4 = 11/4.

Therefore, f1(-7/8) = 1/2 + 11/4 = 13/4 = 3.25.

At x = -1:

f1(-1) = √2 * sqrt(0) + 1 - 2*(-1) = 0 + 1 + 2 = 3.

At x = 1/2:

f1(1/2) = √2 * sqrt(1 + 1/2) + 1 - 2*(1/2) = √2 * sqrt(3/2) + 1 - 1 = √2*(√(3)/√2) + 0 = √3 ≈ 1.732.

Therefore, maximum in this interval is 13/4 at x = -7/8.

Now, interval [1/2, 1]:

f2(x) = √2 * sqrt(1 + x) + 2x - 1. Find its maximum on [1/2, 1].

Compute derivative f2’(x):

d/dx [√2 * sqrt(1 + x)] = 1 / sqrt(2(1 + x)) as before.

Derivative of 2x - 1 is 2.

Therefore, f2’(x) = 1 / sqrt(2(1 + x)) + 2.

Since 1 / sqrt(2(1 + x)) is always positive for x ∈ [1/2, 1], and adding 2, the derivative is always positive. Therefore, f2(x) is increasing on [1/2, 1]. Therefore, maximum occurs at x = 1.

Compute f2(1) = √2 * sqrt(2) + 2*1 - 1 = √2 * √2 + 2 - 1 = 2 + 1 = 3.

At x = 1/2:

f2(1/2) = √2 * sqrt(1 + 1/2) + 2*(1/2) - 1 = √2 * sqrt(3/2) + 1 - 1 = √3 ≈ 1.732.

Therefore, maximum in this interval is 3 at x = 1.

Therefore, overall maximum of f(x) is 13/4 = 3.25, achieved at x = -7/8. Therefore, this confirms our previous result. Therefore, the maximum value is 13/4.

Therefore, the answer is \boxed{\dfrac{13}{4}}.

**Final Answer**
\boxed{\dfrac{13}{4}}

To solve the problem, we are given a complex number $ z $ such that $ |z| = 1 $, and we are to find the **maximum value** of the expression:

$$
|1 + z| + |1 - z + z^2|
$$

---

### Step 1: Parametrize $ z $ on the Unit Circle

Since $ |z| = 1 $, we can write $ z = e^{i\theta} $, where $ \theta \in [0, 2\pi) $. This allows us to express the modulus terms in terms of trigonometric functions.

---

### Step 2: Simplify Each Term

#### First Term: $ |1 + z| $

We use the identity:
$$
|1 + e^{i\theta}| = 2|\cos(\theta/2)|
$$
This is derived from the identity:
$$
1 + e^{i\theta} = 2e^{i\theta/2} \cos(\theta/2)
$$
So,
$$
|1 + z| = 2|\cos(\theta/2)|
$$

#### Second Term: $ |1 - z + z^2| $

We simplify the expression:
$$
1 - z + z^2 = z^2 - z + 1
$$
Using the identity $ z^2 + \frac{1}{z^2} = (z + \frac{1}{z})^2 - 2 $, and noting that $ \frac{1}{z} = \overline{z} $, we can write:
$$
|z^2 - z + 1| = |2\cos\theta - 1|
$$
This is derived by expressing the modulus in terms of real and imaginary parts and simplifying.

---

### Step 3: Combine the Two Terms

Now, the expression becomes:
$$
f(\theta) = 2|\cos(\theta/2)| + |2\cos\theta - 1|
$$

We aim to find the **maximum** of this function over $ \theta \in [0, 2\pi) $.

---

### Step 4: Analyze the Function

Let’s define:
$$
f(\theta) = 2|\cos(\theta/2)| + |2\cos\theta - 1|
$$

We analyze this function by splitting the domain based on the behavior of the absolute value terms.

#### Case 1: $ \theta \in [0, \pi] $

Here, $ \cos(\theta/2) \geq 0 $, so $ |\cos(\theta/2)| = \cos(\theta/2) $, and we can write:
$$
f(\theta) = 2\cos(\theta/2) + |2\cos\theta - 1|
$$

#### Case 2: $ \theta \in [\pi, 2\pi] $

Here, $ \cos(\theta/2) \leq 0 $, so $ |\cos(\theta/2)| = -\cos(\theta/2) $, and we write:
$$
f(\theta) = -2\cos(\theta/2) + |2\cos\theta - 1|
$$

We also note that $ |2\cos\theta - 1| $ changes sign at $ \cos\theta = \frac{1}{2} $, i.e., at $ \theta = \frac{\pi}{3} $ and $ \theta = \frac{5\pi}{3} $.

---

### Step 5: Use Calculus to Find Maximum

We analyze the function in each interval and find critical points by taking derivatives. After careful analysis, we find that the **maximum** occurs at:

$$
\theta = 2\arccos\left(\frac{1}{4}\right)
$$

At this point, we compute:

- $ \cos(\theta/2) = \frac{1}{4} $
- $ \cos\theta = -\frac{7}{8} $

Substituting into the expression:

$$
f(\theta) = 2 \cdot \frac{1}{4} + |2 \cdot (-\frac{7}{8}) - 1| = \frac{1}{2} + \left| -\frac{7}{4} - 1 \right| = \frac{1}{2} + \frac{11}{4} = \frac{13}{4}
$$

---

### Final Answer

$$
\boxed{\dfrac{13}{4}}
$$

[Assistant]
Explanation:
Okay, so I need to find the maximum value of the expression |1 + z| + |1 - z + z²| where z is a complex number with |z| = 1. Hmm, let me start by recalling that if |z| = 1, then z lies on the unit circle in the complex plane. That means I can represent z as e^(iθ) where θ is the argument of z. Maybe that will help me convert the problem into trigonometric terms?

Let me write z = e^(iθ). Then, I can express each modulus in terms of θ. Let me compute each part step by step.

First, compute |1 + z|. Since z = e^(iθ), then 1 + z = 1 + e^(iθ). The modulus of this is |1 + e^(iθ)|. I remember that |1 + e^(iθ)| can be simplified using the formula for the modulus of a complex number. Let me recall that |a + b|² = |a|² + |b|² + 2Re(a\overline{b}). But maybe there's a simpler way. Alternatively, since 1 + e^(iθ) is 2cos(θ/2) e^(iθ/2), right? Because e^(iθ) + 1 = e^(iθ/2)(e^(-iθ/2) + e^(iθ/2)) = 2cos(θ/2) e^(iθ/2). Therefore, the modulus is 2|cos(θ/2)|. So |1 + z| = 2|cos(θ/2)|. Wait, but since modulus is always non-negative, maybe I can drop the absolute value? Wait, no, because cos(θ/2) can be positive or negative. But since modulus is non-negative, it's 2|cos(θ/2)|. Okay, so that's the first term.

Now, the second term is |1 - z + z²|. Let me try to simplify this expression. Let me write it as |z² - z + 1|. Maybe I can factor this or find a way to express it in terms of θ. Let me try substituting z = e^(iθ). Then z² = e^(i2θ), so the expression becomes |e^(i2θ) - e^(iθ) + 1|. Let me compute this modulus. Let me denote this as |1 - e^(iθ) + e^(i2θ)|. Maybe I can write this as |e^(i2θ) - e^(iθ) + 1|. Let me see if there's a way to combine terms or use trigonometric identities.

Alternatively, maybe I can factor this expression. Let me think. Let me consider the polynomial z² - z + 1. The roots of z² - z + 1 = 0 are [1 ± sqrt(1 - 4)] / 2 = [1 ± i√3]/2. So they are complex numbers. But I don't know if that helps here. Alternatively, maybe I can write the expression as z² - z + 1. Let me try to write it in terms of z + 1/z or something? Wait, since |z| = 1, then z̄ = 1/z. Maybe that can help. Let me try multiplying numerator and denominator by something? Wait, but here we have modulus, so maybe I can compute the modulus squared?

Let me compute |1 - z + z²|². Let me expand that. Let me recall that |a + b + c|² = |a|² + |b|² + |c|² + 2Re(ab̄ + ac̄ + bc̄). But maybe it's easier to compute directly. Let me write 1 - z + z² as z² - z + 1. Let me compute its modulus squared:

|z² - z + 1|² = (z² - z + 1)(\overline{z²} - \overline{z} + 1). Since |z| = 1, \overline{z} = 1/z, so \overline{z²} = 1/z². Therefore, substituting:

(z² - z + 1)(1/z² - 1/z + 1). Let me compute this product.

First, multiply term by term:

First term: z² * 1/z² = 1

Second term: z² * (-1/z) = -z

Third term: z² * 1 = z²

Fourth term: (-z) * 1/z² = -1/z

Fifth term: (-z) * (-1/z) = 1

Sixth term: (-z) * 1 = -z

Seventh term: 1 * 1/z² = 1/z²

Eighth term: 1 * (-1/z) = -1/z

Ninth term: 1 * 1 = 1

Wait, that seems complicated. Maybe there's a better way. Alternatively, maybe I can write the expression as z² - z + 1 and then compute its modulus squared. Let me note that for any complex number w, |w|² = w * \overline{w}. So here, w = z² - z + 1, so |w|² = (z² - z + 1)(\overline{z}² - \overline{z} + 1). Since |z| = 1, \overline{z} = 1/z, so \overline{z}² = 1/z². Therefore:

|w|² = (z² - z + 1)(1/z² - 1/z + 1). Let me compute this:

Multiply (z² - z + 1) by (1/z² - 1/z + 1). Let me denote A = z² - z + 1 and B = 1/z² - 1/z + 1. Let me compute A*B.

First, expand A*B:

= z²*(1/z²) + z²*(-1/z) + z²*1 + (-z)*(1/z²) + (-z)*(-1/z) + (-z)*1 + 1*(1/z²) + 1*(-1/z) + 1*1

Wait, that's the same as before. Let me compute term by term:

First term: z² * 1/z² = 1

Second term: z² * (-1/z) = -z

Third term: z² * 1 = z²

Fourth term: (-z) * 1/z² = -1/z

Fifth term: (-z) * (-1/z) = 1

Sixth term: (-z) * 1 = -z

Seventh term: 1 * 1/z² = 1/z²

Eighth term: 1 * (-1/z) = -1/z

Ninth term: 1 * 1 = 1

Now, let me combine all these terms:

1 - z + z² - 1/z + 1 - z + 1/z² - 1/z + 1

Combine like terms:

Constants: 1 + 1 + 1 = 3

Terms with z²: z²

Terms with -z: -z - z = -2z

Terms with 1/z: -1/z -1/z = -2/z

Terms with 1/z²: 1/z²

So altogether:

z² + 1/z² - 2z - 2/z + 3

Hmm, okay. Now, let me note that z² + 1/z² can be written as (z + 1/z)^2 - 2. Similarly, -2z - 2/z = -2(z + 1/z). Let me check:

(z + 1/z)^2 = z² + 2*(z)*(1/z) + 1/z² = z² + 2 + 1/z². Therefore, z² + 1/z² = (z + 1/z)^2 - 2. So substituting:

(z + 1/z)^2 - 2 - 2(z + 1/z) + 3 = (z + 1/z)^2 - 2(z + 1/z) + 1.

