Explanation:
Okay, so I need to solve this equation: $ z^4 + 4z^3i - 6z^2 - 4zi - i = 0 $. The solutions are supposed to be the vertices of a convex polygon, and I need to find the area of that polygon, which can be expressed as $ p^{a/b} $, where p is prime, and a and b are coprime. Then find a + b + p. Alright, let's start.

First, maybe I can try to factor this equation or find some substitution. The equation is a quartic in z, so maybe it's a quadratic in terms of something else? Let me check the coefficients. The coefficients are 1, 4i, -6, -4i, -i. Hmm, they seem to have some pattern. Let me write down the equation again:

$ z^4 + 4i z^3 - 6 z^2 - 4i z - i = 0 $.

Looking at the coefficients, they resemble binomial coefficients. Let me think. The coefficients 1, 4, -6, -4, -1... Wait, if I think of binomial coefficients for (a + b)^4, they are 1, 4, 6, 4, 1. But here, the coefficients are 1, 4i, -6, -4i, -i. Not quite the same, but maybe there's a connection. Maybe if I consider some substitution?

Alternatively, maybe this equation can be rewritten in terms of (z + i)^something? Let me try to see. Let me try substituting w = z + something. Let me think. Let me check if there's a substitution that can make this equation simpler.

Alternatively, maybe factor the equation. Let me try to factor it. Let me try to see if there are obvious roots. Let me try z = i. Let me plug in z = i:

$ (i)^4 + 4i*(i)^3 - 6*(i)^2 - 4i*(i) - i $.

Calculating each term:

i^4 = 1

4i*(i^3) = 4i*(-i) = -4i^2 = -4*(-1) = 4

-6*(i^2) = -6*(-1) = 6

-4i*(i) = -4i^2 = -4*(-1) = 4

Then -i.

So total: 1 + 4 + 6 + 4 - i = 15 - i ≠ 0. Not zero. How about z = -i?

Let me try z = -i:

(-i)^4 + 4i*(-i)^3 - 6*(-i)^2 - 4i*(-i) - i.

Calculating:

(-i)^4 = (i^4) = 1

4i*(-i)^3 = 4i*(-(-i)) = 4i*(i) = 4i^2 = -4

-6*(-i)^2 = -6*(i^2) = -6*(-1) = 6

-4i*(-i) = 4i^2 = -4

Then -i.

Total: 1 - 4 + 6 - 4 - i = -1 - i ≠ 0. Not zero.

How about z = 1? Let me check:

1^4 + 4i*1^3 - 6*1^2 - 4i*1 - i = 1 + 4i - 6 - 4i - i = (1 - 6) + (4i - 4i - i) = -5 - i ≠ 0.

z = -1:

(-1)^4 + 4i*(-1)^3 - 6*(-1)^2 - 4i*(-1) - i = 1 + 4i*(-1) - 6*(1) - 4i*(-1) - i = 1 - 4i - 6 + 4i - i = (1 - 6) + (-4i + 4i - i) = -5 - i ≠ 0.

Hmm. Maybe z = something else? Maybe z = something with i? Let me think. Alternatively, maybe this equation is related to (z + i)^4? Let me expand (z + i)^4:

(z + i)^4 = z^4 + 4z^3 i + 6 z^2 i^2 + 4 z i^3 + i^4 = z^4 + 4i z^3 - 6 z^2 - 4i z + 1.

Wait! Comparing with our equation:

Original equation: z^4 + 4i z^3 - 6 z^2 - 4i z - i = 0.

Compare with (z + i)^4 = z^4 + 4i z^3 - 6 z^2 - 4i z + 1. So if we subtract 1 from (z + i)^4, we get z^4 + 4i z^3 - 6 z^2 - 4i z. Then our equation is that expression minus i equals zero. So:

(z + i)^4 - 1 - i = 0? Wait:

Wait, (z + i)^4 = z^4 + 4i z^3 - 6 z^2 - 4i z + 1. Therefore, if we take (z + i)^4 - 1 = z^4 + 4i z^3 - 6 z^2 - 4i z. Therefore, our equation is (z + i)^4 - 1 - i = 0? Wait, original equation is z^4 + 4i z^3 - 6 z^2 - 4i z - i = 0. Which is equal to [(z + i)^4 - 1] - i = 0. Therefore, (z + i)^4 - 1 - i = 0 => (z + i)^4 = 1 + i.

Oh! That seems promising. So the equation simplifies to (z + i)^4 = 1 + i. Therefore, if I let w = z + i, then w^4 = 1 + i. Therefore, solving for w, then z = w - i. So the solutions for z are the solutions of w^4 = 1 + i, shifted by -i.

So first, let me solve w^4 = 1 + i. Then, once I have the solutions for w, subtract i to get z.

So, solving w^4 = 1 + i. Let me recall that to solve equations like w^n = c, we can write c in polar form and then take the nth roots.

First, let me write 1 + i in polar form. The modulus is sqrt(1^2 + 1^2) = sqrt(2). The argument is π/4, since it's in the first quadrant. Therefore, 1 + i = sqrt(2) * e^{i π/4}.

Therefore, the solutions for w are the fourth roots of sqrt(2) e^{i π/4}. The fourth roots will have modulus (sqrt(2))^{1/4} = (2^{1/2})^{1/4} = 2^{1/8}. The arguments will be (π/4 + 2πk)/4 for k = 0, 1, 2, 3.

Therefore, the four solutions for w are:

w_k = 2^{1/8} e^{i (π/16 + πk/2)} for k = 0, 1, 2, 3.

Therefore, the four solutions for w are:

w_0 = 2^{1/8} e^{i π/16}

w_1 = 2^{1/8} e^{i (π/16 + π/2)} = 2^{1/8} e^{i 9π/16}

w_2 = 2^{1/8} e^{i (π/16 + π)} = 2^{1/8} e^{i 17π/16}

w_3 = 2^{1/8} e^{i (π/16 + 3π/2)} = 2^{1/8} e^{i 25π/16}

Therefore, the solutions for z are w_k - i. So each z_k = w_k - i.

Therefore, the four solutions z are located at points w_k shifted down by i. So to find the polygon formed by these points, we need to plot these four points in the complex plane.

But before that, maybe I can find the coordinates of these points. Let me think. Let me first note that the four roots w_k are equally spaced around a circle of radius 2^{1/8}, starting at angle π/16, then adding π/2 each time. So they are spaced every 90 degrees (π/2 radians). Therefore, they form a square on the complex plane, rotated by π/16 radians from the standard position.

But then, when we subtract i from each of them, we are translating the entire square down by i. So the original square is centered where? Let me think. The original roots w_k are on a circle of radius 2^{1/8}, centered at the origin. So when we subtract i, we are translating the center of the square from the origin to -i. Wait, but the center of the square? Wait, the four roots w_k are symmetrically placed around the origin? Wait, if they are fourth roots of 1 + i, which is not on the real axis, so the roots are not symmetric with respect to the origin. Wait, but they are equally spaced around a circle. So they form a square, but rotated and scaled.

But when we subtract i, we are shifting all points down by i. So the polygon formed by the z's is the square (from the w's) shifted down by i. However, since the original square is not centered at the origin, shifting it down by i might not make it symmetric in any particular way. Wait, but maybe there's a better way to think about this.

Alternatively, maybe I can find coordinates for each z_k. Let me try.

First, let me compute each w_k:

For w_0: angle π/16, modulus 2^{1/8}

So w_0 = 2^{1/8} (cos(π/16) + i sin(π/16))

Similarly, w_1: angle 9π/16, modulus 2^{1/8}

w_1 = 2^{1/8} (cos(9π/16) + i sin(9π/16))

w_2: angle 17π/16, modulus 2^{1/8}

w_2 = 2^{1/8} (cos(17π/16) + i sin(17π/16))

w_3: angle 25π/16, modulus 2^{1/8}

w_3 = 2^{1/8} (cos(25π/16) + i sin(25π/16))

Then z_k = w_k - i, so:

z_0 = 2^{1/8} (cos(π/16) + i sin(π/16)) - i

Similarly for others.

So to find the coordinates of each z_k, we can write them as:

For z_0: x-coordinate is 2^{1/8} cos(π/16), y-coordinate is 2^{1/8} sin(π/16) - 1

Similarly for z_1: x-coordinate is 2^{1/8} cos(9π/16), y-coordinate is 2^{1/8} sin(9π/16) - 1

z_2: x-coordinate is 2^{1/8} cos(17π/16), y-coordinate is 2^{1/8} sin(17π/16) - 1

z_3: x-coordinate is 2^{1/8} cos(25π/16), y-coordinate is 2^{1/8} sin(25π/16) - 1

Now, since the original w's form a square, when we subtract i, we are shifting all points down by 1 unit in the imaginary axis. Therefore, the polygon formed by the z's is a square rotated by π/16, scaled by 2^{1/8}, and then shifted down by 1 unit. However, since the original square is not centered at the origin, shifting it down by 1 unit might not affect the area. Wait, shifting a polygon doesn't change its area. Therefore, the area of the polygon formed by the z's is the same as the area of the square formed by the w's. Therefore, if I can compute the area of the square formed by the w's, that will be the area of the polygon.

But wait, the original w's are the fourth roots of 1 + i, which are four points forming a square. The area of a square can be calculated if we know the side length or the radius of the circumscribed circle. Since all four points lie on a circle of radius 2^{1/8}, which is the distance from the origin to each w_k. For a square inscribed in a circle of radius r, the side length is r * sqrt(2). Wait, no. Wait, for a square inscribed in a circle of radius r (which is the distance from the center to a vertex), the side length s is related by s = r * sqrt(2). Wait, let me recall. The diagonal of the square is 2r. Since the diagonal of a square is s*sqrt(2), so s*sqrt(2) = 2r => s = 2r / sqrt(2) = sqrt(2) r. Therefore, the area of the square is s^2 = (sqrt(2) r)^2 = 2 r^2. Therefore, if the radius is 2^{1/8}, then area is 2*(2^{1/8})^2 = 2*(2^{1/4}) = 2^{1 + 1/4} = 2^{5/4}.

Wait, but hold on. Wait, the radius here is 2^{1/8}, so r = 2^{1/8}, so area is 2*(2^{1/8})^2 = 2*(2^{1/4}) = 2^{1 + 1/4} = 2^{5/4}. Therefore, the area of the square formed by the w's is 2^{5/4}. Therefore, since shifting doesn't change the area, the area of the polygon formed by the z's is also 2^{5/4}. Therefore, the answer would be p^{a/b} where p is prime, and 2^{5/4} is already in that form. So p=2, a=5, b=4. Then a + b + p = 5 + 4 + 2 = 11. But wait, the problem says "the solutions of the equation... are the vertices of a convex polygon". Wait, but if the original w's form a square, then shifting them down by i would still form a square, which is convex. Therefore, the area is 2^{5/4}, so answer is 5 + 4 + 2 = 11? But wait, let me check again.

