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 with alternating signs. Wait, 1, 4, 6, 4, 1... but here it's 1, 4i, -6, -4i, -i. Maybe there's a connection with (z + i)^4? Let me check.

Let me compute $ (z + i)^4 $. Using binomial theorem:

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

Calculating each term:

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

So altogether:

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

Wait, comparing this with our original equation:

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

So if I subtract i from $ (z + i)^4 $, I get:

$ (z + i)^4 - i - 1 = 0 $? Wait, no. Let me check:

Original equation is $ (z + i)^4 - 1 - i = 0 $? Wait:

Original equation is $ (z + i)^4 - 1 - i = 0 $? Let me check:

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

But $ (z + i)^4 = z^4 + 4i z^3 - 6 z^2 - 4i z + 1 $. Therefore, if I subtract 1 and then subtract i, I get:

$ (z + i)^4 - 1 - i = z^4 + 4i z^3 - 6 z^2 - 4i z + 1 - 1 - i = z^4 + 4i z^3 - 6 z^2 - 4i z - i $, which is exactly our original equation. Therefore, the equation can be rewritten as:

$ (z + i)^4 - 1 - i = 0 $, so $ (z + i)^4 = 1 + i $.

Therefore, the solutions are the fourth roots of $ 1 + i $, shifted by -i. So if I let $ w = z + i $, then $ w^4 = 1 + i $, so solving for w gives the fourth roots of $ 1 + i $, and then z = w - i.

So first, I need to find the fourth roots of $ 1 + i $. Let me recall that to find the nth roots of a complex number, we can express the number in polar form, then take the nth root of the modulus and divide the angle by n, adding multiples of 2π/n.

So first, let me write $ 1 + i $ in polar form. The modulus is $ \sqrt{1^2 + 1^2} = \sqrt{2} $. The argument is $ \pi/4 $, since it's in the first quadrant. Therefore, $ 1 + i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) $.

Therefore, the fourth roots of $ 1 + i $ will have modulus $ (\sqrt{2})^{1/4} = 2^{1/8} $, and arguments $ \frac{\pi/4 + 2\pi k}{4} $ for k = 0, 1, 2, 3.

So the four roots are:

For k = 0: $ 2^{1/8} \left( \cos \frac{\pi}{16} + i \sin \frac{\pi}{16} \right) $

For k = 1: $ 2^{1/8} \left( \cos \frac{9\pi}{16} + i \sin \frac{9\pi}{16} \right) $

For k = 2: $ 2^{1/8} \left( \cos \frac{17\pi}{16} + i \sin \frac{17\pi}{16} \right) $

For k = 3: $ 2^{1/8} \left( \cos \frac{25\pi}{16} + i \sin \frac{25\pi}{16} \right) $

But then, since z = w - i, each root w is shifted by -i. Therefore, the solutions z are:

z = w - i, so each of these roots minus i.

Therefore, the four solutions are:

z_k = 2^{1/8} [cos( (π/4 + 2πk)/4 ) + i sin( (π/4 + 2πk)/4 ) ] - i, for k = 0, 1, 2, 3.

Alternatively, maybe I can write them more neatly. Let me compute each root.

Alternatively, maybe there's a better way to think about this. Since the original equation is (z + i)^4 = 1 + i, so the solutions are z = -i + w, where w^4 = 1 + i. Therefore, the four solutions are located at the fourth roots of 1 + i, shifted down by i.

But maybe I can think geometrically. The fourth roots of 1 + i are equally spaced around a circle of radius 2^{1/8}, starting at angle π/16, then π/16 + π/2, etc. Then shifting them down by i would translate all points down by 1 unit in the imaginary axis.

But perhaps I need to find the coordinates of each z_k and then figure out the polygon they form. Since they are the fourth roots of 1 + i shifted by -i, maybe the polygon is a square? Because the fourth roots of a complex number are equally spaced, so if you shift them, they might still form a square, but rotated or translated.

But since the original roots are symmetrically placed, shifting them by -i would translate the entire figure. However, the polygon's shape might still be a square, but its position is shifted. However, the area might be preserved? Wait, translation doesn't change the area. So if the original roots form a square, then shifting them would still form a square with the same area. Therefore, maybe I can compute the area of the square formed by the fourth roots of 1 + i, and that would be the same as the area of the polygon formed by the z's.

But let me check. Let me think: if the roots w are the fourth roots of 1 + i, then they form a square centered at the origin, with radius 2^{1/8}. Then shifting them by -i would translate the square down by 1 unit in the imaginary axis. However, the area of the square would remain the same, since translation doesn't affect area. Therefore, the area of the polygon (which is a square) is the same as the area of the square formed by the roots w.