So |w|² = (z + 1/z)^2 - 2(z + 1/z) + 1. Let me denote y = z + 1/z. Then |w|² = y² - 2y + 1 = (y - 1)^2. Therefore, |w| = |y - 1|. But y = z + 1/z. Since |z| = 1, z + 1/z = z + \overline{z} = 2 Re(z) = 2 cos θ. Therefore, y = 2 cos θ. Therefore, |w| = |2 cos θ - 1|. Wait, so |1 - z + z²| = |2 cos θ - 1|? Wait, but modulus is non-negative, so |w| = |2 cos θ - 1|. Wait, but let me check that again. Because |w|² = (y - 1)^2, so |w| = |y - 1|. Since y = 2 cos θ, then |w| = |2 cos θ - 1|. Therefore, |1 - z + z²| = |2 cos θ - 1|. Wait, is that correct? Let me check with a specific value. Let me take θ = 0. Then z = 1. Then 1 - z + z² = 1 - 1 + 1 = 1, so |1| = 1. On the other hand, 2 cos 0 - 1 = 2*1 - 1 = 1. So that works. Let me take θ = π. Then z = -1. Then 1 - z + z² = 1 - (-1) + 1 = 1 + 1 + 1 = 3. On the other hand, 2 cos π - 1 = 2*(-1) - 1 = -3, absolute value is 3. That works. Another test: θ = π/2. Then z = i. Then 1 - z + z² = 1 - i + (-1) = 0 - i, modulus is 1. On the other hand, 2 cos(π/2) - 1 = 0 - 1 = -1, absolute value is 1. That works too. Okay, so that seems correct. Therefore, |1 - z + z²| = |2 cos θ - 1|.

So putting it all together, the original expression is |1 + z| + |1 - z + z²| = 2|cos(θ/2)| + |2 cos θ - 1|. Therefore, the problem reduces to finding the maximum value of 2|cos(θ/2)| + |2 cos θ - 1| where θ is a real number (since z is on the unit circle).

Now, I need to find the maximum of this function. Let me denote f(θ) = 2|cos(θ/2)| + |2 cos θ - 1|. To find its maximum, perhaps I can analyze it over θ in [0, 2π), since cosine is periodic with period 2π. Also, since cos(θ/2) has period 4π, but since θ is in [0, 2π), θ/2 is in [0, π), so maybe I can consider θ in [0, 2π) and analyze the function accordingly.

But maybe it's easier to consider θ in [0, π] and [π, 2π] separately, since cos(θ/2) is non-negative in [0, π] and non-positive in [π, 2π], but since it's inside absolute value, maybe we can write 2|cos(θ/2)| as 2 cos(θ/2) when θ ∈ [0, π], and 2 cos(θ/2) when θ ∈ [π, 2π] but cos(θ/2) is negative there, so absolute value would make it positive. Wait, θ ∈ [π, 2π], θ/2 ∈ [π/2, π], so cos(θ/2) is negative there. Therefore, |cos(θ/2)| = -cos(θ/2). Therefore, 2|cos(θ/2)| = -2 cos(θ/2) for θ ∈ [π, 2π]. Therefore, maybe I can split the function into two cases:

Case 1: θ ∈ [0, π], so f(θ) = 2 cos(θ/2) + |2 cos θ - 1|

Case 2: θ ∈ [π, 2π], so f(θ) = -2 cos(θ/2) + |2 cos θ - 1|

But maybe even better to consider θ ∈ [0, π] and θ ∈ [π, 2π], but also analyze the absolute value term |2 cos θ - 1|. Let me check when 2 cos θ - 1 is positive or negative. Let me solve 2 cos θ - 1 ≥ 0 => cos θ ≥ 1/2. Which occurs when θ ∈ [-π/3, π/3] modulo 2π. Therefore, in [0, 2π), cos θ ≥ 1/2 when θ ∈ [0, π/3] ∪ [5π/3, 2π). Therefore, in these intervals, |2 cos θ - 1| = 2 cos θ - 1, and in [π/3, 5π/3], it's 1 - 2 cos θ.

Therefore, combining this with the previous cases, maybe I can split the domain into more intervals. Let me try to structure this.

First, let me consider θ ∈ [0, π/3]. In this interval, cos θ ≥ 1/2, so |2 cos θ - 1| = 2 cos θ - 1. Also, θ ∈ [0, π], so 2|cos(θ/2)| = 2 cos(θ/2). Therefore, f(θ) = 2 cos(θ/2) + 2 cos θ - 1.

Next, θ ∈ [π/3, π]. Here, cos θ ≤ 1/2, so |2 cos θ - 1| = 1 - 2 cos θ. Also, θ ∈ [0, π], so 2|cos(θ/2)| = 2 cos(θ/2). Therefore, f(θ) = 2 cos(θ/2) + 1 - 2 cos θ.

Then, θ ∈ [π, 5π/3]. Here, cos θ ≤ 1/2, so |2 cos θ - 1| = 1 - 2 cos θ. Also, θ ∈ [π, 2π], so 2|cos(θ/2)| = -2 cos(θ/2). Therefore, f(θ) = -2 cos(θ/2) + 1 - 2 cos θ.

Finally, θ ∈ [5π/3, 2π]. Here, cos θ ≥ 1/2, so |2 cos θ - 1| = 2 cos θ - 1. Also, θ ∈ [π, 2π], so 2|cos(θ/2)| = -2 cos(θ/2). Therefore, f(θ) = -2 cos(θ/2) + 2 cos θ - 1.

Therefore, the function f(θ) can be broken down into four intervals:

1. [0, π/3]: f(θ) = 2 cos(θ/2) + 2 cos θ - 1

2. [π/3, π]: f(θ) = 2 cos(θ/2) + 1 - 2 cos θ

3. [π, 5π/3]: f(θ) = -2 cos(θ/2) + 1 - 2 cos θ

4. [5π/3, 2π]: f(θ) = -2 cos(θ/2) + 2 cos θ - 1

Now, to find the maximum of f(θ), I need to analyze each interval separately and find the maximum in each, then compare them.

Let me start with interval 1: [0, π/3]. Let me denote f1(θ) = 2 cos(θ/2) + 2 cos θ - 1. Let me compute its derivative to find critical points.

First, compute f1’(θ):

d/dθ [2 cos(θ/2)] = 2 * (-sin(θ/2)) * (1/2) = -sin(θ/2)

d/dθ [2 cos θ] = -2 sin θ

d/dθ [-1] = 0

Therefore, f1’(θ) = -sin(θ/2) - 2 sin θ

Set derivative to zero:

-sin(θ/2) - 2 sin θ = 0 => sin(θ/2) + 2 sin θ = 0

Let me solve this equation for θ in [0, π/3].

Let me express sin θ in terms of sin(θ/2). Recall that sin θ = 2 sin(θ/2) cos(θ/2). Therefore:

sin(θ/2) + 2*(2 sin(θ/2) cos(θ/2)) = 0 => sin(θ/2) + 4 sin(θ/2) cos(θ/2) = 0

Factor out sin(θ/2):

sin(θ/2) [1 + 4 cos(θ/2)] = 0

Therefore, either sin(θ/2) = 0 or 1 + 4 cos(θ/2) = 0.

In the interval θ ∈ [0, π/3], θ/2 ∈ [0, π/6]. Therefore, sin(θ/2) = 0 only when θ = 0. The other equation: 1 + 4 cos(θ/2) = 0 => cos(θ/2) = -1/4. However, θ/2 ∈ [0, π/6], so cos(θ/2) is in [cos(π/6), 1] = [√3/2 ≈ 0.866, 1]. Therefore, cos(θ/2) cannot be -1/4 here. Therefore, the only critical point in this interval is θ = 0.

Therefore, in interval [0, π/3], the critical points are θ = 0 and endpoints θ = π/3. Let me compute f1(θ) at these points.

At θ = 0:

f1(0) = 2 cos(0) + 2 cos(0) - 1 = 2*1 + 2*1 - 1 = 2 + 2 - 1 = 3.

At θ = π/3:

Compute cos(π/6) = √3/2 ≈ 0.866, cos(π/3) = 1/2.

So f1(π/3) = 2*(√3/2) + 2*(1/2) - 1 = √3 + 1 - 1 = √3 ≈ 1.732.

Therefore, in this interval, the maximum is 3 at θ = 0.

Now, moving to interval 2: [π/3, π]. Here, f2(θ) = 2 cos(θ/2) + 1 - 2 cos θ. Let me compute its derivative.

f2’(θ) = derivative of 2 cos(θ/2) is -sin(θ/2), derivative of 1 is 0, derivative of -2 cos θ is 2 sin θ. Therefore:

f2’(θ) = -sin(θ/2) + 2 sin θ

Set derivative to zero:

-sin(θ/2) + 2 sin θ = 0 => 2 sin θ = sin(θ/2)

Again, express sin θ in terms of sin(θ/2):

sin θ = 2 sin(θ/2) cos(θ/2). Therefore:

2*(2 sin(θ/2) cos(θ/2)) = sin(θ/2) => 4 sin(θ/2) cos(θ/2) = sin(θ/2)

Assuming sin(θ/2) ≠ 0, we can divide both sides by sin(θ/2):

4 cos(θ/2) = 1 => cos(θ/2) = 1/4

Therefore, θ/2 = arccos(1/4) => θ = 2 arccos(1/4). Let me compute this value. Let me note that arccos(1/4) is some angle between 0 and π/2, since 1/4 is between 0 and 1. Therefore, θ = 2 arccos(1/4) is between 0 and π. Let me check if this θ lies in [π/3, π]. Since arccos(1/4) ≈ 1.318 radians (since cos(π/3) = 0.5, cos(π/2) = 0, so arccos(1/4) is between π/3 and π/2). Therefore, θ ≈ 2 * 1.318 ≈ 2.636 radians. π is approximately 3.1416, so 2.636 is between π/3 ≈ 1.047 and π. Therefore, this critical point is within the interval [π/3, π].

Therefore, critical points in this interval are θ = 2 arccos(1/4) and endpoints θ = π/3 and θ = π.

Let me compute f2(θ) at these points.

First, compute θ = 2 arccos(1/4). Let me denote α = arccos(1/4), so θ = 2α. Then cos(α) = 1/4, so cos(θ/2) = cos α = 1/4. Also, sin α = sqrt(1 - (1/4)^2) = sqrt(15)/4. Then, sin θ = sin(2α) = 2 sin α cos α = 2*(sqrt(15)/4)*(1/4) = sqrt(15)/8. Wait, but maybe I need to compute f2(θ) = 2 cos(θ/2) + 1 - 2 cos θ. Let me compute each term:

First, 2 cos(θ/2) = 2*(1/4) = 1/2.

Second, 1 is just 1.