Wait, but maybe I made a mistake here. Because the original square is not centered at the origin. Wait, but when you take the fourth roots of a complex number, they are equally spaced around the circle, so they form a square. However, the center of the square is at the origin? Wait, if you have four points equally spaced around a circle, then the center of the square is the center of the circle. So since the w's are fourth roots of 1 + i, which is not zero, so they are on a circle centered at the origin. Therefore, the square is centered at the origin. Therefore, when we shift all points down by i, the center of the square moves from the origin to -i. However, the area remains the same. Therefore, the area of the polygon is indeed 2^{5/4}, so the answer is 2^{5/4}, so p=2, a=5, b=4, so 5 + 4 + 2 = 11. But wait, the problem says "the area of this polygon can be expressed in the form p^{a/b}". However, 2^{5/4} is already in that form. But maybe I need to check if my calculation of the area is correct.

Wait, let me think again. The area of a square inscribed in a circle of radius r is 2r². Wait, let me verify with a simple case. If the square is inscribed in a circle of radius 1, then the diagonal is 2, so the side length is 2 / sqrt(2) = sqrt(2), so area is (sqrt(2))² = 2. Which is 2*(1)^2 = 2. So yes, that formula is correct. Therefore, if the radius is 2^{1/8}, then area is 2*(2^{1/8})² = 2*(2^{1/4}) = 2^{1 + 1/4} = 2^{5/4}. Therefore, that seems correct.

But let me check if the original equation's solutions actually form a square. Since we have four roots, and they are fourth roots of 1 + i, which are equally spaced around a circle, so they form a square. Therefore, the polygon is a square, so area is 2^{5/4}. Therefore, answer is 5 + 4 + 2 = 11. However, I need to check if the problem says "convex polygon", which a square is, so that's okay.

But wait, maybe I made a mistake in assuming that the area is the same as the square. Wait, shifting the square down by i would not change the area. So yes, the area remains the same. Therefore, the answer is 2^{5/4}, so 5 + 4 + 2 = 11. However, I need to check if the problem is expecting something else. Wait, maybe I made a mistake in the equation transformation?

Let me check again. Original equation: z^4 + 4i z^3 - 6 z^2 - 4i z - i = 0. Then I noticed that (z + i)^4 = z^4 + 4i z^3 - 6 z^2 - 4i z + 1. Therefore, original equation is (z + i)^4 - 1 - i = 0 => (z + i)^4 = 1 + i. Therefore, yes, that's correct. Therefore, solving for w = z + i, w^4 = 1 + i. Therefore, the solutions for w are the fourth roots of 1 + i, which are four points on a circle of radius |1 + i|^{1/4} = (sqrt(2))^{1/4} = 2^{1/8}. Therefore, radius is 2^{1/8}, so area of square is 2*(2^{1/8})^2 = 2^{5/4}. Therefore, answer is 2^{5/4}, so 5 + 4 + 2 = 11. But wait, the problem says "the area of this polygon can be expressed in the form p^{a/b}". However, 2^{5/4} is already in that form. However, maybe the problem expects a different form? Let me check if 2^{5/4} can be written as something else. For example, 2^{5/4} is equal to sqrt(2) * 2^{1/4} or something else? But no, it's already in the form p^{a/b} with p=2, a=5, b=4. Since 5 and 4 are coprime, and 2 is prime. Therefore, the answer is 5 + 4 + 2 = 11. However, I need to check if the area is indeed 2^{5/4}.

Alternatively, maybe I need to compute the area using coordinates. Let me try that. Let me compute coordinates of the four points z_0, z_1, z_2, z_3.

First, let me note that the original square (from w's) has vertices at angles π/16, 9π/16, 17π/16, 25π/16. When we subtract i, we are shifting each point down by 1 unit in the imaginary axis. Therefore, the coordinates of z_k are:

For each w_k = 2^{1/8} e^{iθ}, z_k = w_k - i = 2^{1/8} e^{iθ} - i.

But maybe instead of working with angles, I can think of the square. Since the original square is centered at the origin, and after shifting down by i, the center is at -i. However, the area remains the same. Therefore, the area is still 2^{5/4}.

Alternatively, maybe I can compute the distance between two adjacent points and then compute the area. Let me try that. Let me take two adjacent points, say z_0 and z_1. The distance between them would be the same as the distance between w_0 and w_1, since subtracting i from both doesn't change the distance between them. Therefore, the side length of the square is the same as the distance between two adjacent w's.

So let me compute the distance between w_0 and w_1. Since they are two adjacent fourth roots of 1 + i, separated by angle π/2. The distance between two points on a circle of radius r separated by angle θ is 2r sin(θ/2). Here, θ = π/2, so distance is 2r sin(π/4) = 2r*(√2/2) = r√2. Therefore, the side length s = r√2, so area is s² = 2r². Which is the same as before. Therefore, area is 2*(2^{1/8})² = 2^{5/4}. Therefore, same result. Therefore, seems correct.

But let me check with actual coordinates. Let me compute coordinates for z_0 and z_1.

Let me compute coordinates for w_0: angle π/16, modulus 2^{1/8}.

So coordinates of w_0: (2^{1/8} cos(π/16), 2^{1/8} sin(π/16))

Coordinates of z_0: (2^{1/8} cos(π/16), 2^{1/8} sin(π/16) - 1)

Similarly, coordinates of w_1: angle 9π/16, modulus 2^{1/8}

Coordinates of z_1: (2^{1/8} cos(9π/16), 2^{1/8} sin(9π/16) - 1)

So the distance between z_0 and z_1 is sqrt[(2^{1/8} cos(9π/16) - 2^{1/8} cos(π/16))² + (2^{1/8} sin(9π/16) - 1 - (2^{1/8} sin(π/16) - 1))²] = sqrt[(2^{1/8} (cos(9π/16) - cos(π/16)))² + (2^{1/8} (sin(9π/16) - sin(π/16)))²]

Which simplifies to 2^{1/8} * sqrt[(cos(9π/16) - cos(π/16))² + (sin(9π/16) - sin(π/16))²]

Which is 2^{1/8} * sqrt[2 - 2 cos(9π/16) cos(π/16) - 2 sin(9π/16) sin(π/16)] by expanding the squares. Wait, alternatively, using the identity that sqrt[(cos A - cos B)^2 + (sin A - sin B)^2] = 2 |sin((A - B)/2)|. Wait, let me recall:

The distance between two points on the unit circle at angles A and B is 2 sin(|A - B|/2). Wait, more precisely, if you have two points on a circle of radius r, separated by angle θ, then the distance between them is 2r sin(θ/2). In this case, the angle between w_0 and w_1 is π/2, so the distance between w_0 and w_1 is 2 * 2^{1/8} * sin(π/4) = 2^{1 + 1/8} * √2 / 2 = 2^{9/8} * √2 / 2 = 2^{9/8} * 2^{-1/2} = 2^{9/8 - 4/8} = 2^{5/8}. Wait, but earlier I thought the side length was r√2. Wait, if r is 2^{1/8}, then r√2 is 2^{1/8} * 2^{1/2} = 2^{5/8}, which matches. Therefore, the distance between w_0 and w_1 is 2^{5/8}, so the side length is 2^{5/8}, so area is (2^{5/8})² = 2^{5/4}, same as before. Therefore, that's correct.

Therefore, the area is indeed 2^{5/4}, so the answer is 5 + 4 + 2 = 11. However, I need to check if the problem says "the area of this polygon can be expressed in the form p^{a/b}". But 2^{5/4} is already in that form. However, maybe there's a mistake here? Let me check again.

Wait, but maybe the polygon isn't a square? Wait, the original equation is quartic, so four roots. If they are the fourth roots of 1 + i shifted by -i, then they should form a square. But maybe the shifting causes some distortion? Wait, no. Because shifting all points by the same vector doesn't change the shape or the area. So if the original four points form a square, then after shifting, they still form a square. Therefore, the area is the same. Therefore, 2^{5/4} is correct.

But let me check with another approach. Let me think of the original equation again. If I have w^4 = 1 + i, then the four roots are equally spaced, so they form a square. Therefore, the area is 2r² where r is the modulus of each root. Since modulus of each root is |1 + i|^{1/4} = (sqrt(2))^{1/4} = 2^{1/8}, so area is 2*(2^{1/8})² = 2^{1 + 1/4} = 2^{5/4}. Therefore, same result. Therefore, I think this is correct.

But maybe the problem is expecting a different answer? Let me check if I made a mistake in the equation transformation. Let me re-derive that.

Original equation: z^4 + 4i z^3 - 6 z^2 - 4i z - i = 0.

Let me compute (z + i)^4:

(z + i)^4 = z^4 + 4 z^3 i + 6 z^2 i^2 + 4 z i^3 + i^4

= z^4 + 4i z^3 + 6 z^2 (-1) + 4 z (-i) + 1

= z^4 + 4i z^3 - 6 z^2 - 4i z + 1.

Therefore, (z + i)^4 = z^4 + 4i z^3 - 6 z^2 - 4i z + 1.

Comparing with original equation: z^4 + 4i z^3 - 6 z^2 - 4i z - i = 0.

Therefore, original equation is (z + i)^4 - 1 - i = 0 => (z + i)^4 = 1 + i. Correct. Therefore, that's correct.

Therefore, the rest follows. Therefore, the area is 2^{5/4}, so answer is 5 + 4 + 2 = 11. However, I need to check if the problem says "the area of this polygon can be expressed in the form p^{a/b}". But 2^{5/4} is already in that form. However, maybe the problem expects the answer in another form? For example, maybe rationalizing or something else? Let me check.

Alternatively, maybe I made a mistake in assuming that the area is 2r². Let me think again. If the square is inscribed in a circle of radius r, then the area is 2r². But let me think of coordinates. Suppose we have a square centered at the origin, with vertices at (r, 0), (0, r), (-r, 0), (0, -r). Wait, no, that's not a square. Wait, if you have a square inscribed in a circle of radius r, then the coordinates of the vertices are (r, r), (-r, r), (-r, -r), (r, -r)? No, that's a square rotated by 45 degrees. Wait, the distance from the origin to each vertex is sqrt(r² + r²) = r√2. Therefore, if the radius is r, then the side length is sqrt( (2r)^2 + (2r)^2 )? No, wait. Let me think again.

Let me take a square with vertices at (a, a), (-a, a), (-a, -a), (a, -a). The distance between (a, a) and (-a, a) is 2a. The distance between (a, a) and (a, -a) is 2a. Wait, but that's a square with side length 2a√2? No, wait, the distance between (a, a) and (-a, a) is 2a, but that's the length of the side? Wait, no, if you have a square with vertices at (a, a), (-a, a), (-a, -a), (a, -a), then the sides are between (a, a) and (-a, a), which is horizontal line of length 2a, then from (-a, a) to (-a, -a), vertical line of length 2a, etc. Wait, but that's actually a square rotated by 45 degrees relative to the axes? No, actually, that's a square aligned with the axes, but with side length 2a. However, the distance from the origin to each vertex is sqrt(a² + a²) = a√2. Therefore, if the radius is r = a√2, then the side length is 2a = 2r / √2 = sqrt(2) r. Therefore, area is (sqrt(2) r)^2 = 2 r². Therefore, same as before. Therefore, if the radius is 2^{1/8}, then area is 2*(2^{1/8})² = 2^{5/4}. Therefore, correct.