Therefore, maybe I can compute the area of the square formed by the fourth roots of 1 + i. Let me recall that for a square centered at the origin with vertices on a circle of radius r, the area can be calculated. Wait, but the square is not necessarily centered at the origin. Wait, the fourth roots of 1 + i are equally spaced around a circle, so they form a square. However, the center of the square is at the origin? Wait, if you take the fourth roots of a complex number, they are symmetrically placed around the origin. So yes, the square is centered at the origin. Therefore, if we shift all points by -i, the square is translated down, but the area remains the same. Therefore, the area of the polygon is the same as the area of the square formed by the fourth roots of 1 + i.

Therefore, I need to compute the area of that square. How?

Well, for a square centered at the origin with vertices on a circle of radius r, the distance between adjacent vertices is the side length. However, the area can be calculated if we know the side length or the radius.

Alternatively, since the square is inscribed in a circle of radius r, the diagonal of the square is equal to 2r. For a square, diagonal d = side length * sqrt(2). Therefore, side length s = d / sqrt(2) = 2r / sqrt(2) = sqrt(2) r. Then area is s^2 = 2 r^2.

Wait, let me check. If the diagonal is 2r, then area is (diagonal^2)/2 = (4r^2)/2 = 2r^2. Yes, that's correct. So area is 2r^2. Therefore, if the square is inscribed in a circle of radius r, then area is 2r^2.

In our case, the radius r is 2^{1/8}, so area would be 2*(2^{1/8})^2 = 2*(2^{1/4}) = 2^{1 + 1/4} = 2^{5/4}.

Wait, but hold on. Wait, the square is formed by the fourth roots of 1 + i. But is the square inscribed in a circle of radius |1 + i|^{1/4}? Wait, yes. Because each root w has modulus |1 + i|^{1/4} = (sqrt(2))^{1/4} = 2^{1/8}, as we had earlier. Therefore, the square is inscribed in a circle of radius 2^{1/8}, so area is 2*(2^{1/8})^2 = 2*2^{1/4} = 2^{1 + 1/4} = 2^{5/4}.

But wait, the problem says the polygon is convex, which a square is, so that seems okay. Therefore, the area would be 2^{5/4}, which is p^{a/b} where p is 2, a = 5, b = 4. Then a + b + p = 5 + 4 + 2 = 11. But wait, before I get too confident, let me check if my reasoning is correct.

Wait, but hold on. Is the square actually inscribed in a circle of radius 2^{1/8}? Let me think again. The fourth roots of 1 + i are equally spaced around a circle of radius |1 + i|^{1/4} = (sqrt(2))^{1/4} = 2^{1/8}. Therefore, yes, they lie on a circle of radius 2^{1/8}, so the square is inscribed in that circle. Therefore, the area is 2*(radius)^2 = 2*(2^{1/8})^2 = 2*2^{1/4} = 2^{5/4}. Therefore, the area is 2^{5/4}, so p = 2, a = 5, b = 4. Then a + b + p = 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 made a mistake here?

Wait, but let me check again. Maybe the square isn't inscribed in a circle of radius 2^{1/8}? Wait, the fourth roots of 1 + i are points on the circle of radius |1 + i|^{1/4} = 2^{1/8}, so yes. Therefore, the square is inscribed in that circle. Therefore, the area is 2*(radius)^2. Wait, but let me think again. For a square inscribed in a circle of radius r, the diagonal is 2r. Therefore, the side length is 2r / sqrt(2) = sqrt(2) r. Therefore, area is (sqrt(2) r)^2 = 2 r^2. So yes, area is 2r^2. Therefore, with r = 2^{1/8}, area is 2*(2^{1/8})^2 = 2*2^{1/4} = 2^{1 + 1/4} = 2^{5/4}. Therefore, 2^{5/4} is the area. Therefore, p = 2, a = 5, b = 4. Then a + b + p = 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 the polygon is indeed a square? Because if the roots are shifted, maybe the polygon is not a square? Wait, but shifting doesn't change the shape, just the position. So if the original roots form a square, then shifting them would still form a square. Therefore, the area is the same. Therefore, the answer is 11? But I need to check again.

Wait, but maybe my assumption that the square is inscribed in a circle of radius 2^{1/8} is wrong? Let me think again. Let me take specific roots. Let me compute the four roots of w^4 = 1 + i. Let me compute them numerically.

First, modulus of 1 + i is sqrt(2), so modulus of each root is (sqrt(2))^{1/4} = 2^{1/8} ≈ 1.0905. The arguments are π/16, π/16 + π/2, π/16 + π, π/16 + 3π/2. So angles are π/16, 9π/16, 17π/16, 25π/16. So plotting these points on the complex plane, they are equally spaced every π/2 radians, so they form a square. Therefore, the square is centered at the origin, with vertices on the circle of radius 2^{1/8}. Therefore, the area is 2*(2^{1/8})^2 = 2^{5/4} as before. Therefore, the area is 2^{5/4}, so answer is 5 + 4 + 2 = 11. But wait, let me check if the problem says "the solutions are the vertices of a convex polygon". Since it's a square, which is convex, so that's okay.