Third, -2 cos θ. Let me compute cos θ. Since θ = 2α, cos θ = cos(2α) = 2 cos² α - 1 = 2*(1/4)^2 - 1 = 2*(1/16) - 1 = 1/8 - 1 = -7/8. Therefore, -2 cos θ = -2*(-7/8) = 14/8 = 7/4.

Therefore, f2(θ) = 1/2 + 1 + 7/4 = (2/4 + 4/4 + 7/4) = 13/4 = 3.25. So 3.25 is 13/4.

Now, check endpoints:

At θ = π/3:

f2(π/3) = 2 cos(π/6) + 1 - 2 cos(π/3) = 2*(√3/2) + 1 - 2*(1/2) = √3 + 1 - 1 = √3 ≈ 1.732.

At θ = π:

f2(π) = 2 cos(π/2) + 1 - 2 cos π = 2*0 + 1 - 2*(-1) = 0 + 1 + 2 = 3.

Therefore, in interval [π/3, π], the maximum is 13/4 = 3.25 at θ = 2 arccos(1/4).

Moving on to interval 3: [π, 5π/3]. Here, f3(θ) = -2 cos(θ/2) + 1 - 2 cos θ. Let me compute its derivative.

f3’(θ) = derivative of -2 cos(θ/2) is -2*(-sin(θ/2))*(1/2) = sin(θ/2)

Derivative of 1 is 0, derivative of -2 cos θ is 2 sin θ. Therefore:

f3’(θ) = sin(θ/2) + 2 sin θ

Set derivative to zero:

sin(θ/2) + 2 sin θ = 0

Again, express sin θ as 2 sin(θ/2) cos(θ/2):

sin(θ/2) + 2*(2 sin(θ/2) cos(θ/2)) = 0 => sin(θ/2) + 4 sin(θ/2) cos(θ/2) = 0

Factor out sin(θ/2):

sin(θ/2) [1 + 4 cos(θ/2)] = 0

Therefore, either sin(θ/2) = 0 or 1 + 4 cos(θ/2) = 0.

In interval θ ∈ [π, 5π/3], θ/2 ∈ [π/2, 5π/6]. Therefore, sin(θ/2) is non-negative here (since θ/2 is between π/2 and 5π/6, where sine is positive). Therefore, sin(θ/2) = 0 only when θ/2 = π, i.e., θ = 2π, which is the endpoint of the interval [π, 5π/3]. The other equation: 1 + 4 cos(θ/2) = 0 => cos(θ/2) = -1/4. Let me check if this is possible. θ/2 ∈ [π/2, 5π/6], so cos(θ/2) ∈ [cos(5π/6), cos(π/2)] = [-√3/2, 0]. Since -1/4 is within [-√3/2 ≈ -0.866, 0], so yes, there is a solution here. Let me find θ such that cos(θ/2) = -1/4. Let me denote β = θ/2, so β ∈ [π/2, 5π/6], and cos β = -1/4. Therefore, β = arccos(-1/4). Let me compute arccos(-1/4). Since cos(π - arccos(1/4)) = -cos(arccos(1/4)) = -1/4. Therefore, β = π - arccos(1/4). Therefore, θ = 2β = 2π - 2 arccos(1/4). Let me check if this θ is in [π, 5π/3]. Let me compute 2π - 2 arccos(1/4). Since arccos(1/4) ≈ 1.318 radians, so 2π - 2*1.318 ≈ 6.283 - 2.636 ≈ 3.647 radians. Now, π ≈ 3.1416, 5π/3 ≈ 5.236. Therefore, 3.647 is between π and 5π/3. Therefore, this critical point is within the interval.

Therefore, critical points in this interval are θ = 2π - 2 arccos(1/4) and endpoints θ = π and θ = 5π/3.

Let me compute f3(θ) at these points.

First, at θ = 2π - 2 arccos(1/4). Let me denote β = arccos(1/4), so θ = 2π - 2β. Then θ/2 = π - β. Therefore, cos(θ/2) = cos(π - β) = -cos β = -1/4. Also, sin(θ/2) = sin(π - β) = sin β = sqrt(1 - (1/4)^2) = sqrt(15)/4. Also, cos θ = cos(2π - 2β) = cos(2β) = 2 cos² β - 1 = 2*(1/4)^2 - 1 = 2*(1/16) - 1 = 1/8 - 1 = -7/8. Therefore, compute f3(θ):

-2 cos(θ/2) + 1 - 2 cos θ = -2*(-1/4) + 1 - 2*(-7/8) = 1/2 + 1 + 14/8 = 1/2 + 1 + 7/4 = (2/4 + 4/4 + 7/4) = 13/4 = 3.25. Same as before.

Now, check endpoints:

At θ = π:

f3(π) = -2 cos(π/2) + 1 - 2 cos π = -2*0 + 1 - 2*(-1) = 0 + 1 + 2 = 3.

At θ = 5π/3:

Compute cos(θ/2) = cos(5π/6) = -√3/2. Therefore, -2 cos(θ/2) = -2*(-√3/2) = √3.

Compute cos θ = cos(5π/3) = cos(π/3) = 1/2. Therefore, -2 cos θ = -2*(1/2) = -1. Therefore, f3(5π/3) = √3 + 1 - 1 = √3 ≈ 1.732.

Therefore, in interval [π, 5π/3], the maximum is 13/4 = 3.25 at θ = 2π - 2 arccos(1/4).

Now, moving to interval 4: [5π/3, 2π]. Here, f4(θ) = -2 cos(θ/2) + 2 cos θ - 1. Let me compute its derivative.

f4’(θ) = derivative of -2 cos(θ/2) is -2*(-sin(θ/2))*(1/2) = sin(θ/2)

Derivative of 2 cos θ is -2 sin θ

Derivative of -1 is 0. Therefore:

f4’(θ) = sin(θ/2) - 2 sin θ

Set derivative to zero:

sin(θ/2) - 2 sin θ = 0 => sin(θ/2) = 2 sin θ

Again, express sin θ as 2 sin(θ/2) cos(θ/2):

sin(θ/2) = 2*(2 sin(θ/2) cos(θ/2)) => sin(θ/2) = 4 sin(θ/2) cos(θ/2)

Assuming sin(θ/2) ≠ 0, divide both sides by sin(θ/2):

1 = 4 cos(θ/2) => cos(θ/2) = 1/4

Therefore, θ/2 = arccos(1/4) or θ/2 = -arccos(1/4). But θ ∈ [5π/3, 2π], so θ/2 ∈ [5π/6, π]. Therefore, arccos(1/4) is in [0, π/2], so θ/2 = arccos(1/4) would be in [0, π/2], which is not in [5π/6, π]. Therefore, no solution here. Alternatively, maybe θ/2 = 2π - arccos(1/4)? Wait, cos(θ/2) = 1/4. Since θ/2 ∈ [5π/6, π], cos(θ/2) is negative there (since cos(π) = -1, cos(5π/6) = -√3/2 ≈ -0.866). But 1/4 is positive, so there is no solution here. Therefore, the only critical points are when sin(θ/2) = 0. In θ ∈ [5π/3, 2π], θ/2 ∈ [5π/6, π]. sin(θ/2) is non-negative here (since θ/2 is between 5π/6 and π, where sine is positive). Therefore, sin(θ/2) = 0 only when θ/2 = π, i.e., θ = 2π, which is the endpoint. Therefore, the critical points in this interval are θ = 2π and endpoints θ = 5π/3 and θ = 2π. Wait, but θ = 2π is the same as θ = 0, but since we are considering θ ∈ [5π/3, 2π], the endpoints are θ = 5π/3 and θ = 2π.

Compute f4(θ) at these points.

At θ = 5π/3:

f4(5π/3) = -2 cos(5π/6) + 2 cos(5π/3) - 1 = -2*(-√3/2) + 2*(1/2) - 1 = √3 + 1 - 1 = √3 ≈ 1.732.

At θ = 2π:

f4(2π) = -2 cos(π) + 2 cos(2π) - 1 = -2*(-1) + 2*1 - 1 = 2 + 2 - 1 = 3.

Therefore, in interval [5π/3, 2π], the maximum is 3 at θ = 2π.

So summarizing the maximum values in each interval:

- Interval 1: 3

- Interval 2: 13/4 = 3.25

- Interval 3: 13/4 = 3.25

- Interval 4: 3

Therefore, the overall maximum is 13/4, which is 3.25. However, let me check if there are any other critical points or if I missed something. Wait, but in intervals 2 and 3, we found critical points with value 13/4, which is higher than 3. So 13/4 is 3.25, which is higher than 3. Therefore, the maximum value is 13/4. However, let me check if there are any other critical points or if I made a mistake in calculations.

Wait, but 13/4 is 3.25, which is 3 + 1/4. Let me check if that's correct. Let me recheck the calculation for the critical point in interval 2. When θ = 2 arccos(1/4), we had:

f2(θ) = 2 cos(θ/2) + 1 - 2 cos θ. We found that cos(θ/2) = 1/4, so 2*(1/4) = 1/2. Then cos θ = -7/8, so -2 cos θ = 14/8 = 7/4. Then total is 1/2 + 1 + 7/4 = 1/2 + 4/4 + 7/4 = (2/4 + 4/4 + 7/4) = 13/4. Yes, that's correct. Similarly for interval 3, same value. So 13/4 is indeed the maximum.

But let me check if there are other points where the maximum could be higher. For example, maybe at θ = π/2? Let me compute f(π/2). Let me compute |1 + z| + |1 - z + z²| when θ = π/2. z = i.

Compute |1 + i| = sqrt(1² + 1²) = √2 ≈ 1.414.

Compute |1 - i + i²| = |1 - i -1| = |-i| = 1. Therefore, total is √2 + 1 ≈ 2.414, which is less than 3.25. So no.

Another test: θ = π/2. Wait, already did that. How about θ = π/3? As before, gives √3 ≈ 1.732. How about θ = 0? Gives 3. How about θ = π? Gives 3. So 3.25 is higher. So seems like 13/4 is the maximum.

But let me check another angle. Let me take θ where cos θ = 1/2, which is θ = π/3 or 5π/3. Wait, already checked those. How about θ where cos θ = something else?

Alternatively, maybe I can think geometrically. Let me think about the original problem: |1 + z| + |1 - z + z²|. Since |z| = 1, maybe there's a geometric interpretation?

Alternatively, maybe I can parametrize z as e^(iθ) and write the expression in terms of θ, then find maximum. But I think the previous analysis is thorough. However, let me check if 13/4 is indeed the maximum. Let me compute 13/4 = 3.25. Let me check if there's a way to get higher.

Alternatively, maybe I can consider specific values. For example, suppose θ is such that z is a root of some equation? Not sure. Alternatively, think of the expression as |1 + z| + |z² - z + 1|. Maybe there's a way to relate these terms?

Alternatively, note that z² - z + 1 = (z³ + 1)/(z + 1) when z ≠ -1. Wait, let me check:

(z³ - 1) = (z - 1)(z² + z + 1). But here we have z² - z + 1. Not sure. Alternatively, maybe not helpful.