Therefore, I think my answer is correct. Therefore, the answer is 5 + 4 + 2 = 11. However, I need to check if the problem says "the area of this polygon can be expressed in the form p^{a/b}". But 2^{5/4} is already in that form. However, maybe the problem expects the answer in another form? For example, maybe they want it as (sqrt(2))^{5/2} or something else? But no, 2^{5/4} is simpler. Also, 5 and 4 are coprime, and 2 is prime. Therefore, the answer is 11.

But wait, I just thought of something. The original equation is quartic, so four roots. The problem says "the solutions... are the vertices of a convex polygon". A square is convex, so that's okay. However, maybe the polygon is not a square? Wait, if the four roots are equally spaced on a circle, then they form a square. Therefore, yes. Therefore, the area is 2^{5/4}. Therefore, answer is 11. However, I need to check if there is an error in my calculation. Let me check with specific numbers.

Let me compute 2^{5/4}. 2^{1/4} is sqrt(sqrt(2)) ≈ 1.1892, so 2^{5/4} is 2 * 2^{1/4} ≈ 2.3784. Alternatively, if I compute the area using coordinates. Let me take specific values.

Let me compute coordinates of z_0, z_1, z_2, z_3. Let me compute approximate values.

First, compute 2^{1/8}. Let me compute 2^{1/8} ≈ e^{(ln 2)/8} ≈ e^{0.0866} ≈ 1.089. Let me check with calculator:

2^(1/8) = 2^(0.125) ≈ 1.0905.

So modulus of each w is approx 1.0905.

Now, angles:

For w_0: π/16 ≈ 0.19635 radians ≈ 11.25 degrees.

cos(π/16) ≈ cos(11.25°) ≈ 0.98078

sin(π/16) ≈ 0.19509

Therefore, coordinates of w_0 ≈ (1.0905 * 0.98078, 1.0905 * 0.19509) ≈ (1.070, 0.2126)

Then z_0 ≈ (1.070, 0.2126 - 1) ≈ (1.070, -0.7874)

Similarly, w_1: angle 9π/16 ≈ 1.767 radians ≈ 101.25 degrees.

cos(9π/16) ≈ cos(101.25°) ≈ -0.19509

sin(9π/16) ≈ sin(101.25°) ≈ 0.98078

Therefore, coordinates of w_1 ≈ (1.0905 * (-0.19509), 1.0905 * 0.98078) ≈ (-0.2126, 1.070)

Then z_1 ≈ (-0.2126, 1.070 - 1) ≈ (-0.2126, 0.070)

Similarly, w_2: angle 17π/16 ≈ 3.334 radians ≈ 191.25 degrees.

cos(17π/16) ≈ cos(191.25°) ≈ -cos(11.25°) ≈ -0.98078

sin(17π/16) ≈ sin(191.25°) ≈ -sin(11.25°) ≈ -0.19509

Coordinates of w_2 ≈ (1.0905 * (-0.98078), 1.0905 * (-0.19509)) ≈ (-1.070, -0.2126)

Then z_2 ≈ (-1.070, -0.2126 - 1) ≈ (-1.070, -1.2126)

Similarly, w_3: angle 25π/16 ≈ 4.909 radians ≈ 281.25 degrees.

cos(25π/16) ≈ cos(281.25°) ≈ cos(-78.75°) ≈ cos(78.75°) ≈ 0.19509

sin(25π/16) ≈ sin(281.25°) ≈ sin(-78.75°) ≈ -sin(78.75°) ≈ -0.98078

Coordinates of w_3 ≈ (1.0905 * 0.19509, 1.0905 * (-0.98078)) ≈ (0.2126, -1.070)

Then z_3 ≈ (0.2126, -1.070 - 1) ≈ (0.2126, -2.070)

So the four points are approximately:

z_0 ≈ (1.070, -0.7874)

z_1 ≈ (-0.2126, 0.070)

z_2 ≈ (-1.070, -1.2126)

z_3 ≈ (0.2126, -2.070)

Wait, let me check if these points form a square. Let me compute the distances between consecutive points.

First, between z_0 and z_1:

Distance between (1.070, -0.7874) and (-0.2126, 0.070):

Δx = -0.2126 - 1.070 = -1.2826

Δy = 0.070 - (-0.7874) = 0.8574

Distance squared: (-1.2826)^2 + (0.8574)^2 ≈ 1.645 + 0.735 ≈ 2.38

Distance ≈ sqrt(2.38) ≈ 1.543

Between z_1 and z_2:

(-0.2126, 0.070) to (-1.070, -1.2126):

Δx = -1.070 - (-0.2126) = -0.8574

Δy = -1.2126 - 0.070 = -1.2826

Distance squared: (-0.8574)^2 + (-1.2826)^2 ≈ 0.735 + 1.645 ≈ 2.38, same as before. Distance ≈ 1.543

Between z_2 and z_3:

(-1.070, -1.2126) to (0.2126, -2.070):

Δx = 0.2126 - (-1.070) = 1.2826

Δy = -2.070 - (-1.2126) = -0.8574

Distance squared: (1.2826)^2 + (-0.8574)^2 ≈ same as before, 2.38. Distance ≈ 1.543

Between z_3 and z_0:

(0.2126, -2.070) to (1.070, -0.7874):

Δx = 1.070 - 0.2126 = 0.8574

Δy = -0.7874 - (-2.070) = 1.2826

Distance squared: (0.8574)^2 + (1.2826)^2 ≈ same as before, 2.38. Distance ≈ 1.543

So all sides are equal. Now check the diagonals. Let me check between z_0 and z_2, and between z_1 and z_3.

Between z_0 (1.070, -0.7874) and z_2 (-1.070, -1.2126):

Δx = -1.070 - 1.070 = -2.14

Δy = -1.2126 - (-0.7874) = -0.4252

Distance squared: (-2.14)^2 + (-0.4252)^2 ≈ 4.5796 + 0.1808 ≈ 4.7604

Distance ≈ 2.18

Between z_1 (-0.2126, 0.070) and z_3 (0.2126, -2.070):

Δx = 0.2126 - (-0.2126) = 0.4252

Δy = -2.070 - 0.070 = -2.14

Distance squared: (0.4252)^2 + (-2.14)^2 ≈ same as above, 0.1808 + 4.5796 ≈ 4.7604. Same distance. Therefore, diagonals are equal. Therefore, the figure is a square. Therefore, the area can be calculated as (side length)^2. The side length squared is approx 2.38, so area is approx 2.38. Let me compute 2^{5/4} ≈ 2^(1.25) ≈ 2.378, which matches the approximate value. Therefore, this confirms that the area is indeed 2^{5/4}.

Therefore, the answer is 2^{5/4}, so p=2, a=5, b=4, so a + b + p = 5 + 4 + 2 = 11. Therefore, the answer is 11.

But wait, just to be thorough, let me check if there's another way the problem could be interpreted. For example, maybe the polygon is not a square? But with four points forming a square, it's convex. The problem says "the solutions... are the vertices of a convex polygon", which is satisfied. Therefore, I think my answer is correct.

Therefore, I think the answer is 11.

**Final Answer**
\boxed{11}

To solve the equation:

$$
z^4 + 4z^3i - 6z^2 - 4zi - i = 0
$$

we begin by recognizing a pattern in the coefficients. This suggests a possible substitution or transformation. Let us consider the expression:

$$
(z + i)^4 = z^4 + 4i z^3 - 6 z^2 - 4i z + 1
$$

Comparing this with the original equation, we see that:

$$
z^4 + 4i z^3 - 6 z^2 - 4i z - i = (z + i)^4 - 1 - i
$$

Thus, the original equation becomes:

$$
(z + i)^4 = 1 + i
$$

Letting $ w = z + i $, we now solve:

$$
w^4 = 1 + i
$$

---

### Step 1: Express $ 1 + i $ in polar form

The modulus of $ 1 + i $ is:

$$
|1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}
$$

The argument is:

$$
\arg(1 + i) = \frac{\pi}{4}
$$

So, in polar form:

$$
1 + i = \sqrt{2} \cdot e^{i\pi/4}
$$

---

### Step 2: Find the fourth roots of $ 1 + i $

The fourth roots of $ \sqrt{2} \cdot e^{i\pi/4} $ are:

$$
w_k = \left( \sqrt{2} \right)^{1/4} \cdot e^{i(\pi/16 + k\pi/2)}, \quad k = 0, 1, 2, 3
$$

Note that $ \left( \sqrt{2} \right)^{1/4} = 2^{1/8} $, so:

$$
w_k = 2^{1/8} \cdot e^{i(\pi/16 + k\pi/2)}
$$

These four roots are equally spaced on a circle of radius $ 2^{1/8} $, forming a square in the complex plane.

---

### Step 3: Find the corresponding $ z $ values

Recall that $ z = w - i $, so the four solutions for $ z $ are:

$$
z_k = 2^{1/8} \cdot e^{i(\pi/16 + k\pi/2)} - i
$$

These four points form a square in the complex plane, centered at $ -i $, with side length $ 2^{5/8} $, as derived from the distance between adjacent roots.

---

### Step 4: Compute the area of the square

The area of a square with side length $ s $ is $ s^2 $. Here, the side length is:

$$
s = 2^{5/8}
$$

So the area is:

$$
\text{Area} = (2^{5/8})^2 = 2^{5/4}
$$

---

### Step 5: Express the area in the required form

We are told the area can be expressed as $ p^{a/b} $, where $ p $ is prime, and $ a $ and $ b $ are relatively prime. Here:

$$
2^{5/4} = 2^{5/4}
$$

So $ p = 2 $, $ a = 5 $, $ b = 4 $, and since $ \gcd(5, 4) = 1 $, this satisfies the conditions.

---

### Final Step: Compute $ a + b + p $

$$
a + b + p = 5 + 4 + 2 = \boxed{11}
$$

[Assistant]
Explanation:
Okay, so I need to solve this equation: $ z^4 + 4z^3i - 6z^2 - 4zi - i = 0 $. The solutions are supposed to be the vertices of a convex polygon, and I need to find the area of that polygon, which can be expressed as $ p^{a/b} $, where p is prime, and a and b are coprime. Then find a + b + p. Alright, let's start.

First, maybe I can try to factor this equation or find some substitution. The equation is a quartic in z, so maybe it's a quadratic in terms of something else? Let me check the coefficients. The coefficients are 1, 4i, -6, -4i, -i. Hmm, they seem to have some pattern. Let me write down the equation again:

$ z^4 + 4i z^3 - 6 z^2 - 4i z - i = 0 $.

Looking at the coefficients, they resemble binomial coefficients. Let me think. The coefficients 1, 4, -6, -4, -1... Wait, if I think of binomial coefficients for (a + b)^4, they are 1, 4, 6, 4, 1. But here, the coefficients are 1, 4i, -6, -4i, -i. Not quite the same, but maybe there's a connection. Maybe if I consider some substitution?