But wait, maybe I made a mistake in assuming that the square's area is 2r^2. Let me check with coordinates. Let me take two adjacent roots and compute the distance between them, then compute the area.

Let me take two adjacent roots. Let me compute the coordinates of two adjacent roots. Let me take w_0 = 2^{1/8} [cos(π/16) + i sin(π/16)] and w_1 = 2^{1/8} [cos(9π/16) + i sin(9π/16)]. The distance between w_0 and w_1 is the side length of the square. Let me compute that.

The distance squared between w_0 and w_1 is |w_0 - w_1|^2. Let me compute that. Since they are two adjacent vertices of a square inscribed in a circle of radius r, the angle between them is π/2. Therefore, the distance between them is sqrt(2) r sin(π/4) = sqrt(2) * r * sqrt(2)/2 = r. Wait, no. Wait, the chord length between two points on a circle separated by angle θ is 2r sin(θ/2). Here, θ is π/2, so chord length is 2r sin(π/4) = 2r*(√2/2) = r√2. Therefore, the side length is r√2, so area is (r√2)^2 = 2r^2. Which matches the previous result. Therefore, area is 2r^2. Therefore, with r = 2^{1/8}, area is 2*(2^{1/8})^2 = 2*2^{1/4} = 2^{5/4}. Therefore, that seems correct.

But let me check with coordinates. Let me compute coordinates of w_0 and w_1. Let me take r = 2^{1/8} ≈ 1.0905. Let me compute cos(π/16) and sin(π/16). π/16 is 11.25 degrees. cos(π/16) ≈ 0.98078, sin(π/16) ≈ 0.19509. Therefore, w_0 ≈ 1.0905*(0.98078 + 0.19509i) ≈ 1.070 + 0.213i. Similarly, w_1 is at angle 9π/16, which is 101.25 degrees. cos(9π/16) ≈ -0.19509, sin(9π/16) ≈ 0.98078. Therefore, w_1 ≈ 1.0905*(-0.19509 + 0.98078i) ≈ -0.213 + 1.070i. Then the distance between w_0 and w_1 is sqrt[ (1.070 - (-0.213))^2 + (0.213 - 1.070)^2 ] = sqrt[ (1.283)^2 + (-0.857)^2 ] ≈ sqrt[1.646 + 0.734] ≈ sqrt[2.38] ≈ 1.543. Now, if r = 2^{1/8} ≈ 1.0905, then r√2 ≈ 1.0905 * 1.414 ≈ 1.543, which matches. Therefore, the side length is indeed r√2, so area is 2r^2. Therefore, 2*(2^{1/8})^2 = 2^{5/4} ≈ 2.378. Therefore, seems correct.

But then, the problem says the solutions are the vertices of a convex polygon. So if the original roots form a square, then shifting them by -i would translate the square down by 1 unit in the imaginary axis. However, the area remains the same. Therefore, the area is 2^{5/4}, so answer is 5 + 4 + 2 = 11. However, I need to check if the problem is expecting something else. Wait, maybe I made a mistake in assuming that the polygon is a square? Let me think again.

Wait, the original equation is (z + i)^4 = 1 + i. Therefore, the solutions are z = -i + w, where w^4 = 1 + i. Therefore, the four solutions are the four fourth roots of 1 + i shifted down by i. Therefore, if the original roots w form a square centered at the origin, then shifting them down by i would translate the square down. However, the square is not necessarily centered at the origin anymore. Wait, but does that affect the area? No, translation doesn't change the area. Therefore, the area is still 2^{5/4}. Therefore, answer is 11. But maybe I need to check if the polygon is actually a square? Let me think.

Alternatively, maybe the roots are not forming a square? Wait, if you take the fourth roots of a complex number, they are equally spaced around a circle, so they form a square. Therefore, yes, they form a square. Therefore, shifting them doesn't change the shape, just the position. Therefore, the area is same. Therefore, answer is 11. However, I need to check if the problem is perhaps expecting a different answer. Maybe I made a mistake in the modulus?

Wait, let me check modulus again. The modulus of 1 + i is sqrt(2). Therefore, modulus of each root is (sqrt(2))^{1/4} = 2^{1/8}. Therefore, radius is 2^{1/8}. Therefore, area is 2*(2^{1/8})^2 = 2^{1 + 1/4} = 2^{5/4}. Therefore, that's correct. Therefore, 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}". 2^{5/4} is already in that form, with p=2, a=5, b=4. Since 5 and 4 are coprime, and 2 is prime. Therefore, answer is 5 + 4 + 2 = 11. Therefore, maybe the answer is 11. However, I need to check again if the polygon is indeed a square. Let me think of another approach.