Alternatively, think of z² - z + 1. Let me write it as z² + 1 - z. Since |z| = 1, z² = \overline{z}^{-2}? Wait, no. Wait, z² = z * z, and since |z| = 1, z̄ = 1/z. Therefore, z² = z * z, but not sure.

Alternatively, maybe use triangle inequality? But since we have sum of two moduli, not sure.

Alternatively, think of |1 + z| + |z² - z + 1|. Maybe express z² - z + 1 in terms of z + 1/z? Wait, since z² + 1/z² = (z + 1/z)^2 - 2. Wait, earlier we had |z² - z + 1| = |2 cos θ - 1|. But maybe not helpful here.

Alternatively, since we already did the calculus approach and found 13/4 as maximum, which is 3.25, and verified with specific points, maybe that's the answer. However, let me check if 13/4 is indeed the maximum. Let me compute 13/4 = 3.25. Let me check if there's a way to get higher.

Alternatively, think of θ where the two terms are maximized. For example, |1 + z| is maximized when z = 1, giving 2, and |1 - z + z²| when z = 1 is |1 - 1 + 1| = 1, so total 3. When z = -1, |1 + (-1)| = 0, |1 - (-1) + 1| = |1 + 1 + 1| = 3, total 3. So maximum of 3. But we found 3.25, which is higher. Therefore, that seems possible.

Alternatively, maybe check θ where the critical point occurs. Let me compute θ = 2 arccos(1/4). Let me compute arccos(1/4) ≈ 1.318 radians, so θ ≈ 2.636 radians. Let me compute z = e^(iθ). Let me compute |1 + z| and |1 - z + z²| numerically.

First, compute θ ≈ 2.636 radians. Let me compute cos θ and sin θ. Since θ ≈ 2.636 radians, which is approximately 151 degrees (since π radians is 180 degrees, so 2.636 radians ≈ 151 degrees). Let me compute cos θ ≈ cos(2.636). Let me use calculator:

cos(2.636) ≈ cos(π - 0.505) ≈ -cos(0.505) ≈ -0.876. Wait, but earlier we had cos θ = -7/8 = -0.875. So that's consistent. Therefore, cos θ ≈ -0.875, sin θ ≈ sin(2.636) ≈ sin(π - 0.505) ≈ sin(0.505) ≈ 0.486. Therefore, z ≈ cos θ + i sin θ ≈ -0.875 + i*0.486.

Compute |1 + z|: 1 + z ≈ 1 - 0.875 + i*0.486 = 0.125 + i*0.486. The modulus is sqrt(0.125² + 0.486²) ≈ sqrt(0.0156 + 0.236) ≈ sqrt(0.2516) ≈ 0.5016. Wait, but according to our previous calculation, |1 + z| = 2|cos(θ/2)|. Let me compute θ/2 ≈ 1.318 radians, cos(θ/2) = 1/4 = 0.25, so 2*0.25 = 0.5. Which matches the approximate value. So |1 + z| ≈ 0.5.

Then |1 - z + z²|. Let me compute z². Since z ≈ -0.875 + i*0.486, z² = (-0.875)^2 - (0.486)^2 + 2*(-0.875)*(0.486)i ≈ 0.7656 - 0.236 - 0.848i ≈ 0.5296 - 0.848i. Then 1 - z + z² ≈ 1 - (-0.875 + i*0.486) + (0.5296 - 0.848i) ≈ 1 + 0.875 - i*0.486 + 0.5296 - 0.848i ≈ (1 + 0.875 + 0.5296) + (-0.486 - 0.848)i ≈ 2.4046 - 1.334i. The modulus is sqrt(2.4046² + 1.334²) ≈ sqrt(5.78 + 1.78) ≈ sqrt(7.56) ≈ 2.75. Which is approximately 13/4 = 3.25? Wait, no, 2.75 is 11/4. Wait, but according to our previous calculation, |1 - z + z²| = |2 cos θ - 1|. Let me compute 2 cos θ - 1 ≈ 2*(-0.875) - 1 = -1.75 - 1 = -2.75, so absolute value is 2.75. Therefore, total expression is |1 + z| + |1 - z + z²| ≈ 0.5 + 2.75 = 3.25, which is 13/4. So that checks out. Therefore, the calculation is correct.

Therefore, the maximum value is 13/4. However, let me check if 13/4 is indeed the maximum. Let me think if there's a way to get higher. For example, suppose θ is such that both terms are maximized? But since they are dependent on θ, it's not straightforward. However, since we did the calculus and found critical points, and verified with specific examples, I think 13/4 is correct.

But let me check another critical point. For example, suppose θ where cos θ = something else. Wait, but according to our analysis, the maximum occurs at those critical points. Therefore, I think 13/4 is the correct answer.

But let me check if 13/4 is the maximum. Let me convert 13/4 to decimal, which is 3.25. Let me check if there's a case where the sum is higher. For example, suppose θ where |1 + z| is large and |1 - z + z²| is also large. For example, suppose z is such that both terms are large. However, since |1 + z| is maximized at 2 (when z = 1), and |1 - z + z²| is 1 at that point. If z is somewhere else, maybe both terms can be larger? For example, if z is such that 1 + z is large in magnitude and 1 - z + z² is also large. But since |z| = 1, maybe not. Let me think.

Alternatively, think of the expression as |1 + z| + |z² - z + 1|. Let me try to write z² - z + 1 in terms of z + 1/z. Wait, since z² + 1/z² = (z + 1/z)^2 - 2. But z² - z + 1 = z² + 1 - z. Not sure. Alternatively, maybe write z² - z + 1 = z² + 1 - z. Since |z| = 1, z² = \overline{z}^{-2}? Wait, no. Wait, z² = z * z, and since |z| = 1, z̄ = 1/z. Therefore, z² = z * z, and z̄² = 1/z². Not sure.

Alternatively, maybe consider writing the entire expression in terms of real and imaginary parts. Let me try that. Let me let z = x + iy, with x² + y² = 1. Then compute |1 + z| + |1 - z + z²|.

First, |1 + z| = sqrt((1 + x)^2 + y^2). Since x² + y² = 1, this becomes sqrt(1 + 2x + x² + y²) = sqrt(1 + 2x + 1) = sqrt(2 + 2x) = sqrt(2(1 + x)) = √2 * sqrt(1 + x). Wait, that's interesting. Wait, let me check:

(1 + x)^2 + y^2 = 1 + 2x + x² + y² = 1 + 2x + (x² + y²) = 1 + 2x + 1 = 2 + 2x. Yes, so |1 + z| = sqrt(2 + 2x) = √(2(1 + x)) = √2 * sqrt(1 + x). Since x = Re(z), which is cos θ, so this matches our previous result of 2|cos(θ/2)|. Because sqrt(2(1 + cos θ)) = sqrt(2 * 2 cos²(θ/2)) = sqrt(4 cos²(θ/2)) = 2|cos(θ/2)|. Yes, that's correct.

Now, compute |1 - z + z²|. Let me compute 1 - z + z². Let me write z = x + iy. Then z² = (x + iy)^2 = x² - y² + 2ixy. Therefore, 1 - z + z² = 1 - x - iy + x² - y² + 2ixy. Combine real and imaginary parts:

Real part: 1 - x + x² - y²

Imaginary part: -y + 2xy

Since x² + y² = 1, we can replace y² with 1 - x². Therefore, real part becomes 1 - x + x² - (1 - x²) = 1 - x + x² - 1 + x² = 2x² - x.

Imaginary part: -y + 2xy = y(-1 + 2x).

Therefore, |1 - z + z²| = sqrt[(2x² - x)^2 + (y(-1 + 2x))^2]. Let me compute this:

= sqrt{(2x² - x)^2 + y²(2x - 1)^2}

But since y² = 1 - x², substitute:

= sqrt{(2x² - x)^2 + (1 - x²)(2x - 1)^2}

Let me expand this expression:

First, compute (2x² - x)^2:

= 4x^4 - 4x^3 + x²

Then compute (1 - x²)(2x - 1)^2:

First compute (2x - 1)^2 = 4x² - 4x + 1

Multiply by (1 - x²):

= (1 - x²)(4x² - 4x + 1) = 4x²(1 - x²) - 4x(1 - x²) + 1*(1 - x²)

= 4x² - 4x^4 - 4x + 4x^3 + 1 - x²

Combine like terms:

-4x^4 + 4x^3 + (4x² - x²) + (-4x) + 1

= -4x^4 + 4x^3 + 3x² - 4x + 1

Therefore, total expression under sqrt is:

(4x^4 - 4x^3 + x²) + (-4x^4 + 4x^3 + 3x² - 4x + 1) =

Let me add term by term:

4x^4 - 4x^4 = 0

-4x^3 + 4x^3 = 0

x² + 3x² = 4x²

-4x remains

+1 remains

Therefore, total expression under sqrt is 4x² - 4x + 1. Therefore, |1 - z + z²| = sqrt(4x² - 4x + 1). Let me note that 4x² - 4x + 1 = (2x - 1)^2. Therefore, |1 - z + z²| = |2x - 1|. Since modulus is non-negative, it's |2x - 1|. Therefore, this matches our previous result where we had |2 cos θ - 1|, since x = cos θ. Therefore, this confirms that |1 - z + z²| = |2x - 1| = |2 cos θ - 1|.

Therefore, the original expression is √2 * sqrt(1 + x) + |2x - 1|, where x = cos θ and x ∈ [-1, 1]. Therefore, we can rephrase the problem as finding the maximum of f(x) = √2 * sqrt(1 + x) + |2x - 1| for x ∈ [-1, 1].

This seems like a simpler one-variable calculus problem. Let me try this approach.

So f(x) = √2 * sqrt(1 + x) + |2x - 1|. Let me analyze this function over x ∈ [-1, 1].

First, note that 2x - 1 changes sign at x = 1/2. Therefore, we can split the domain into two intervals:

1. x ∈ [-1, 1/2], where 2x - 1 ≤ 0, so |2x - 1| = 1 - 2x.

2. x ∈ [1/2, 1], where 2x - 1 ≥ 0, so |2x - 1| = 2x - 1.

Therefore, f(x) can be written as:

For x ∈ [-1, 1/2]: f(x) = √2 * sqrt(1 + x) + 1 - 2x

For x ∈ [1/2, 1]: f(x) = √2 * sqrt(1 + x) + 2x - 1

Now, let's find the maximum of f(x) in each interval.

First, interval [-1, 1/2]:

Let me denote f1(x) = √2 * sqrt(1 + x) + 1 - 2x. Find its maximum on [-1, 1/2].

Compute derivative f1’(x):

d/dx [√2 * sqrt(1 + x)] = √2 * (1/(2 sqrt(1 + x))) = √2 / (2 sqrt(1 + x)) = 1 / sqrt(2(1 + x))

Derivative of 1 - 2x is -2.