Alternatively, maybe this equation can be rewritten in terms of (z + i)^something? Let me try to see. Let me try substituting w = z + something. Let me think. Let me check if there's a substitution that can make this equation simpler.

Alternatively, maybe factor the equation. Let me try to factor it. Let me try to see if there are obvious roots. Let me try z = i. Let me plug in z = i:

$ (i)^4 + 4i*(i)^3 - 6*(i)^2 - 4i*(i) - i $.

Calculating each term:

i^4 = 1

4i*(i^3) = 4i*(-i) = -4i^2 = -4*(-1) = 4

-6*(i^2) = -6*(-1) = 6

-4i*(i) = -4i^2 = -4*(-1) = 4

Then -i.

So total: 1 + 4 + 6 + 4 - i = 15 - i ≠ 0. Not zero. How about z = -i?

Let me try z = -i:

(-i)^4 + 4i*(-i)^3 - 6*(-i)^2 - 4i*(-i) - i.

Calculating:

(-i)^4 = (i^4) = 1

4i*(-i)^3 = 4i*(-(-i)) = 4i*(i) = 4i^2 = -4

-6*(-i)^2 = -6*(i^2) = -6*(-1) = 6

-4i*(-i) = 4i^2 = -4

Then -i.

Total: 1 - 4 + 6 - 4 - i = -1 - i ≠ 0. Not zero.

How about z = 1? Let me check:

1^4 + 4i*1^3 - 6*1^2 - 4i*1 - i = 1 + 4i - 6 - 4i - i = (1 - 6) + (4i - 4i - i) = -5 - i ≠ 0.

z = -1:

(-1)^4 + 4i*(-1)^3 - 6*(-1)^2 - 4i*(-1) - i = 1 + 4i*(-1) - 6*(1) - 4i*(-1) - i = 1 - 4i - 6 + 4i - i = (1 - 6) + (-4i + 4i - i) = -5 - i ≠ 0.

Hmm. Maybe z = something else? Maybe z = something with i? Let me think. Alternatively, maybe this equation is related to (z + i)^4? Let me expand (z + i)^4:

(z + i)^4 = z^4 + 4z^3 i + 6 z^2 i^2 + 4 z i^3 + i^4 = z^4 + 4i z^3 - 6 z^2 - 4i z + 1.

Wait! Comparing with our equation:

Original equation: z^4 + 4i z^3 - 6 z^2 - 4i z - i = 0.

Compare with (z + i)^4 = z^4 + 4i z^3 - 6 z^2 - 4i z + 1. So if we subtract 1 from (z + i)^4, we get z^4 + 4i z^3 - 6 z^2 - 4i z. Then our equation is that expression minus i equals zero. So:

(z + i)^4 - 1 - i = 0? Wait:

Wait, (z + i)^4 = z^4 + 4i z^3 - 6 z^2 - 4i z + 1. Therefore, if we take (z + i)^4 - 1 = z^4 + 4i z^3 - 6 z^2 - 4i z. Therefore, our equation is (z + i)^4 - 1 - i = 0? Wait, original equation is z^4 + 4i z^3 - 6 z^2 - 4i z - i = 0. Which is equal to [(z + i)^4 - 1] - i = 0. Therefore, (z + i)^4 - 1 - i = 0 => (z + i)^4 = 1 + i.

Oh! That seems promising. So the equation simplifies to (z + i)^4 = 1 + i. Therefore, if I let w = z + i, then w^4 = 1 + i. Therefore, solving for w, then z = w - i. So the solutions for z are the solutions of w^4 = 1 + i, shifted by -i.

So first, let me solve w^4 = 1 + i. Then, once I have the solutions for w, subtract i to get z.

So, solving w^4 = 1 + i. Let me recall that to solve equations like w^n = c, we can write c in polar form and then take the nth roots.

First, let me write 1 + i in polar form. The modulus is sqrt(1^2 + 1^2) = sqrt(2). The argument is π/4, since it's in the first quadrant. Therefore, 1 + i = sqrt(2) * e^{i π/4}.

Therefore, the solutions for w are the fourth roots of sqrt(2) e^{i π/4}. The fourth roots will have modulus (sqrt(2))^{1/4} = (2^{1/2})^{1/4} = 2^{1/8}. The arguments will be (π/4 + 2πk)/4 for k = 0, 1, 2, 3.

Therefore, the four solutions for w are:

w_k = 2^{1/8} e^{i (π/16 + πk/2)} for k = 0, 1, 2, 3.

Therefore, the four solutions for w are:

w_0 = 2^{1/8} e^{i π/16}

w_1 = 2^{1/8} e^{i (π/16 + π/2)} = 2^{1/8} e^{i 9π/16}

w_2 = 2^{1/8} e^{i (π/16 + π)} = 2^{1/8} e^{i 17π/16}

w_3 = 2^{1/8} e^{i (π/16 + 3π/2)} = 2^{1/8} e^{i 25π/16}

Therefore, the solutions for z are w_k - i. So each z_k = w_k - i.

Therefore, the four solutions z are located at points w_k shifted down by i. So to find the polygon formed by these points, we need to plot these four points in the complex plane.

But before that, maybe I can find the coordinates of these points. Let me think. Let me first note that the four roots w_k are equally spaced around a circle of radius 2^{1/8}, starting at angle π/16, then adding π/2 each time. So they are spaced every 90 degrees (π/2 radians). Therefore, they form a square on the complex plane, rotated by π/16 radians from the standard position.

But then, when we subtract i from each of them, we are translating the entire square down by i. So the original square is centered where? Let me think. The original roots w_k are on a circle of radius 2^{1/8}, centered at the origin. So when we subtract i, we are translating the center of the square from the origin to -i. Wait, but the center of the square? Wait, the four roots w_k are symmetrically placed around the origin? Wait, if they are fourth roots of 1 + i, which is not on the real axis, so the roots are not symmetric with respect to the origin. Wait, but they are equally spaced around a circle. So they form a square, but rotated and scaled.

But when we subtract i, we are shifting all points down by i. So the polygon formed by the z's is the square (from the w's) shifted down by i. However, since the original square is not centered at the origin, shifting it down by i might not make it symmetric in any particular way. Wait, but maybe there's a better way to think about this.

Alternatively, maybe I can find coordinates for each z_k. Let me try.

First, let me compute each w_k:

For w_0: angle π/16, modulus 2^{1/8}

So w_0 = 2^{1/8} (cos(π/16) + i sin(π/16))

Similarly, w_1: angle 9π/16, modulus 2^{1/8}

w_1 = 2^{1/8} (cos(9π/16) + i sin(9π/16))

w_2: angle 17π/16, modulus 2^{1/8}

w_2 = 2^{1/8} (cos(17π/16) + i sin(17π/16))

w_3: angle 25π/16, modulus 2^{1/8}

w_3 = 2^{1/8} (cos(25π/16) + i sin(25π/16))

Then z_k = w_k - i, so:

z_0 = 2^{1/8} (cos(π/16) + i sin(π/16)) - i

Similarly for others.

So to find the coordinates of each z_k, we can write them as:

For z_0: x-coordinate is 2^{1/8} cos(π/16), y-coordinate is 2^{1/8} sin(π/16) - 1

Similarly for z_1: x-coordinate is 2^{1/8} cos(9π/16), y-coordinate is 2^{1/8} sin(9π/16) - 1

z_2: x-coordinate is 2^{1/8} cos(17π/16), y-coordinate is 2^{1/8} sin(17π/16) - 1

z_3: x-coordinate is 2^{1/8} cos(25π/16), y-coordinate is 2^{1/8} sin(25π/16) - 1

Now, since the original w's form a square, when we subtract i, we are shifting all points down by 1 unit in the imaginary axis. Therefore, the polygon formed by the z's is a square rotated by π/16, scaled by 2^{1/8}, and then shifted down by 1 unit. However, since the original square is not centered at the origin, shifting it down by 1 unit might not affect the area. Wait, shifting a polygon doesn't change its area. Therefore, the area of the polygon formed by the z's is the same as the area of the square formed by the w's. Therefore, if I can compute the area of the square formed by the w's, that will be the area of the polygon.

But wait, the original w's are the fourth roots of 1 + i, which are four points forming a square. The area of a square can be calculated if we know the side length or the radius of the circumscribed circle. Since all four points lie on a circle of radius 2^{1/8}, which is the distance from the origin to each w_k. For a square inscribed in a circle of radius r, the side length is r * sqrt(2). Wait, no. Wait, for a square inscribed in a circle of radius r (which is the distance from the center to a vertex), the side length s is related by s = r * sqrt(2). Wait, let me recall. The diagonal of the square is 2r. Since the diagonal of a square is s*sqrt(2), so s*sqrt(2) = 2r => s = 2r / sqrt(2) = sqrt(2) r. Therefore, the area of the square is s^2 = (sqrt(2) r)^2 = 2 r^2. Therefore, if the radius is 2^{1/8}, then area is 2*(2^{1/8})^2 = 2*(2^{1/4}) = 2^{1 + 1/4} = 2^{5/4}.

Wait, but hold on. Wait, the radius here is 2^{1/8}, so r = 2^{1/8}, so area is 2*(2^{1/8})^2 = 2*(2^{1/4}) = 2^{1 + 1/4} = 2^{5/4}. Therefore, the area of the square formed by the w's is 2^{5/4}. Therefore, since shifting doesn't change the area, the area of the polygon formed by the z's is also 2^{5/4}. Therefore, the answer would be p^{a/b} where p is prime, and 2^{5/4} is already in that form. So p=2, a=5, b=4. Then a + b + p = 5 + 4 + 2 = 11. But wait, the problem says "the solutions of the equation... are the vertices of a convex polygon". Wait, but if the original w's form a square, then shifting them down by i would still form a square, which is convex. Therefore, the area is 2^{5/4}, so answer is 5 + 4 + 2 = 11? But wait, let me check again.

Wait, but maybe I made a mistake here. Because the original square is not centered at the origin. Wait, but when you take the fourth roots of a complex number, they are equally spaced around the circle, so they form a square. However, the center of the square is at the origin? Wait, if you have four points equally spaced around a circle, then the center of the square is the center of the circle. So since the w's are fourth roots of 1 + i, which is not zero, so they are on a circle centered at the origin. Therefore, the square is centered at the origin. Therefore, when we shift all points down by i, the center of the square moves from the origin to -i. However, the area remains the same. Therefore, the area of the polygon is indeed 2^{5/4}, so the answer is 2^{5/4}, so p=2, a=5, b=4, so 5 + 4 + 2 = 11. But wait, the problem says "the area of this polygon can be expressed in the form p^{a/b}". However, 2^{5/4} is already in that form. But maybe I need to check if my calculation of the area is correct.