Therefore, f1’(x) = 1 / sqrt(2(1 + x)) - 2.

Set derivative to zero:

1 / sqrt(2(1 + x)) - 2 = 0 => 1 / sqrt(2(1 + x)) = 2 => sqrt(2(1 + x)) = 1/2 => 2(1 + x) = 1/4 => 1 + x = 1/8 => x = 1/8 - 1 = -7/8.

Check if x = -7/8 is in [-1, 1/2]. Yes, since -7/8 ≈ -0.875 is between -1 and 0.5.

Therefore, critical points at x = -7/8 and endpoints x = -1 and x = 1/2.

Compute f1(-7/8):

First, compute sqrt(1 + x) = sqrt(1 - 7/8) = sqrt(1/8) = 1/(2√2). Therefore, √2 * sqrt(1 + x) = √2 * 1/(2√2) = 1/2.

Then, 1 - 2x = 1 - 2*(-7/8) = 1 + 14/8 = 1 + 7/4 = 11/4.

Therefore, f1(-7/8) = 1/2 + 11/4 = 13/4 = 3.25.

At x = -1:

f1(-1) = √2 * sqrt(0) + 1 - 2*(-1) = 0 + 1 + 2 = 3.

At x = 1/2:

f1(1/2) = √2 * sqrt(1 + 1/2) + 1 - 2*(1/2) = √2 * sqrt(3/2) + 1 - 1 = √2*(√(3)/√2) + 0 = √3 ≈ 1.732.

Therefore, maximum in this interval is 13/4 at x = -7/8.

Now, interval [1/2, 1]:

f2(x) = √2 * sqrt(1 + x) + 2x - 1. Find its maximum on [1/2, 1].

Compute derivative f2’(x):

d/dx [√2 * sqrt(1 + x)] = 1 / sqrt(2(1 + x)) as before.

Derivative of 2x - 1 is 2.

Therefore, f2’(x) = 1 / sqrt(2(1 + x)) + 2.

Since 1 / sqrt(2(1 + x)) is always positive for x ∈ [1/2, 1], and adding 2, the derivative is always positive. Therefore, f2(x) is increasing on [1/2, 1]. Therefore, maximum occurs at x = 1.

Compute f2(1) = √2 * sqrt(2) + 2*1 - 1 = √2 * √2 + 2 - 1 = 2 + 1 = 3.

At x = 1/2:

f2(1/2) = √2 * sqrt(1 + 1/2) + 2*(1/2) - 1 = √2 * sqrt(3/2) + 1 - 1 = √3 ≈ 1.732.

Therefore, maximum in this interval is 3 at x = 1.

Therefore, overall maximum of f(x) is 13/4 = 3.25, achieved at x = -7/8. Therefore, this confirms our previous result. Therefore, the maximum value is 13/4.

Therefore, the answer is \boxed{\dfrac{13}{4}}.

**Final Answer**
\boxed{\dfrac{13}{4}}

To solve the problem, we are given a complex number $ z $ such that $ |z| = 1 $, and we are to find the **maximum value** of the expression:

$$
|1 + z| + |1 - z + z^2|
$$

---

### Step 1: Parametrize $ z $ on the Unit Circle

Since $ |z| = 1 $, we can write $ z = e^{i\theta} $, where $ \theta \in [0, 2\pi) $. This allows us to express the modulus terms in terms of trigonometric functions.

---

### Step 2: Simplify Each Term

#### First Term: $ |1 + z| $

We use the identity:
$$
|1 + e^{i\theta}| = 2|\cos(\theta/2)|
$$
This is derived from the identity:
$$
1 + e^{i\theta} = 2e^{i\theta/2} \cos(\theta/2)
$$
So,
$$
|1 + z| = 2|\cos(\theta/2)|
$$

#### Second Term: $ |1 - z + z^2| $

We simplify the expression:
$$
1 - z + z^2 = z^2 - z + 1
$$
Using the identity $ z^2 + \frac{1}{z^2} = (z + \frac{1}{z})^2 - 2 $, and noting that $ \frac{1}{z} = \overline{z} $, we can write:
$$
|z^2 - z + 1| = |2\cos\theta - 1|
$$
This is derived by expressing the modulus in terms of real and imaginary parts and simplifying.

---

### Step 3: Combine the Two Terms

Now, the expression becomes:
$$
f(\theta) = 2|\cos(\theta/2)| + |2\cos\theta - 1|
$$

We aim to find the **maximum** of this function over $ \theta \in [0, 2\pi) $.

---

### Step 4: Analyze the Function

Let’s define:
$$
f(\theta) = 2|\cos(\theta/2)| + |2\cos\theta - 1|
$$

We analyze this function by splitting the domain based on the behavior of the absolute value terms.

#### Case 1: $ \theta \in [0, \pi] $

Here, $ \cos(\theta/2) \geq 0 $, so $ |\cos(\theta/2)| = \cos(\theta/2) $, and we can write:
$$
f(\theta) = 2\cos(\theta/2) + |2\cos\theta - 1|
$$

#### Case 2: $ \theta \in [\pi, 2\pi] $

Here, $ \cos(\theta/2) \leq 0 $, so $ |\cos(\theta/2)| = -\cos(\theta/2) $, and we write:
$$
f(\theta) = -2\cos(\theta/2) + |2\cos\theta - 1|
$$

We also note that $ |2\cos\theta - 1| $ changes sign at $ \cos\theta = \frac{1}{2} $, i.e., at $ \theta = \frac{\pi}{3} $ and $ \theta = \frac{5\pi}{3} $.

---

### Step 5: Use Calculus to Find Maximum

We analyze the function in each interval and find critical points by taking derivatives. After careful analysis, we find that the **maximum** occurs at:

$$
\theta = 2\arccos\left(\frac{1}{4}\right)
$$

At this point, we compute:

- $ \cos(\theta/2) = \frac{1}{4} $
- $ \cos\theta = -\frac{7}{8} $

Substituting into the expression:

$$
f(\theta) = 2 \cdot \frac{1}{4} + |2 \cdot (-\frac{7}{8}) - 1| = \frac{1}{2} + \left| -\frac{7}{4} - 1 \right| = \frac{1}{2} + \frac{11}{4} = \frac{13}{4}
$$

---

### Final Answer

$$
\boxed{\dfrac{13}{4}}
$$

Solution:
Let $z = x + yi,$ where $x$ and $y$ are real numbers.  Since $|z| = 1,$ $x^2 + y^2 = 1.$  Then
\begin{align*}
|1 + z| + |1 - z + z^2| &= |1 + x + yi| + |1 - x - yi + x^2 + 2xyi - y^2| \\
&= |(1 + x) + yi| + |(1 - x + x^2 - 1 + x^2) + (-y + 2xy)i| \\
&= |(1 + x) + yi| + |(-x + 2x^2) + (-y + 2xy)i| \\
&= \sqrt{(1 + x)^2 + y^2} + \sqrt{(-x + 2x^2)^2 + (-y + 2xy)^2} \\
&= \sqrt{(1 + x)^2 + y^2} + \sqrt{(-x + 2x^2)^2 + y^2 (1 - 2x)^2} \\
&= \sqrt{(1 + x)^2 + 1 - x^2} + \sqrt{(-x + 2x^2)^2 + (1 - x^2) (1 - 2x)^2} \\
&= \sqrt{2 + 2x} + \sqrt{1 - 4x + 4x^2} \\
&= \sqrt{2 + 2x} + |1 - 2x|.
\end{align*}Let $u = \sqrt{2 + 2x}.$  Then $u^2 = 2 + 2x,$ so
\[\sqrt{2 + 2x} + |1 - 2x| = u + |3 - u^2|.\]Since $-1 \le x \le 1,$ $0 \le u \le 2.$

If $0 \le u \le \sqrt{3},$ then
\[u + |3 - u^2| = u + 3 - u^2 = \frac{13}{4} - \left( u - \frac{1}{2} \right)^2 \le \frac{13}{4}.\]Equality occurs when $u = \frac{1}{2},$ or $x = -\frac{7}{8}.$

If $\sqrt{3} \le u \le 2,$ then
\[u + u^2 - 3 = \left( u + \frac{1}{2} \right)^2 - \frac{13}{4} \le \left( 2 + \frac{1}{2} \right)^2 - \frac{13}{4} = 3 < \frac{13}{4}.\]Therefore, the maximum value is $\boxed{\frac{13}{4}}.$


Explanation:
Okay, so I need to figure out where the reflection takes the vector $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$. The problem says that the same reflection takes $\begin{pmatrix} 5 \\ 0 \end{pmatrix}$ to $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$. Hmm, reflections in linear algebra... Let me recall. A reflection is a type of linear transformation, right? So it should be represented by a matrix. If I can find that matrix, then I can apply it to any vector, including $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$, to find its image under the reflection.

First, let me remember that a reflection matrix in 2D can be represented in different ways depending on the line of reflection. The most common ones are reflections over the x-axis, y-axis, or the line y = x. But since the problem doesn't specify the line, I need to figure out which line is being reflected over based on the given information.

So, given that reflecting $\begin{pmatrix} 5 \\ 0 \end{pmatrix}$ gives $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$, I can use this information to determine the reflection line. Let me think. If I can find the line of reflection, then I can write the reflection matrix accordingly.

Alternatively, since reflections are linear transformations, maybe I can set up equations to find the reflection matrix. Let me try that approach.

Let me denote the reflection matrix as $ R = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $. Since it's a reflection, it should satisfy certain properties. For example, it should be its own inverse, meaning $ R^2 = I $, and its determinant should be -1. Also, the trace of the matrix should be 0 because for a reflection matrix, trace is 0. Wait, is that true?

Wait, let me check. For a reflection over a line making an angle θ with the x-axis, the reflection matrix is:

$$
R = \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}
$$

So, the trace of this matrix is $ \cos 2\theta - \cos 2\theta = 0 $. So yes, the trace is 0. Also, determinant is $ -\cos^2 2\theta - \sin^2 2\theta $? Wait, no. Let me compute determinant:

Determinant of R is $ (a)(d) - (b)(c) $. For the matrix above, that would be $ \cos 2\theta \times (-\cos 2\theta) - \sin 2\theta \times \sin 2\theta = -\cos^2 2\theta - \sin^2 2\theta = -(\cos^2 2\theta + \sin^2 2\theta) = -1 $. So determinant is -1. So, for any reflection matrix, determinant is -1 and trace is 0. So, if I can find such a matrix R that satisfies R * v = w, where v is $\begin{pmatrix} 5 \\ 0 \end{pmatrix}$ and w is $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$, then I can find the matrix.

So, let me write the equation:

$$
\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 5 \\ 0 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}
$$

Multiplying the matrix with the vector:

First component: 5a + 0*b = 5a = 4 ⇒ a = 4/5

Second component: 5c + 0*d = 5c = 3 ⇒ c = 3/5

So, we have a = 4/5 and c = 3/5. Now, since the matrix is a reflection matrix, we know that trace is 0, so a + d = 0 ⇒ d = -a = -4/5.

Also, determinant is -1, so ad - bc = -1. Let me check that.

We have a = 4/5, d = -4/5, so ad = (4/5)(-4/5) = -16/25. Then bc is b*c. But we don't know b and c yet. Wait, we already found c = 3/5. Wait, but we need to find b. Wait, but we have another equation from the determinant?

Wait, but we have only used one vector so far. But we need more information. Wait, but maybe we can use the fact that the reflection matrix must satisfy that the line of reflection is the set of fixed points. So, if we can find another vector that is fixed by the reflection, or maybe use another condition?

Alternatively, since we have a = 4/5, c = 3/5, d = -4/5, and we need to find b. Let me use the determinant condition.

Determinant is ad - bc = -1. So:

(4/5)(-4/5) - b*(3/5) = -1 ⇒ (-16/25) - (3b)/5 = -1 ⇒

Let me solve for b:

- (3b)/5 = -1 + 16/25 = (-25/25 + 16/25) = -9/25 ⇒

Multiply both sides by -1: (3b)/5 = 9/25 ⇒

Multiply both sides by 5: 3b = 9/5 ⇒ b = 3/5.

So, b = 3/5. Therefore, the reflection matrix is:

$$
R = \begin{pmatrix} 4/5 & 3/5 \\ 3/5 & -4/5 \end{pmatrix}
$$

Wait, let me check if this matrix satisfies the determinant condition:

ad - bc = (4/5)(-4/5) - (3/5)(3/5) = (-16/25) - (9/25) = -25/25 = -1. Yes, that works. And trace is 4/5 + (-4/5) = 0. So that's good.

So, the reflection matrix is R = [[4/5, 3/5], [3/5, -4/5]]. Let me check if this matrix actually reflects the vector [5, 0] to [4, 3]. Let me compute R * [5, 0]:

First component: (4/5)*5 + (3/5)*0 = 4 + 0 = 4. Second component: (3/5)*5 + (-4/5)*0 = 3 + 0 = 3. Perfect, that matches the given information. So this matrix is correct.

Now, the question is, what is R * [-2, 3]^T?

Let me compute that. Let me write the vector as $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$.

First component: (4/5)*(-2) + (3/5)*(3) = (-8/5) + (9/5) = (1/5)

Second component: (3/5)*(-2) + (-4/5)*(3) = (-6/5) + (-12/5) = (-18/5)

So, the result is $\begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix}$. Let me check if this makes sense.

Alternatively, maybe I can think about the reflection line. Let me see if I can find the line of reflection, and then use that to compute the reflection of the vector. Maybe that's another way to approach the problem, and verify my answer.

So, the reflection matrix we found is [[4/5, 3/5], [3/5, -4/5]]. Let me recall that a reflection matrix over a line with angle θ has the form:

$$
\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}
$$

Comparing this with our matrix:

cos 2θ = 4/5, sin 2θ = 3/5. Let me check if that's consistent. Since cos²2θ + sin²2θ = (16/25) + (9/25) = 25/25 = 1, which is correct. So 2θ is an angle whose cosine is 4/5 and sine is 3/5. Therefore, θ is half of that angle. So, the line of reflection is at angle θ, where tan 2θ = sin 2θ / cos 2θ = (3/5)/(4/5) = 3/4. Therefore, tan 2θ = 3/4. So θ is (1/2) arctan(3/4). But maybe I don't need to find θ explicitly.

Alternatively, since the line of reflection is the set of points fixed by the reflection. So, if I can find a vector that is fixed by the reflection, that would be a direction along the line of reflection. Let me think. Suppose v is a vector on the line of reflection, then Rv = v. So, solving Rv = v.

So, (R - I)v = 0. Let me compute R - I:

$$
R - I = \begin{pmatrix} 4/5 - 1 & 3/5 \\ 3/5 & -4/5 - 1 \end{pmatrix} = \begin{pmatrix} -1/5 & 3/5 \\ 3/5 & -9/5 \end{pmatrix}
$$

So, the system is:

-1/5 * x + 3/5 * y = 0

3/5 * x - 9/5 * y = 0

Let me simplify the first equation: Multiply both sides by 5: -x + 3y = 0 ⇒ x = 3y. So, the direction vector of the line of reflection is (3, 1). Because if x = 3y, then for y = 1, x = 3. So, the line of reflection is in the direction of vector (3, 1). Therefore, the line of reflection is the line through the origin in the direction of (3, 1). So, the line can be represented as y = (1/3)x.

Alternatively, since the line of reflection is y = (1/3)x. Let me check if this makes sense. Let me take a point on the line, say (3, 1). Applying the reflection matrix R to (3, 1):

First component: (4/5)(3) + (3/5)(1) = 12/5 + 3/5 = 15/5 = 3

Second component: (3/5)(3) + (-4/5)(1) = 9/5 - 4/5 = 5/5 = 1. So, R*(3,1) = (3,1). So yes, that's fixed. So the line of reflection is y = (1/3)x. So, that's good.

Alternatively, if I wanted to find the reflection of a vector over the line y = (1/3)x, I could use the formula for reflection over a line. But since I already have the matrix, maybe it's easier to just use the matrix. However, maybe I can verify my answer using another method.

Alternatively, since I know the line of reflection is y = (1/3)x, I can use the formula for reflecting a point over a line. Let me recall that formula.

The formula for reflecting a point (x, y) over the line ax + by + c = 0 is:

The reflected point (x', y') is given by:

x' = x - 2a(ax + by + c)/(a² + b²)

y' = y - 2b(ax + by + c)/(a² + b²)

But in our case, the line is y = (1/3)x, which can be rewritten as (1/3)x - y = 0, or multiplying both sides by 3: x - 3y = 0. So, a = 1, b = -3, c = 0.

So, the formula becomes:

x' = x - 2*(1)*(1*x + (-3)*y + 0)/(1² + (-3)²) = x - 2*(x - 3y)/10

Similarly,

y' = y - 2*(-3)*(x - 3y)/10 = y + 6*(x - 3y)/10

Let me compute that for the vector (-2, 3):

First, compute x - 3y: (-2) - 3*(3) = -2 - 9 = -11

Then,

x' = -2 - 2*(1)*(-11)/10 = -2 + 22/10 = -2 + 11/5 = (-10/5 + 11/5) = 1/5

Similarly,

y' = 3 + 6*(-11)/10 = 3 - 66/10 = 3 - 33/5 = (15/5 - 33/5) = -18/5

So, the reflected point is (1/5, -18/5), which matches the result I got earlier using the matrix. So that's reassuring. Therefore, the answer is $\begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix}$.

But let me check once more with another approach. Let me think about the reflection formula. Another way to think about reflection over a line is that it's the composition of two reflections over the axes, but maybe that's more complicated. Alternatively, since we know the line of reflection is y = (1/3)x, we can use vector projections.

The formula for reflecting a vector v over a line with unit vector u is:

Reflection of v is 2 proj_u(v) - v.

So, let me try that. First, find the unit vector u in the direction of the line y = (1/3)x. The direction vector is (3, 1), as before. Let me compute its magnitude: sqrt(3² + 1²) = sqrt(10). Therefore, the unit vector u is (3/sqrt(10), 1/sqrt(10)).

So, for a vector v = (-2, 3), the projection onto u is:

proj_u(v) = [(v ⋅ u)/||u||²] * u. But since u is a unit vector, ||u||² = 1, so proj_u(v) = (v ⋅ u) * u.

Compute v ⋅ u:

(-2)(3/sqrt(10)) + 3*(1/sqrt(10)) = (-6 + 3)/sqrt(10) = (-3)/sqrt(10)

Therefore, proj_u(v) = (-3/sqrt(10)) * (3/sqrt(10), 1/sqrt(10)) = (-9/10, -3/10)

Then, the reflection of v is 2*proj_u(v) - v:

2*(-9/10, -3/10) - (-2, 3) = (-18/10, -6/10) + (2, -3) = (-9/5, -3/5) + (10/5, -15/5) = (1/5, -18/5). Which is the same result as before. So that's consistent. Therefore, I can be confident that the answer is correct.

Therefore, the reflection takes $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ to $\begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix}$.

But let me check once again if I didn't make any calculation errors. Let me recompute the matrix multiplication:

Original matrix R is [[4/5, 3/5], [3/5, -4/5]]

Vector is [-2, 3]^T.

First component: (4/5)(-2) + (3/5)(3) = (-8/5) + 9/5 = 1/5. Correct.

Second component: (3/5)(-2) + (-4/5)(3) = (-6/5) + (-12/5) = (-18/5). Correct.

Yes, that's right. So, the answer is (1/5, -18/5). So, written as a vector, that's $\begin{pmatrix} \frac{1}{5} \\ -\frac{18}{5} \end{pmatrix}$.

Alternatively, if the problem expects the answer in fractions, that's fine. So, I think that's the correct answer.

But just to make sure, let me think if there's another way someone might approach this problem, maybe by using the fact that reflections are involutions (i.e., applying them twice gives the identity), but I don't think that's necessary here. Alternatively, maybe using the midpoint between the original vector and its image lies on the line of reflection. Let me try that.

Given that reflecting vector v = [5, 0] gives w = [4, 3]. The midpoint between v and w should lie on the line of reflection. Let me compute the midpoint:

Midpoint M = (v + w)/2 = ([5 + 4]/2, [0 + 3]/2) = (9/2, 3/2). So, the point (9/2, 3/2) lies on the line of reflection y = (1/3)x. Let me check: y = (1/3)x. If x = 9/2, then y should be (1/3)(9/2) = 3/2. Which matches. So that's correct. Therefore, the line of reflection is indeed y = (1/3)x. So that's another confirmation.

Alternatively, if I wanted to find the line of reflection using this midpoint and the fact that the line is perpendicular to the vector connecting v and w. Wait, the line of reflection is the perpendicular bisector of the segment joining v and w. So, since the reflection maps v to w, the line of reflection is the perpendicular bisector of the segment from v to w.

So, let me compute that. The vector from v to w is w - v = [4 - 5, 3 - 0] = [-1, 3]. The direction of this vector is (-1, 3). Therefore, the line of reflection should be perpendicular to this vector. The slope of the vector (-1, 3) is 3/(-1) = -3. Therefore, the slope of the line of reflection should be the negative reciprocal, which is 1/3. Which matches our previous result that the line of reflection is y = (1/3)x. So that's another confirmation. Therefore, the line of reflection is y = (1/3)x, which is consistent with all methods.

Therefore, I can be confident that the reflection matrix is correct, and the image of [-2, 3] is indeed [1/5, -18/5].

So, the answer is $\boxed{\begin{pmatrix} \dfrac{1}{5} \\ -\dfrac{18}{5} \end{pmatrix}}$.

**Final Answer**
The reflection takes $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ to \boxed{\begin{pmatrix} \dfrac{1}{5} \\ -\dfrac{18}{5} \end{pmatrix}}.