Wait, let me think again. The area of a square inscribed in a circle of radius r is 2r². Wait, let me verify with a simple case. If the square is inscribed in a circle of radius 1, then the diagonal is 2, so the side length is 2 / sqrt(2) = sqrt(2), so area is (sqrt(2))² = 2. Which is 2*(1)^2 = 2. So yes, that formula is correct. Therefore, if the radius is 2^{1/8}, then area is 2*(2^{1/8})² = 2*(2^{1/4}) = 2^{1 + 1/4} = 2^{5/4}. Therefore, that seems correct.

But let me check if the original equation's solutions actually form a square. Since we have four roots, and they are fourth roots of 1 + i, which are equally spaced around a circle, so they form a square. Therefore, the polygon is a square, so area is 2^{5/4}. Therefore, answer is 5 + 4 + 2 = 11. However, I need to check if the problem says "convex polygon", which a square is, so that's okay.

But wait, maybe I made a mistake in assuming that the area is the same as the square. Wait, shifting the square down by i would not change the area. So yes, the area remains the same. Therefore, the answer is 2^{5/4}, so 5 + 4 + 2 = 11. However, I need to check if the problem is expecting something else. Wait, maybe I made a mistake in the equation transformation?

Let me check again. Original equation: z^4 + 4i z^3 - 6 z^2 - 4i z - i = 0. Then I noticed that (z + i)^4 = z^4 + 4i z^3 - 6 z^2 - 4i z + 1. Therefore, original equation is (z + i)^4 - 1 - i = 0 => (z + i)^4 = 1 + i. Therefore, yes, that's correct. Therefore, solving for w = z + i, w^4 = 1 + i. Therefore, the solutions for w are the fourth roots of 1 + i, which are four points on a circle of radius |1 + i|^{1/4} = (sqrt(2))^{1/4} = 2^{1/8}. Therefore, radius is 2^{1/8}, so area of square is 2*(2^{1/8})^2 = 2^{5/4}. Therefore, answer is 2^{5/4}, so 5 + 4 + 2 = 11. But wait, the problem says "the area of this polygon can be expressed in the form p^{a/b}". However, 2^{5/4} is already in that form. However, maybe the problem expects a different form? Let me check if 2^{5/4} can be written as something else. For example, 2^{5/4} is equal to sqrt(2) * 2^{1/4} or something else? But no, it's already in the form p^{a/b} with p=2, a=5, b=4. Since 5 and 4 are coprime, and 2 is prime. Therefore, the answer is 5 + 4 + 2 = 11. However, I need to check if the area is indeed 2^{5/4}.

Alternatively, maybe I need to compute the area using coordinates. Let me try that. Let me compute coordinates of the four points z_0, z_1, z_2, z_3.

First, let me note that the original square (from w's) has vertices at angles π/16, 9π/16, 17π/16, 25π/16. When we subtract i, we are shifting each point down by 1 unit in the imaginary axis. Therefore, the coordinates of z_k are:

For each w_k = 2^{1/8} e^{iθ}, z_k = w_k - i = 2^{1/8} e^{iθ} - i.

But maybe instead of working with angles, I can think of the square. Since the original square is centered at the origin, and after shifting down by i, the center is at -i. However, the area remains the same. Therefore, the area is still 2^{5/4}.

Alternatively, maybe I can compute the distance between two adjacent points and then compute the area. Let me try that. Let me take two adjacent points, say z_0 and z_1. The distance between them would be the same as the distance between w_0 and w_1, since subtracting i from both doesn't change the distance between them. Therefore, the side length of the square is the same as the distance between two adjacent w's.

So let me compute the distance between w_0 and w_1. Since they are two adjacent fourth roots of 1 + i, separated by angle π/2. The distance between two points on a circle of radius r separated by angle θ is 2r sin(θ/2). Here, θ = π/2, so distance is 2r sin(π/4) = 2r*(√2/2) = r√2. Therefore, the side length s = r√2, so area is s² = 2r². Which is the same as before. Therefore, area is 2*(2^{1/8})² = 2^{5/4}. Therefore, same result. Therefore, seems correct.

But let me check with actual coordinates. Let me compute coordinates for z_0 and z_1.

Let me compute coordinates for w_0: angle π/16, modulus 2^{1/8}.

So coordinates of w_0: (2^{1/8} cos(π/16), 2^{1/8} sin(π/16))

Coordinates of z_0: (2^{1/8} cos(π/16), 2^{1/8} sin(π/16) - 1)

Similarly, coordinates of w_1: angle 9π/16, modulus 2^{1/8}

Coordinates of z_1: (2^{1/8} cos(9π/16), 2^{1/8} sin(9π/16) - 1)

So the distance between z_0 and z_1 is sqrt[(2^{1/8} cos(9π/16) - 2^{1/8} cos(π/16))² + (2^{1/8} sin(9π/16) - 1 - (2^{1/8} sin(π/16) - 1))²] = sqrt[(2^{1/8} (cos(9π/16) - cos(π/16)))² + (2^{1/8} (sin(9π/16) - sin(π/16)))²]

Which simplifies to 2^{1/8} * sqrt[(cos(9π/16) - cos(π/16))² + (sin(9π/16) - sin(π/16))²]

Which is 2^{1/8} * sqrt[2 - 2 cos(9π/16) cos(π/16) - 2 sin(9π/16) sin(π/16)] by expanding the squares. Wait, alternatively, using the identity that sqrt[(cos A - cos B)^2 + (sin A - sin B)^2] = 2 |sin((A - B)/2)|. Wait, let me recall:

The distance between two points on the unit circle at angles A and B is 2 sin(|A - B|/2). Wait, more precisely, if you have two points on a circle of radius r, separated by angle θ, then the distance between them is 2r sin(θ/2). In this case, the angle between w_0 and w_1 is π/2, so the distance between w_0 and w_1 is 2 * 2^{1/8} * sin(π/4) = 2^{1 + 1/8} * √2 / 2 = 2^{9/8} * √2 / 2 = 2^{9/8} * 2^{-1/2} = 2^{9/8 - 4/8} = 2^{5/8}. Wait, but earlier I thought the side length was r√2. Wait, if r is 2^{1/8}, then r√2 is 2^{1/8} * 2^{1/2} = 2^{5/8}, which matches. Therefore, the distance between w_0 and w_1 is 2^{5/8}, so the side length is 2^{5/8}, so area is (2^{5/8})² = 2^{5/4}, same as before. Therefore, that's correct.

Therefore, the area is indeed 2^{5/4}, so the answer is 5 + 4 + 2 = 11. However, I need to check if the problem says "the area of this polygon can be expressed in the form p^{a/b}". But 2^{5/4} is already in that form. However, maybe there's a mistake here? Let me check again.

Wait, but maybe the polygon isn't a square? Wait, the original equation is quartic, so four roots. If they are the fourth roots of 1 + i shifted by -i, then they should form a square. But maybe the shifting causes some distortion? Wait, no. Because shifting all points by the same vector doesn't change the shape or the area. So if the original four points form a square, then after shifting, they still form a square. Therefore, the area is the same. Therefore, 2^{5/4} is correct.

But let me check with another approach. Let me think of the original equation again. If I have w^4 = 1 + i, then the four roots are equally spaced, so they form a square. Therefore, the area is 2r² where r is the modulus of each root. Since modulus of each root is |1 + i|^{1/4} = (sqrt(2))^{1/4} = 2^{1/8}, so area is 2*(2^{1/8})² = 2^{1 + 1/4} = 2^{5/4}. Therefore, same result. Therefore, I think this is correct.

But maybe the problem is expecting a different answer? Let me check if I made a mistake in the equation transformation. Let me re-derive that.

Original equation: z^4 + 4i z^3 - 6 z^2 - 4i z - i = 0.

Let me compute (z + i)^4:

(z + i)^4 = z^4 + 4 z^3 i + 6 z^2 i^2 + 4 z i^3 + i^4

= z^4 + 4i z^3 + 6 z^2 (-1) + 4 z (-i) + 1

= z^4 + 4i z^3 - 6 z^2 - 4i z + 1.

Therefore, (z + i)^4 = z^4 + 4i z^3 - 6 z^2 - 4i z + 1.

Comparing with original equation: z^4 + 4i z^3 - 6 z^2 - 4i z - i = 0.

Therefore, original equation is (z + i)^4 - 1 - i = 0 => (z + i)^4 = 1 + i. Correct. Therefore, that's correct.

Therefore, the rest follows. Therefore, the area is 2^{5/4}, so answer is 5 + 4 + 2 = 11. However, I need to check if the problem says "the area of this polygon can be expressed in the form p^{a/b}". But 2^{5/4} is already in that form. However, maybe the problem expects the answer in another form? For example, maybe rationalizing or something else? Let me check.

Alternatively, maybe I made a mistake in assuming that the area is 2r². Let me think again. If the square is inscribed in a circle of radius r, then the area is 2r². But let me think of coordinates. Suppose we have a square centered at the origin, with vertices at (r, 0), (0, r), (-r, 0), (0, -r). Wait, no, that's not a square. Wait, if you have a square inscribed in a circle of radius r, then the coordinates of the vertices are (r, r), (-r, r), (-r, -r), (r, -r)? No, that's a square rotated by 45 degrees. Wait, the distance from the origin to each vertex is sqrt(r² + r²) = r√2. Therefore, if the radius is r, then the side length is sqrt( (2r)^2 + (2r)^2 )? No, wait. Let me think again.

Let me take a square with vertices at (a, a), (-a, a), (-a, -a), (a, -a). The distance between (a, a) and (-a, a) is 2a. The distance between (a, a) and (a, -a) is 2a. Wait, but that's a square with side length 2a√2? No, wait, the distance between (a, a) and (-a, a) is 2a, but that's the length of the side? Wait, no, if you have a square with vertices at (a, a), (-a, a), (-a, -a), (a, -a), then the sides are between (a, a) and (-a, a), which is horizontal line of length 2a, then from (-a, a) to (-a, -a), vertical line of length 2a, etc. Wait, but that's actually a square rotated by 45 degrees relative to the axes? No, actually, that's a square aligned with the axes, but with side length 2a. However, the distance from the origin to each vertex is sqrt(a² + a²) = a√2. Therefore, if the radius is r = a√2, then the side length is 2a = 2r / √2 = sqrt(2) r. Therefore, area is (sqrt(2) r)^2 = 2 r². Therefore, same as before. Therefore, if the radius is 2^{1/8}, then area is 2*(2^{1/8})² = 2^{5/4}. Therefore, correct.

Therefore, I think my answer is correct. Therefore, the answer is 5 + 4 + 2 = 11. However, I need to check if the problem says "the area of this polygon can be expressed in the form p^{a/b}". But 2^{5/4} is already in that form. However, maybe the problem expects the answer in another form? For example, maybe they want it as (sqrt(2))^{5/2} or something else? But no, 2^{5/4} is simpler. Also, 5 and 4 are coprime, and 2 is prime. Therefore, the answer is 11.