To determine where the reflection takes the vector
$$
\begin{pmatrix} -2 \\ 3 \end{pmatrix},
$$
we are given that the same reflection maps
$$
\begin{pmatrix} 5 \\ 0 \end{pmatrix} \mapsto \begin{pmatrix} 4 \\ 3 \end{pmatrix}.
$$

---

### Step 1: Determine the Reflection Matrix

Let the reflection matrix be
$$
R = \begin{pmatrix} a & b \\ c & d \end{pmatrix}.
$$
Since reflection is a linear transformation, we can use the given mapping to find the matrix.

Apply the matrix to the vector $\begin{pmatrix} 5 \\ 0 \end{pmatrix}$:

$$
R \begin{pmatrix} 5 \\ 0 \end{pmatrix} = \begin{pmatrix} 5a \\ 5c \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}.
$$

From this, we get:
- $5a = 4 \Rightarrow a = \frac{4}{5}$
- $5c = 3 \Rightarrow c = \frac{3}{5}$

Since reflection matrices have trace 0 (i.e., $a + d = 0$), we find:
- $d = -a = -\frac{4}{5}$

Now, use the determinant condition for a reflection matrix: $\det(R) = -1$. Compute:

$$
\det(R) = ad - bc = \left(\frac{4}{5}\right)\left(-\frac{4}{5}\right) - b\left(\frac{3}{5}\right) = -\frac{16}{25} - \frac{3b}{5}.
$$

Set this equal to $-1$:

$$
-\frac{16}{25} - \frac{3b}{5} = -1 \Rightarrow -\frac{3b}{5} = -1 + \frac{16}{25} = -\frac{9}{25} \Rightarrow b = \frac{3}{5}.
$$

So the reflection matrix is:

$$
R = \begin{pmatrix} \frac{4}{5} & \frac{3}{5} \\ \frac{3}{5} & -\frac{4}{5} \end{pmatrix}.
$$

---

### Step 2: Apply the Matrix to the Given Vector

Now apply this matrix to the vector $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$:

$$
R \begin{pmatrix} -2 \\ 3 \end{pmatrix} =
\begin{pmatrix} \frac{4}{5} & \frac{3}{5} \\ \frac{3}{5} & -\frac{4}{5} \end{pmatrix}
\begin{pmatrix} -2 \\ 3 \end{pmatrix}
=
\begin{pmatrix}
\frac{4}{5}(-2) + \frac{3}{5}(3) \\
\frac{3}{5}(-2) + \left(-\frac{4}{5}\right)(3)
\end{pmatrix}
=
\begin{pmatrix}
-\frac{8}{5} + \frac{9}{5} \\
-\frac{6}{5} - \frac{12}{5}
\end{pmatrix}
=
\begin{pmatrix}
\frac{1}{5} \\
-\frac{18}{5}
\end{pmatrix}.
$$

---

### Final Answer

$$
\boxed{\begin{pmatrix} \dfrac{1}{5} \\ -\dfrac{18}{5} \end{pmatrix}}
$$

[Assistant]
Explanation:
Okay, so I need to figure out where the reflection takes the vector $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$. The problem says that the same reflection takes $\begin{pmatrix} 5 \\ 0 \end{pmatrix}$ to $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$. Hmm, reflections in linear algebra... Let me recall. A reflection is a type of linear transformation, right? So it should be represented by a matrix. If I can find that matrix, then I can apply it to any vector, including $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$, to find its image under the reflection.

First, let me remember that a reflection matrix in 2D can be represented in different ways depending on the line of reflection. The most common ones are reflections over the x-axis, y-axis, or the line y = x. But since the problem doesn't specify the line, I need to figure out which line is being reflected over based on the given information.

So, given that reflecting $\begin{pmatrix} 5 \\ 0 \end{pmatrix}$ gives $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$, I can use this information to determine the reflection line. Let me think. If I can find the line of reflection, then I can write the reflection matrix accordingly.

Alternatively, since reflections are linear transformations, maybe I can set up equations to find the reflection matrix. Let me try that approach.

Let me denote the reflection matrix as $ R = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $. Since it's a reflection, it should satisfy certain properties. For example, it should be its own inverse, meaning $ R^2 = I $, and its determinant should be -1. Also, the trace of the matrix should be 0 because for a reflection matrix, trace is 0. Wait, is that true?

Wait, let me check. For a reflection over a line making an angle θ with the x-axis, the reflection matrix is:

$$
R = \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}
$$

So, the trace of this matrix is $ \cos 2\theta - \cos 2\theta = 0 $. So yes, the trace is 0. Also, determinant is $ -\cos^2 2\theta - \sin^2 2\theta $? Wait, no. Let me compute determinant:

Determinant of R is $ (a)(d) - (b)(c) $. For the matrix above, that would be $ \cos 2\theta \times (-\cos 2\theta) - \sin 2\theta \times \sin 2\theta = -\cos^2 2\theta - \sin^2 2\theta = -(\cos^2 2\theta + \sin^2 2\theta) = -1 $. So determinant is -1. So, for any reflection matrix, determinant is -1 and trace is 0. So, if I can find such a matrix R that satisfies R * v = w, where v is $\begin{pmatrix} 5 \\ 0 \end{pmatrix}$ and w is $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$, then I can find the matrix.

So, let me write the equation:

$$
\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 5 \\ 0 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}
$$

Multiplying the matrix with the vector:

First component: 5a + 0*b = 5a = 4 ⇒ a = 4/5

Second component: 5c + 0*d = 5c = 3 ⇒ c = 3/5

So, we have a = 4/5 and c = 3/5. Now, since the matrix is a reflection matrix, we know that trace is 0, so a + d = 0 ⇒ d = -a = -4/5.

Also, determinant is -1, so ad - bc = -1. Let me check that.

We have a = 4/5, d = -4/5, so ad = (4/5)(-4/5) = -16/25. Then bc is b*c. But we don't know b and c yet. Wait, we already found c = 3/5. Wait, but we need to find b. Wait, but we have another equation from the determinant?

Wait, but we have only used one vector so far. But we need more information. Wait, but maybe we can use the fact that the reflection matrix must satisfy that the line of reflection is the set of fixed points. So, if we can find another vector that is fixed by the reflection, or maybe use another condition?

Alternatively, since we have a = 4/5, c = 3/5, d = -4/5, and we need to find b. Let me use the determinant condition.

Determinant is ad - bc = -1. So:

(4/5)(-4/5) - b*(3/5) = -1 ⇒ (-16/25) - (3b)/5 = -1 ⇒

Let me solve for b:

- (3b)/5 = -1 + 16/25 = (-25/25 + 16/25) = -9/25 ⇒

Multiply both sides by -1: (3b)/5 = 9/25 ⇒

Multiply both sides by 5: 3b = 9/5 ⇒ b = 3/5.

So, b = 3/5. Therefore, the reflection matrix is:

$$
R = \begin{pmatrix} 4/5 & 3/5 \\ 3/5 & -4/5 \end{pmatrix}
$$

Wait, let me check if this matrix satisfies the determinant condition:

ad - bc = (4/5)(-4/5) - (3/5)(3/5) = (-16/25) - (9/25) = -25/25 = -1. Yes, that works. And trace is 4/5 + (-4/5) = 0. So that's good.

So, the reflection matrix is R = [[4/5, 3/5], [3/5, -4/5]]. Let me check if this matrix actually reflects the vector [5, 0] to [4, 3]. Let me compute R * [5, 0]:

First component: (4/5)*5 + (3/5)*0 = 4 + 0 = 4. Second component: (3/5)*5 + (-4/5)*0 = 3 + 0 = 3. Perfect, that matches the given information. So this matrix is correct.

Now, the question is, what is R * [-2, 3]^T?

Let me compute that. Let me write the vector as $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$.

First component: (4/5)*(-2) + (3/5)*(3) = (-8/5) + (9/5) = (1/5)

Second component: (3/5)*(-2) + (-4/5)*(3) = (-6/5) + (-12/5) = (-18/5)

So, the result is $\begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix}$. Let me check if this makes sense.

Alternatively, maybe I can think about the reflection line. Let me see if I can find the line of reflection, and then use that to compute the reflection of the vector. Maybe that's another way to approach the problem, and verify my answer.

So, the reflection matrix we found is [[4/5, 3/5], [3/5, -4/5]]. Let me recall that a reflection matrix over a line with angle θ has the form:

$$
\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}
$$

Comparing this with our matrix:

cos 2θ = 4/5, sin 2θ = 3/5. Let me check if that's consistent. Since cos²2θ + sin²2θ = (16/25) + (9/25) = 25/25 = 1, which is correct. So 2θ is an angle whose cosine is 4/5 and sine is 3/5. Therefore, θ is half of that angle. So, the line of reflection is at angle θ, where tan 2θ = sin 2θ / cos 2θ = (3/5)/(4/5) = 3/4. Therefore, tan 2θ = 3/4. So θ is (1/2) arctan(3/4). But maybe I don't need to find θ explicitly.

Alternatively, since the line of reflection is the set of points fixed by the reflection. So, if I can find a vector that is fixed by the reflection, that would be a direction along the line of reflection. Let me think. Suppose v is a vector on the line of reflection, then Rv = v. So, solving Rv = v.

So, (R - I)v = 0. Let me compute R - I:

$$
R - I = \begin{pmatrix} 4/5 - 1 & 3/5 \\ 3/5 & -4/5 - 1 \end{pmatrix} = \begin{pmatrix} -1/5 & 3/5 \\ 3/5 & -9/5 \end{pmatrix}
$$

So, the system is:

-1/5 * x + 3/5 * y = 0

3/5 * x - 9/5 * y = 0

Let me simplify the first equation: Multiply both sides by 5: -x + 3y = 0 ⇒ x = 3y. So, the direction vector of the line of reflection is (3, 1). Because if x = 3y, then for y = 1, x = 3. So, the line of reflection is in the direction of vector (3, 1). Therefore, the line of reflection is the line through the origin in the direction of (3, 1). So, the line can be represented as y = (1/3)x.

Alternatively, since the line of reflection is y = (1/3)x. Let me check if this makes sense. Let me take a point on the line, say (3, 1). Applying the reflection matrix R to (3, 1):

First component: (4/5)(3) + (3/5)(1) = 12/5 + 3/5 = 15/5 = 3

Second component: (3/5)(3) + (-4/5)(1) = 9/5 - 4/5 = 5/5 = 1. So, R*(3,1) = (3,1). So yes, that's fixed. So the line of reflection is y = (1/3)x. So, that's good.

Alternatively, if I wanted to find the reflection of a vector over the line y = (1/3)x, I could use the formula for reflection over a line. But since I already have the matrix, maybe it's easier to just use the matrix. However, maybe I can verify my answer using another method.

Alternatively, since I know the line of reflection is y = (1/3)x, I can use the formula for reflecting a point over a line. Let me recall that formula.

The formula for reflecting a point (x, y) over the line ax + by + c = 0 is:

The reflected point (x', y') is given by:

x' = x - 2a(ax + by + c)/(a² + b²)

y' = y - 2b(ax + by + c)/(a² + b²)

But in our case, the line is y = (1/3)x, which can be rewritten as (1/3)x - y = 0, or multiplying both sides by 3: x - 3y = 0. So, a = 1, b = -3, c = 0.

So, the formula becomes:

x' = x - 2*(1)*(1*x + (-3)*y + 0)/(1² + (-3)²) = x - 2*(x - 3y)/10

Similarly,

y' = y - 2*(-3)*(x - 3y)/10 = y + 6*(x - 3y)/10

Let me compute that for the vector (-2, 3):

First, compute x - 3y: (-2) - 3*(3) = -2 - 9 = -11

Then,

x' = -2 - 2*(1)*(-11)/10 = -2 + 22/10 = -2 + 11/5 = (-10/5 + 11/5) = 1/5

Similarly,

y' = 3 + 6*(-11)/10 = 3 - 66/10 = 3 - 33/5 = (15/5 - 33/5) = -18/5

So, the reflected point is (1/5, -18/5), which matches the result I got earlier using the matrix. So that's reassuring. Therefore, the answer is $\begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix}$.

But let me check once more with another approach. Let me think about the reflection formula. Another way to think about reflection over a line is that it's the composition of two reflections over the axes, but maybe that's more complicated. Alternatively, since we know the line of reflection is y = (1/3)x, we can use vector projections.

The formula for reflecting a vector v over a line with unit vector u is:

Reflection of v is 2 proj_u(v) - v.

So, let me try that. First, find the unit vector u in the direction of the line y = (1/3)x. The direction vector is (3, 1), as before. Let me compute its magnitude: sqrt(3² + 1²) = sqrt(10). Therefore, the unit vector u is (3/sqrt(10), 1/sqrt(10)).

So, for a vector v = (-2, 3), the projection onto u is:

proj_u(v) = [(v ⋅ u)/||u||²] * u. But since u is a unit vector, ||u||² = 1, so proj_u(v) = (v ⋅ u) * u.

Compute v ⋅ u:

(-2)(3/sqrt(10)) + 3*(1/sqrt(10)) = (-6 + 3)/sqrt(10) = (-3)/sqrt(10)

Therefore, proj_u(v) = (-3/sqrt(10)) * (3/sqrt(10), 1/sqrt(10)) = (-9/10, -3/10)

Then, the reflection of v is 2*proj_u(v) - v:

2*(-9/10, -3/10) - (-2, 3) = (-18/10, -6/10) + (2, -3) = (-9/5, -3/5) + (10/5, -15/5) = (1/5, -18/5). Which is the same result as before. So that's consistent. Therefore, I can be confident that the answer is correct.

Therefore, the reflection takes $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ to $\begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix}$.

But let me check once again if I didn't make any calculation errors. Let me recompute the matrix multiplication:

Original matrix R is [[4/5, 3/5], [3/5, -4/5]]

Vector is [-2, 3]^T.

First component: (4/5)(-2) + (3/5)(3) = (-8/5) + 9/5 = 1/5. Correct.

Second component: (3/5)(-2) + (-4/5)(3) = (-6/5) + (-12/5) = (-18/5). Correct.

Yes, that's right. So, the answer is (1/5, -18/5). So, written as a vector, that's $\begin{pmatrix} \frac{1}{5} \\ -\frac{18}{5} \end{pmatrix}$.

Alternatively, if the problem expects the answer in fractions, that's fine. So, I think that's the correct answer.

But just to make sure, let me think if there's another way someone might approach this problem, maybe by using the fact that reflections are involutions (i.e., applying them twice gives the identity), but I don't think that's necessary here. Alternatively, maybe using the midpoint between the original vector and its image lies on the line of reflection. Let me try that.

Given that reflecting vector v = [5, 0] gives w = [4, 3]. The midpoint between v and w should lie on the line of reflection. Let me compute the midpoint:

Midpoint M = (v + w)/2 = ([5 + 4]/2, [0 + 3]/2) = (9/2, 3/2). So, the point (9/2, 3/2) lies on the line of reflection y = (1/3)x. Let me check: y = (1/3)x. If x = 9/2, then y should be (1/3)(9/2) = 3/2. Which matches. So that's correct. Therefore, the line of reflection is indeed y = (1/3)x. So that's another confirmation.

Alternatively, if I wanted to find the line of reflection using this midpoint and the fact that the line is perpendicular to the vector connecting v and w. Wait, the line of reflection is the perpendicular bisector of the segment joining v and w. So, since the reflection maps v to w, the line of reflection is the perpendicular bisector of the segment from v to w.

So, let me compute that. The vector from v to w is w - v = [4 - 5, 3 - 0] = [-1, 3]. The direction of this vector is (-1, 3). Therefore, the line of reflection should be perpendicular to this vector. The slope of the vector (-1, 3) is 3/(-1) = -3. Therefore, the slope of the line of reflection should be the negative reciprocal, which is 1/3. Which matches our previous result that the line of reflection is y = (1/3)x. So that's another confirmation. Therefore, the line of reflection is y = (1/3)x, which is consistent with all methods.

Therefore, I can be confident that the reflection matrix is correct, and the image of [-2, 3] is indeed [1/5, -18/5].

So, the answer is $\boxed{\begin{pmatrix} \dfrac{1}{5} \\ -\dfrac{18}{5} \end{pmatrix}}$.

**Final Answer**
The reflection takes $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ to \boxed{\begin{pmatrix} \dfrac{1}{5} \\ -\dfrac{18}{5} \end{pmatrix}}.

To determine where the reflection takes the vector
$$
\begin{pmatrix} -2 \\ 3 \end{pmatrix},
$$
we are given that the same reflection maps
$$
\begin{pmatrix} 5 \\ 0 \end{pmatrix} \mapsto \begin{pmatrix} 4 \\ 3 \end{pmatrix}.
$$

---

### Step 1: Determine the Reflection Matrix

Let the reflection matrix be
$$
R = \begin{pmatrix} a & b \\ c & d \end{pmatrix}.
$$
Since reflection is a linear transformation, we can use the given mapping to find the matrix.

Apply the matrix to the vector $\begin{pmatrix} 5 \\ 0 \end{pmatrix}$:

$$
R \begin{pmatrix} 5 \\ 0 \end{pmatrix} = \begin{pmatrix} 5a \\ 5c \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}.
$$

From this, we get:
- $5a = 4 \Rightarrow a = \frac{4}{5}$
- $5c = 3 \Rightarrow c = \frac{3}{5}$

Since reflection matrices have trace 0 (i.e., $a + d = 0$), we find:
- $d = -a = -\frac{4}{5}$

Now, use the determinant condition for a reflection matrix: $\det(R) = -1$. Compute:

$$
\det(R) = ad - bc = \left(\frac{4}{5}\right)\left(-\frac{4}{5}\right) - b\left(\frac{3}{5}\right) = -\frac{16}{25} - \frac{3b}{5}.
$$

Set this equal to $-1$:

$$
-\frac{16}{25} - \frac{3b}{5} = -1 \Rightarrow -\frac{3b}{5} = -1 + \frac{16}{25} = -\frac{9}{25} \Rightarrow b = \frac{3}{5}.
$$

So the reflection matrix is:

$$
R = \begin{pmatrix} \frac{4}{5} & \frac{3}{5} \\ \frac{3}{5} & -\frac{4}{5} \end{pmatrix}.
$$

---

### Step 2: Apply the Matrix to the Given Vector

Now apply this matrix to the vector $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$:

$$
R \begin{pmatrix} -2 \\ 3 \end{pmatrix} =
\begin{pmatrix} \frac{4}{5} & \frac{3}{5} \\ \frac{3}{5} & -\frac{4}{5} \end{pmatrix}
\begin{pmatrix} -2 \\ 3 \end{pmatrix}
=
\begin{pmatrix}
\frac{4}{5}(-2) + \frac{3}{5}(3) \\
\frac{3}{5}(-2) + \left(-\frac{4}{5}\right)(3)
\end{pmatrix}
=
\begin{pmatrix}
-\frac{8}{5} + \frac{9}{5} \\
-\frac{6}{5} - \frac{12}{5}
\end{pmatrix}
=
\begin{pmatrix}
\frac{1}{5} \\
-\frac{18}{5}
\end{pmatrix}.
$$

---

### Final Answer

$$
\boxed{\begin{pmatrix} \dfrac{1}{5} \\ -\dfrac{18}{5} \end{pmatrix}}
$$

Solution:
The midpoint of $(5,0)$ and $(4,3)$ is
\[\left( \frac{5 + 4}{2}, \frac{0 + 3}{2} \right) = \left( \frac{9}{2}, \frac{3}{2} \right).\]This tells us that the vector being reflected over is a scalar multiple of $\begin{pmatrix} \frac{9}{2} \\ \frac{3}{2} \end{pmatrix}.$  We can then assume that the vector being reflected over is $\begin{pmatrix} 3 \\ 1 \end{pmatrix}.$

[asy]
usepackage("amsmath");

unitsize(1 cm);

pair A, B, M, O, R, S;

O = (0,0);
A = (5,0);
R = (4,3);
B = (-2,3);
S = (1/5,-18/5);
M = (A + R)/2;

draw((-3,-1)--(5,5/3),red + dashed);
draw(O--M,red,Arrow(6));
draw((-4,0)--(6,0));
draw((0,-4)--(0,4));
draw(O--A,Arrow(6));
draw(O--R,Arrow(6));
draw(A--R,dashed,Arrow(6));
draw(O--B,Arrow(6));
draw(O--S,Arrow(6));
draw(B--S,dashed,Arrow(6));
label("$\begin{pmatrix} 5 \\ 0 \end{pmatrix}$", A, S);
label("$\begin{pmatrix} 4 \\ 3 \end{pmatrix}$", R, NE);
label("$\begin{pmatrix} -2 \\ 3 \end{pmatrix}$", B, NW);
label("$\begin{pmatrix} \frac{9}{2} \\ \frac{3}{2} \end{pmatrix}$", M, N);
[/asy]

The projection of $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ onto $\begin{pmatrix} 3 \\ 1 \end{pmatrix}$ is
\[\operatorname{proj}_{\begin{pmatrix} 3 \\ 1 \end{pmatrix}} \begin{pmatrix} -2 \\ 3 \end{pmatrix} = \frac{\begin{pmatrix} -2 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 1 \end{pmatrix}}{\begin{pmatrix} 3 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 1 \end{pmatrix}} \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \frac{-3}{10} \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} -\frac{9}{10} \\ -\frac{3}{10} \end{pmatrix}.\]Hence, the reflection of $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ is $2 \begin{pmatrix} -\frac{9}{10} \\ -\frac{3}{10} \end{pmatrix} - \begin{pmatrix} -2 \\ 3 \end{pmatrix} = \boxed{\begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix}}.$