But wait, I just thought of something. The original equation is quartic, so four roots. The problem says "the solutions... are the vertices of a convex polygon". A square is convex, so that's okay. However, maybe the polygon is not a square? Wait, if the four roots are equally spaced on a circle, then they form a square. Therefore, yes. Therefore, the area is 2^{5/4}. Therefore, answer is 11. However, I need to check if there is an error in my calculation. Let me check with specific numbers.

Let me compute 2^{5/4}. 2^{1/4} is sqrt(sqrt(2)) ≈ 1.1892, so 2^{5/4} is 2 * 2^{1/4} ≈ 2.3784. Alternatively, if I compute the area using coordinates. Let me take specific values.

Let me compute coordinates of z_0, z_1, z_2, z_3. Let me compute approximate values.

First, compute 2^{1/8}. Let me compute 2^{1/8} ≈ e^{(ln 2)/8} ≈ e^{0.0866} ≈ 1.089. Let me check with calculator:

2^(1/8) = 2^(0.125) ≈ 1.0905.

So modulus of each w is approx 1.0905.

Now, angles:

For w_0: π/16 ≈ 0.19635 radians ≈ 11.25 degrees.

cos(π/16) ≈ cos(11.25°) ≈ 0.98078

sin(π/16) ≈ 0.19509

Therefore, coordinates of w_0 ≈ (1.0905 * 0.98078, 1.0905 * 0.19509) ≈ (1.070, 0.2126)

Then z_0 ≈ (1.070, 0.2126 - 1) ≈ (1.070, -0.7874)

Similarly, w_1: angle 9π/16 ≈ 1.767 radians ≈ 101.25 degrees.

cos(9π/16) ≈ cos(101.25°) ≈ -0.19509

sin(9π/16) ≈ sin(101.25°) ≈ 0.98078

Therefore, coordinates of w_1 ≈ (1.0905 * (-0.19509), 1.0905 * 0.98078) ≈ (-0.2126, 1.070)

Then z_1 ≈ (-0.2126, 1.070 - 1) ≈ (-0.2126, 0.070)

Similarly, w_2: angle 17π/16 ≈ 3.334 radians ≈ 191.25 degrees.

cos(17π/16) ≈ cos(191.25°) ≈ -cos(11.25°) ≈ -0.98078

sin(17π/16) ≈ sin(191.25°) ≈ -sin(11.25°) ≈ -0.19509

Coordinates of w_2 ≈ (1.0905 * (-0.98078), 1.0905 * (-0.19509)) ≈ (-1.070, -0.2126)

Then z_2 ≈ (-1.070, -0.2126 - 1) ≈ (-1.070, -1.2126)

Similarly, w_3: angle 25π/16 ≈ 4.909 radians ≈ 281.25 degrees.

cos(25π/16) ≈ cos(281.25°) ≈ cos(-78.75°) ≈ cos(78.75°) ≈ 0.19509

sin(25π/16) ≈ sin(281.25°) ≈ sin(-78.75°) ≈ -sin(78.75°) ≈ -0.98078

Coordinates of w_3 ≈ (1.0905 * 0.19509, 1.0905 * (-0.98078)) ≈ (0.2126, -1.070)

Then z_3 ≈ (0.2126, -1.070 - 1) ≈ (0.2126, -2.070)

So the four points are approximately:

z_0 ≈ (1.070, -0.7874)

z_1 ≈ (-0.2126, 0.070)

z_2 ≈ (-1.070, -1.2126)

z_3 ≈ (0.2126, -2.070)

Wait, let me check if these points form a square. Let me compute the distances between consecutive points.

First, between z_0 and z_1:

Distance between (1.070, -0.7874) and (-0.2126, 0.070):

Δx = -0.2126 - 1.070 = -1.2826

Δy = 0.070 - (-0.7874) = 0.8574

Distance squared: (-1.2826)^2 + (0.8574)^2 ≈ 1.645 + 0.735 ≈ 2.38

Distance ≈ sqrt(2.38) ≈ 1.543

Between z_1 and z_2:

(-0.2126, 0.070) to (-1.070, -1.2126):

Δx = -1.070 - (-0.2126) = -0.8574

Δy = -1.2126 - 0.070 = -1.2826

Distance squared: (-0.8574)^2 + (-1.2826)^2 ≈ 0.735 + 1.645 ≈ 2.38, same as before. Distance ≈ 1.543

Between z_2 and z_3:

(-1.070, -1.2126) to (0.2126, -2.070):

Δx = 0.2126 - (-1.070) = 1.2826

Δy = -2.070 - (-1.2126) = -0.8574

Distance squared: (1.2826)^2 + (-0.8574)^2 ≈ same as before, 2.38. Distance ≈ 1.543

Between z_3 and z_0:

(0.2126, -2.070) to (1.070, -0.7874):

Δx = 1.070 - 0.2126 = 0.8574

Δy = -0.7874 - (-2.070) = 1.2826

Distance squared: (0.8574)^2 + (1.2826)^2 ≈ same as before, 2.38. Distance ≈ 1.543

So all sides are equal. Now check the diagonals. Let me check between z_0 and z_2, and between z_1 and z_3.

Between z_0 (1.070, -0.7874) and z_2 (-1.070, -1.2126):

Δx = -1.070 - 1.070 = -2.14

Δy = -1.2126 - (-0.7874) = -0.4252

Distance squared: (-2.14)^2 + (-0.4252)^2 ≈ 4.5796 + 0.1808 ≈ 4.7604

Distance ≈ 2.18

Between z_1 (-0.2126, 0.070) and z_3 (0.2126, -2.070):

Δx = 0.2126 - (-0.2126) = 0.4252

Δy = -2.070 - 0.070 = -2.14

Distance squared: (0.4252)^2 + (-2.14)^2 ≈ same as above, 0.1808 + 4.5796 ≈ 4.7604. Same distance. Therefore, diagonals are equal. Therefore, the figure is a square. Therefore, the area can be calculated as (side length)^2. The side length squared is approx 2.38, so area is approx 2.38. Let me compute 2^{5/4} ≈ 2^(1.25) ≈ 2.378, which matches the approximate value. Therefore, this confirms that the area is indeed 2^{5/4}.

Therefore, the answer is 2^{5/4}, so p=2, a=5, b=4, so a + b + p = 5 + 4 + 2 = 11. Therefore, the answer is 11.

But wait, just to be thorough, let me check if there's another way the problem could be interpreted. For example, maybe the polygon is not a square? But with four points forming a square, it's convex. The problem says "the solutions... are the vertices of a convex polygon", which is satisfied. Therefore, I think my answer is correct.

Therefore, I think the answer is 11.

**Final Answer**
\boxed{11}

To solve the equation:

$$
z^4 + 4z^3i - 6z^2 - 4zi - i = 0
$$

we begin by recognizing a pattern in the coefficients. This suggests a possible substitution or transformation. Let us consider the expression:

$$
(z + i)^4 = z^4 + 4i z^3 - 6 z^2 - 4i z + 1
$$

Comparing this with the original equation, we see that:

$$
z^4 + 4i z^3 - 6 z^2 - 4i z - i = (z + i)^4 - 1 - i
$$

Thus, the original equation becomes:

$$
(z + i)^4 = 1 + i
$$

Letting $ w = z + i $, we now solve:

$$
w^4 = 1 + i
$$

---

### Step 1: Express $ 1 + i $ in polar form

The modulus of $ 1 + i $ is:

$$
|1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}
$$

The argument is:

$$
\arg(1 + i) = \frac{\pi}{4}
$$

So, in polar form:

$$
1 + i = \sqrt{2} \cdot e^{i\pi/4}
$$

---

### Step 2: Find the fourth roots of $ 1 + i $

The fourth roots of $ \sqrt{2} \cdot e^{i\pi/4} $ are:

$$
w_k = \left( \sqrt{2} \right)^{1/4} \cdot e^{i(\pi/16 + k\pi/2)}, \quad k = 0, 1, 2, 3
$$

Note that $ \left( \sqrt{2} \right)^{1/4} = 2^{1/8} $, so:

$$
w_k = 2^{1/8} \cdot e^{i(\pi/16 + k\pi/2)}
$$

These four roots are equally spaced on a circle of radius $ 2^{1/8} $, forming a square in the complex plane.

---

### Step 3: Find the corresponding $ z $ values

Recall that $ z = w - i $, so the four solutions for $ z $ are:

$$
z_k = 2^{1/8} \cdot e^{i(\pi/16 + k\pi/2)} - i
$$

These four points form a square in the complex plane, centered at $ -i $, with side length $ 2^{5/8} $, as derived from the distance between adjacent roots.

---

### Step 4: Compute the area of the square

The area of a square with side length $ s $ is $ s^2 $. Here, the side length is:

$$
s = 2^{5/8}
$$

So the area is:

$$
\text{Area} = (2^{5/8})^2 = 2^{5/4}
$$

---

### Step 5: Express the area in the required form

We are told the area can be expressed as $ p^{a/b} $, where $ p $ is prime, and $ a $ and $ b $ are relatively prime. Here:

$$
2^{5/4} = 2^{5/4}
$$

So $ p = 2 $, $ a = 5 $, $ b = 4 $, and since $ \gcd(5, 4) = 1 $, this satisfies the conditions.

---

### Final Step: Compute $ a + b + p $

$$
a + b + p = 5 + 4 + 2 = \boxed{11}
$$

Solution:
By the Binomial Theorem,
\begin{align*}
(z + i)^4 &= z^4 + 4z^3 i + 6z^2 i^2 + 4zi^3 + 1 \\
&= z^4 + 4iz^3 - 6z^2 - 4iz + 1.
\end{align*}So, if $z^4 + 4z^3 i - 6z^2 - 4zi - i = 0,$ then
\[(z + i)^4 = z^4 + 4iz^3 - 6z^2 - 4iz + 1 = 1 + i.\]Let $w = z + i,$ so $w^4 = 1 + i.$  (If we plot the solutions $w$ in the complex plane, we obtain the same area as from the solutions $z$ in the complex plane, because the substitution $w = z + i$ simply translates the polygon.)

If $w^4 = 1 + i,$ then
\[(wi)^4 = w^4 i^4 = w^4 = 1 + i.\]Thus, if $w$ is a solution, then so are $iw,$ $i^2 w = -w,$ and $i^3 w = -iw,$ which form a square in the complex plane.

[asy]
unitsize(2 cm);

pair A, B, C, D;

A = 2^(1/8)*dir(45/4);
B = 2^(1/8)*dir(45/4 + 90);
C = 2^(1/8)*dir(45/4 + 180);
D = 2^(1/8)*dir(45/4 + 270);

draw(A--B--C--D--cycle);
draw((-1.5,0)--(1.5,0));
draw((0,-1.5)--(0,1.5));

dot("$w$", A, E);
dot("$iw$", B, N);
dot("$-w$", C, W);
dot("$-iw$", D, S);
[/asy]

From the equation $w^4 = 1 + i,$ $|w^4| = |1 + i|.$  Then $|w|^4 = \sqrt{2},$ so $|w| = 2^{1/8}.$  Therefore, the side length of the square is
\[|w - iw| = |w||1 - i| = 2^{1/8} \sqrt{2} = 2^{5/8},\]so the area of the square is $(2^{5/8})^2 = 2^{5/4}.$  The final answer is $5 + 4 + 2 = \boxed{11}.$


Explanation:
Okay, so I need to find the minimum value of the function $\frac{(x + 5)(x + 2)}{x + 1}$ where $x > 0$. Hmm, let me think about how to approach this.

First, maybe I should simplify the expression. Let me expand the numerator. The numerator is $(x + 5)(x + 2)$. Let me multiply that out:

$$
(x + 5)(x + 2) = x^2 + 2x + 5x + 10 = x^2 + 7x + 10
$$

So the function becomes:

$$
\frac{x^2 + 7x + 10}{x + 1}
$$

Hmm, maybe I can perform polynomial division here to simplify it further. Let me try dividing $x^2 + 7x + 10$ by $x + 1$.

Dividing $x^2$ by $x$ gives $x$. Multiply $x + 1$ by $x$ to get $x^2 + x$. Subtract that from the original polynomial:

$$
(x^2 + 7x + 10) - (x^2 + x) = 6x + 10
$$

Now, divide $6x$ by $x$ to get 6. Multiply $x + 1$ by 6 to get $6x + 6$. Subtract that from the previous remainder:

$$
(6x + 10) - (6x + 6) = 4
$$

So the division gives:

$$
x + 6 + \frac{4}{x + 1}
$$

Therefore, the original function simplifies to:

$$
x + 6 + \frac{4}{x + 1}
$$

So now, the problem is to find the minimum of $f(x) = x + 6 + \frac{4}{x + 1}$ for $x > 0$.

Hmm, maybe I can take the derivative and set it to zero to find critical points. Let me try that.

First, let me write $f(x)$ again:

$$
f(x) = x + 6 + \frac{4}{x + 1}
$$

Compute the derivative $f'(x)$:

The derivative of $x$ is 1, the derivative of 6 is 0, and the derivative of $\frac{4}{x + 1}$ is $-4/(x + 1)^2$. So:

$$
f'(x) = 1 - \frac{4}{(x + 1)^2}
$$

To find critical points, set $f'(x) = 0$:

$$
1 - \frac{4}{(x + 1)^2} = 0 \\
\Rightarrow \frac{4}{(x + 1)^2} = 1 \\
\Rightarrow (x + 1)^2 = 4 \\
\Rightarrow x + 1 = \pm 2
$$

But since $x > 0$, $x + 1$ is definitely positive, so we take the positive root:

$$
x + 1 = 2 \Rightarrow x = 1
$$

So the critical point is at $x = 1$. Now, I need to check if this is a minimum. Let me use the second derivative test.

Compute the second derivative $f''(x)$:

The derivative of $f'(x) = 1 - 4/(x + 1)^2$ is:

$$
f''(x) = 0 - (-8)/(x + 1)^3) = \frac{8}{(x + 1)^3}
$$

At $x = 1$, $f''(1) = 8/(1 + 1)^3 = 8/8 = 1 > 0$. Since the second derivative is positive, this critical point is a local minimum. Since the function tends to infinity as $x$ approaches 0 and as $x$ approaches infinity, this local minimum is actually the global minimum for $x > 0$.

Therefore, the minimum occurs at $x = 1$. Let me compute the value of the function at $x = 1$:

Original function: $\frac{(1 + 5)(1 + 2)}{1 + 1} = \frac{6 \times 3}{2} = \frac{18}{2} = 9$.

Alternatively, using the simplified expression $x + 6 + 4/(x + 1)$:

At $x = 1$, that's $1 + 6 + 4/(1 + 1) = 7 + 4/2 = 7 + 2 = 9$. Yep, same result.

Wait, but just to make sure there isn't a lower value somewhere else... Let me check another point. For example, take $x = 2$:

Original function: $\frac{(2 + 5)(2 + 2)}{2 + 1} = \frac{7 \times 4}{3} = 28/3 ≈ 9.333$, which is higher than 9. How about $x = 0.5$:

$\frac{(0.5 + 5)(0.5 + 2)}{0.5 + 1} = \frac{5.5 \times 2.5}{1.5}$. Let me compute that:

5.5 * 2.5 = 13.75; 13.75 / 1.5 ≈ 9.166..., which is still higher than 9. How about $x = 3$:

$\frac{(3 + 5)(3 + 2)}{3 + 1} = \frac{8 * 5}{4} = 40/4 = 10$, which is higher. So seems like 9 is indeed the minimum.

Alternatively, maybe I can use the AM-GM inequality? Let me think. The simplified expression is $x + 6 + \frac{4}{x + 1}$. Let me make substitution to make it easier. Let me set $t = x + 1$, since $x > 0$, then $t > 1$. Then $x = t - 1$. Substitute into the expression:

Original expression becomes:

$$
(t - 1) + 6 + \frac{4}{t} = t - 1 + 6 + \frac{4}{t} = t + 5 + \frac{4}{t}
$$

So now, the problem reduces to minimizing $t + 5 + \frac{4}{t}$ where $t > 1$.

Hmm, maybe I can apply AM-GM on $t$ and $\frac{4}{t}$. But AM-GM requires positive numbers, which they are here since $t > 1 > 0$. However, AM-GM for two terms says that $a + b \geq 2\sqrt{ab}$, with equality when $a = b$. Let me try that.

Let me consider $t + \frac{4}{t}$. Then:

$$
t + \frac{4}{t} \geq 2\sqrt{t \times \frac{4}{t}} = 2\sqrt{4} = 4
$$

So the minimum of $t + \frac{4}{t}$ is 4, achieved when $t = \frac{4}{t}$, i.e., $t^2 = 4$, so $t = 2$ (since $t > 1$). Therefore, the expression $t + 5 + \frac{4}{t}$ has minimum value $4 + 5 = 9$, achieved when $t = 2$. Which corresponds to $x + 1 = 2$ so $x = 1$, same as before.

Therefore, this confirms that the minimum value is indeed 9.

Alternatively, maybe I can use calculus again, but since I already did that and confirmed with substitution and AM-GM, I think that's solid.

Wait, but just to make sure, let me check if there's another way. Maybe completing the square or something else? Let me think.

Looking back at the simplified expression $x + 6 + \frac{4}{x + 1}$. Let me set $y = x + 1$, so $x = y - 1$, and since $x > 0$, $y > 1$. Then substituting:

$$
(y - 1) + 6 + \frac{4}{y} = y + 5 + \frac{4}{y}
$$

Which is the same as before. So again, same expression. So, same conclusion.

Alternatively, if I didn't think of substitution, maybe I could have considered the original function and tried to write it as something squared plus something. But I think the methods I used are sufficient.

Another check: Let me consider the function $f(x) = \frac{(x + 5)(x + 2)}{x + 1}$. Maybe I can write it as:

Let me try to perform substitution. Let me set $x + 1 = t$, so $x = t - 1$, then:

Numerator becomes $(t - 1 + 5)(t - 1 + 2) = (t + 4)(t + 1)$. Therefore, the function becomes:

$$
\frac{(t + 4)(t + 1)}{t} = \frac{t^2 + 5t + 4}{t} = t + 5 + \frac{4}{t}
$$

Which is the same as before. So again, same expression. Therefore, the minimum is 9 when t = 2, so x = 1.

Therefore, regardless of the method, the minimum value is 9.

Wait, but just to make sure, let me check if there's a possibility that the minimum occurs at the boundary. However, since $x > 0$, the domain is open interval (0, ∞). So there's no boundary at x approaching 0 or infinity. But let me check the limits:

As $x \to 0^+$, the original function is $\frac{(0 + 5)(0 + 2)}{0 + 1} = \frac{5 * 2}{1} = 10$. So approaching 10.

As $x \to \infty$, the function behaves like $\frac{x^2}{x} = x$, which tends to infinity. Therefore, the function approaches infinity as x approaches infinity. Therefore, the minimum must be somewhere in between, which we found at x = 1 with value 9.

Therefore, I think that's solid.

Alternatively, if I didn't use calculus, but tried to use AM-GM, as I did earlier, that also gives the same answer.

So, conclusion: the minimum value is 9, achieved at x = 1.

**Final Answer**
The minimum value is \boxed{9}.

To find the minimum value of the expression

$$
\frac{(x + 5)(x + 2)}{x + 1}
$$

for $ x > 0 $, we begin by simplifying the expression.

---

### Step 1: Expand the Numerator

$$
(x + 5)(x + 2) = x^2 + 7x + 10
$$

So the function becomes:

$$
f(x) = \frac{x^2 + 7x + 10}{x + 1}
$$

---

### Step 2: Perform Polynomial Division

Divide $ x^2 + 7x + 10 $ by $ x + 1 $:

- Divide $ x^2 $ by $ x $: $ x $
- Multiply $ x + 1 $ by $ x $: $ x^2 + x $
- Subtract: $ (x^2 + 7x + 10) - (x^2 + x) = 6x + 10 $
- Divide $ 6x $ by $ x $: $ 6 $
- Multiply $ x + 1 $ by $ 6 $: $ 6x + 6 $
- Subtract: $ (6x + 10) - (6x + 6) = 4 $

So the division yields:

$$
f(x) = x + 6 + \frac{4}{x + 1}
$$

---

### Step 3: Minimize the Simplified Function

We now minimize:

$$
f(x) = x + 6 + \frac{4}{x + 1}
$$

Let’s define:

$$
f(x) = x + 6 + \frac{4}{x + 1}
$$

Take the derivative:

$$
f'(x) = 1 - \frac{4}{(x + 1)^2}
$$

Set the derivative equal to zero to find critical points:

$$
1 - \frac{4}{(x + 1)^2} = 0 \Rightarrow (x + 1)^2 = 4 \Rightarrow x + 1 = 2 \Rightarrow x = 1
$$

Since $ x > 0 $, this is valid.

---

### Step 4: Confirm Minimum with Second Derivative

Compute the second derivative:

$$
f''(x) = \frac{8}{(x + 1)^3}
$$

At $ x = 1 $:

$$
f''(1) = \frac{8}{(1 + 1)^3} = \frac{8}{8} = 1 > 0
$$

So the function has a local minimum at $ x = 1 $, and since the function tends to infinity as $ x \to 0^+ $ and $ x \to \infty $, this is the **global minimum** on $ x > 0 $.

---

### Step 5: Compute the Minimum Value

Substitute $ x = 1 $ into the original function:

$$
f(1) = \frac{(1 + 5)(1 + 2)}{1 + 1} = \frac{6 \cdot 3}{2} = \frac{18}{2} = 9
$$

Alternatively, using the simplified expression:

$$
f(1) = 1 + 6 + \frac{4}{1 + 1} = 7 + 2 = 9
$$

---

### Final Answer

$$
\boxed{9}
$$

[Assistant]
Explanation:
Okay, so I need to find the minimum value of the function $\frac{(x + 5)(x + 2)}{x + 1}$ where $x > 0$. Hmm, let me think about how to approach this.

First, maybe I should simplify the expression. Let me expand the numerator. The numerator is $(x + 5)(x + 2)$. Let me multiply that out:

$$
(x + 5)(x + 2) = x^2 + 2x + 5x + 10 = x^2 + 7x + 10
$$

So the function becomes:

$$
\frac{x^2 + 7x + 10}{x + 1}
$$

Hmm, maybe I can perform polynomial division here to simplify it further. Let me try dividing $x^2 + 7x + 10$ by $x + 1$.

Dividing $x^2$ by $x$ gives $x$. Multiply $x + 1$ by $x$ to get $x^2 + x$. Subtract that from the original polynomial:

$$
(x^2 + 7x + 10) - (x^2 + x) = 6x + 10
$$

Now, divide $6x$ by $x$ to get 6. Multiply $x + 1$ by 6 to get $6x + 6$. Subtract that from the previous remainder:

$$
(6x + 10) - (6x + 6) = 4
$$

So the division gives:

$$
x + 6 + \frac{4}{x + 1}
$$

Therefore, the original function simplifies to:

$$
x + 6 + \frac{4}{x + 1}
$$

So now, the problem is to find the minimum of $f(x) = x + 6 + \frac{4}{x + 1}$ for $x > 0$.

Hmm, maybe I can take the derivative and set it to zero to find critical points. Let me try that.

First, let me write $f(x)$ again:

$$
f(x) = x + 6 + \frac{4}{x + 1}
$$

Compute the derivative $f'(x)$:

The derivative of $x$ is 1, the derivative of 6 is 0, and the derivative of $\frac{4}{x + 1}$ is $-4/(x + 1)^2$. So:

$$
f'(x) = 1 - \frac{4}{(x + 1)^2}
$$

To find critical points, set $f'(x) = 0$:

$$
1 - \frac{4}{(x + 1)^2} = 0 \\
\Rightarrow \frac{4}{(x + 1)^2} = 1 \\
\Rightarrow (x + 1)^2 = 4 \\
\Rightarrow x + 1 = \pm 2
$$

But since $x > 0$, $x + 1$ is definitely positive, so we take the positive root:

$$
x + 1 = 2 \Rightarrow x = 1
$$

So the critical point is at $x = 1$. Now, I need to check if this is a minimum. Let me use the second derivative test.

Compute the second derivative $f''(x)$:

The derivative of $f'(x) = 1 - 4/(x + 1)^2$ is:

$$
f''(x) = 0 - (-8)/(x + 1)^3) = \frac{8}{(x + 1)^3}
$$

At $x = 1$, $f''(1) = 8/(1 + 1)^3 = 8/8 = 1 > 0$. Since the second derivative is positive, this critical point is a local minimum. Since the function tends to infinity as $x$ approaches 0 and as $x$ approaches infinity, this local minimum is actually the global minimum for $x > 0$.

Therefore, the minimum occurs at $x = 1$. Let me compute the value of the function at $x = 1$:

Original function: $\frac{(1 + 5)(1 + 2)}{1 + 1} = \frac{6 \times 3}{2} = \frac{18}{2} = 9$.

Alternatively, using the simplified expression $x + 6 + 4/(x + 1)$:

At $x = 1$, that's $1 + 6 + 4/(1 + 1) = 7 + 4/2 = 7 + 2 = 9$. Yep, same result.

Wait, but just to make sure there isn't a lower value somewhere else... Let me check another point. For example, take $x = 2$:

Original function: $\frac{(2 + 5)(2 + 2)}{2 + 1} = \frac{7 \times 4}{3} = 28/3 ≈ 9.333$, which is higher than 9. How about $x = 0.5$:

$\frac{(0.5 + 5)(0.5 + 2)}{0.5 + 1} = \frac{5.5 \times 2.5}{1.5}$. Let me compute that:

5.5 * 2.5 = 13.75; 13.75 / 1.5 ≈ 9.166..., which is still higher than 9. How about $x = 3$:

$\frac{(3 + 5)(3 + 2)}{3 + 1} = \frac{8 * 5}{4} = 40/4 = 10$, which is higher. So seems like 9 is indeed the minimum.

Alternatively, maybe I can use the AM-GM inequality? Let me think. The simplified expression is $x + 6 + \frac{4}{x + 1}$. Let me make substitution to make it easier. Let me set $t = x + 1$, since $x > 0$, then $t > 1$. Then $x = t - 1$. Substitute into the expression:

Original expression becomes:

$$
(t - 1) + 6 + \frac{4}{t} = t - 1 + 6 + \frac{4}{t} = t + 5 + \frac{4}{t}
$$

So now, the problem reduces to minimizing $t + 5 + \frac{4}{t}$ where $t > 1$.

Hmm, maybe I can apply AM-GM on $t$ and $\frac{4}{t}$. But AM-GM requires positive numbers, which they are here since $t > 1 > 0$. However, AM-GM for two terms says that $a + b \geq 2\sqrt{ab}$, with equality when $a = b$. Let me try that.

Let me consider $t + \frac{4}{t}$. Then:

$$
t + \frac{4}{t} \geq 2\sqrt{t \times \frac{4}{t}} = 2\sqrt{4} = 4
$$

So the minimum of $t + \frac{4}{t}$ is 4, achieved when $t = \frac{4}{t}$, i.e., $t^2 = 4$, so $t = 2$ (since $t > 1$). Therefore, the expression $t + 5 + \frac{4}{t}$ has minimum value $4 + 5 = 9$, achieved when $t = 2$. Which corresponds to $x + 1 = 2$ so $x = 1$, same as before.

Therefore, this confirms that the minimum value is indeed 9.

Alternatively, maybe I can use calculus again, but since I already did that and confirmed with substitution and AM-GM, I think that's solid.

Wait, but just to make sure, let me check if there's another way. Maybe completing the square or something else? Let me think.

Looking back at the simplified expression $x + 6 + \frac{4}{x + 1}$. Let me set $y = x + 1$, so $x = y - 1$, and since $x > 0$, $y > 1$. Then substituting:

$$
(y - 1) + 6 + \frac{4}{y} = y + 5 + \frac{4}{y}
$$

Which is the same as before. So again, same expression. So, same conclusion.

Alternatively, if I didn't think of substitution, maybe I could have considered the original function and tried to write it as something squared plus something. But I think the methods I used are sufficient.

Another check: Let me consider the function $f(x) = \frac{(x + 5)(x + 2)}{x + 1}$. Maybe I can write it as:

Let me try to perform substitution. Let me set $x + 1 = t$, so $x = t - 1$, then:

Numerator becomes $(t - 1 + 5)(t - 1 + 2) = (t + 4)(t + 1)$. Therefore, the function becomes:

$$
\frac{(t + 4)(t + 1)}{t} = \frac{t^2 + 5t + 4}{t} = t + 5 + \frac{4}{t}
$$

Which is the same as before. So again, same expression. Therefore, the minimum is 9 when t = 2, so x = 1.

Therefore, regardless of the method, the minimum value is 9.

Wait, but just to make sure, let me check if there's a possibility that the minimum occurs at the boundary. However, since $x > 0$, the domain is open interval (0, ∞). So there's no boundary at x approaching 0 or infinity. But let me check the limits:

As $x \to 0^+$, the original function is $\frac{(0 + 5)(0 + 2)}{0 + 1} = \frac{5 * 2}{1} = 10$. So approaching 10.

As $x \to \infty$, the function behaves like $\frac{x^2}{x} = x$, which tends to infinity. Therefore, the function approaches infinity as x approaches infinity. Therefore, the minimum must be somewhere in between, which we found at x = 1 with value 9.

Therefore, I think that's solid.

Alternatively, if I didn't use calculus, but tried to use AM-GM, as I did earlier, that also gives the same answer.

So, conclusion: the minimum value is 9, achieved at x = 1.

**Final Answer**
The minimum value is \boxed{9}.

To find the minimum value of the expression

$$
\frac{(x + 5)(x + 2)}{x + 1}
$$

for $ x > 0 $, we begin by simplifying the expression.

---

### Step 1: Expand the Numerator

$$
(x + 5)(x + 2) = x^2 + 7x + 10
$$

So the function becomes:

$$
f(x) = \frac{x^2 + 7x + 10}{x + 1}
$$

---

### Step 2: Perform Polynomial Division

Divide $ x^2 + 7x + 10 $ by $ x + 1 $:

- Divide $ x^2 $ by $ x $: $ x $
- Multiply $ x + 1 $ by $ x $: $ x^2 + x $
- Subtract: $ (x^2 + 7x + 10) - (x^2 + x) = 6x + 10 $
- Divide $ 6x $ by $ x $: $ 6 $
- Multiply $ x + 1 $ by $ 6 $: $ 6x + 6 $
- Subtract: $ (6x + 10) - (6x + 6) = 4 $

So the division yields:

$$
f(x) = x + 6 + \frac{4}{x + 1}
$$

---

### Step 3: Minimize the Simplified Function

We now minimize:

$$
f(x) = x + 6 + \frac{4}{x + 1}
$$

Let’s define:

$$
f(x) = x + 6 + \frac{4}{x + 1}
$$

Take the derivative:

$$
f'(x) = 1 - \frac{4}{(x + 1)^2}
$$

Set the derivative equal to zero to find critical points:

$$
1 - \frac{4}{(x + 1)^2} = 0 \Rightarrow (x + 1)^2 = 4 \Rightarrow x + 1 = 2 \Rightarrow x = 1
$$

Since $ x > 0 $, this is valid.

---

### Step 4: Confirm Minimum with Second Derivative

Compute the second derivative:

$$
f''(x) = \frac{8}{(x + 1)^3}
$$

At $ x = 1 $:

$$
f''(1) = \frac{8}{(1 + 1)^3} = \frac{8}{8} = 1 > 0
$$

So the function has a local minimum at $ x = 1 $, and since the function tends to infinity as $ x \to 0^+ $ and $ x \to \infty $, this is the **global minimum** on $ x > 0 $.

---

### Step 5: Compute the Minimum Value

Substitute $ x = 1 $ into the original function:

$$
f(1) = \frac{(1 + 5)(1 + 2)}{1 + 1} = \frac{6 \cdot 3}{2} = \frac{18}{2} = 9
$$

Alternatively, using the simplified expression:

$$
f(1) = 1 + 6 + \frac{4}{1 + 1} = 7 + 2 = 9
$$

---

### Final Answer

$$
\boxed{9}
$$

Solution:
Expanding, we get
\[\frac{(x + 5)(x + 2)}{x + 1} = \frac{x^2 + 7x + 10}{x + 1}.\]By long division,
\[\frac{x^2 + 7x + 10}{x + 1} = x + 6 + \frac{4}{x + 1} = (x + 1) + \frac{4}{x + 1} + 5.\]By AM-GM,
\[(x + 1) + \frac{4}{x + 1} \ge 2 \sqrt{(x + 1) \cdot \frac{4}{x + 1}} = 4,\]so $(x + 1) + \frac{4}{x + 1} + 5 \ge 9.$

Equality occurs when $x = 1,$ so the minimum value is $\boxed{9}.$