Alternatively, maybe I can find the coordinates of the four solutions z and compute the area using coordinates. Let me try that. Let me compute each z.

First, let me recall that w^4 = 1 + i, so w = 2^{1/8} e^{i(π/16 + kπ/2)}, for k = 0, 1, 2, 3. Therefore, z = w - i. Therefore, each z is:

z_k = 2^{1/8} e^{i(π/16 + kπ/2)} - i.

Let me compute coordinates for each z_k.

Let me compute for k = 0:

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

Similarly for k = 1:

z_1 = 2^{1/8} [cos(9π/16) + i sin(9π/16)] - i.

k = 2:

z_2 = 2^{1/8} [cos(17π/16) + i sin(17π/16)] - i.

k = 3:

z_3 = 2^{1/8} [cos(25π/16) + i sin(25π/16)] - i.

Alternatively, since cos(17π/16) = cos(π + π/16) = -cos(π/16), and sin(17π/16) = sin(π + π/16) = -sin(π/16). Similarly, cos(25π/16) = cos(2π - 7π/16) = cos(7π/16), and sin(25π/16) = sin(2π - 7π/16) = -sin(7π/16). Wait, maybe I need to compute each coordinate.

Alternatively, maybe I can think of the four points z_0, z_1, z_2, z_3. Since they are the fourth roots of 1 + i shifted down by i, maybe they form a square. However, if I shift a square down by i, it's still a square. Therefore, the area is same as before. Therefore, 2^{5/4} is correct. Therefore, answer is 11. However, maybe I need to check with coordinates.

Alternatively, maybe I can compute the coordinates of the four points and use the shoelace formula to compute the area. Let me try that. Let me denote the four points as z_0, z_1, z_2, z_3. Let me compute their coordinates.

First, let me compute the coordinates of w_k, then subtract i.

Let me compute w_0:

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

Similarly, w_1 = 2^{1/8} [cos(9π/16) + i sin(9π/16)].

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

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

Therefore, z_k = w_k - i. Therefore, coordinates:

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

For z_1: x = 2^{1/8} cos(9π/16), y = 2^{1/8} sin(9π/16) - 1.

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

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

Now, to compute the area using shoelace formula, I need coordinates of all four points. However, since the points are symmetric, maybe I can find the coordinates more easily.

Alternatively, since the original square is centered at the origin, and we are shifting it down by 1 unit in the imaginary axis, the center of the square is now at (0, -1). However, the area remains the same. Therefore, regardless of the position, the area is 2^{5/4}. Therefore, answer is 11. However, maybe I need to check with coordinates.

Alternatively, maybe I can think of the square as having vertices at points (a, b), (-b, a), (-a, -b), (b, -a), or something like that. Wait, if the original square is centered at the origin, with vertices on the circle of radius r, then coordinates can be written as (r cos θ, r sin θ), where θ = π/16, 9π/16, 17π/16, 25π/16. Then shifting down by i (i.e., subtracting 1 from the imaginary part) would make the coordinates:

For each vertex (x, y), new coordinates are (x, y - 1). Therefore, the square is translated down by 1 unit. However, the area remains the same. Therefore, the area is still 2r^2 = 2*(2^{1/8})^2 = 2^{5/4}.

Alternatively, maybe I can compute the distance between two adjacent points and then compute the area. Let me take two adjacent points, say z_0 and z_1. The distance between them is the same as the distance between w_0 and w_1, which we already calculated as r√2. Therefore, the side length is r√2, so area is (r√2)^2 = 2r^2. Therefore, same result. Therefore, answer is 2^{5/4}.

But let me check if the problem says "the area of this polygon can be expressed in the form p^{a/b}". 2^{5/4} is already in that form. However, maybe the problem expects a different form? For example, maybe I made a mistake in the modulus?

Wait, let me check modulus again. The original equation is (z + i)^4 = 1 + i. Therefore, modulus of both sides: |z + i|^4 = |1 + i| = sqrt(2). Therefore, |z + i| = (sqrt(2))^{1/4} = 2^{1/8}. Therefore, the distance from each solution z to -i is 2^{1/8}. Therefore, the four solutions lie on a circle of radius 2^{1/8} centered at -i. Therefore, the polygon is a square inscribed in a circle of radius 2^{1/8}, centered at -i. Therefore, the area of the square is still 2*(2^{1/8})^2 = 2^{5/4}. Therefore, same result. Therefore, answer is 11.

But maybe I need to check with coordinates. Let me try to compute coordinates of z_0 and z_1 and compute the area.

Let me take specific values. Let me compute coordinates numerically.

First, compute 2^{1/8} ≈ e^{(ln 2)/8} ≈ e^{0.0866} ≈ 1.0905.

Compute π/16 ≈ 0.19635 radians.

cos(π/16) ≈ 0.98078, sin(π/16) ≈ 0.19509.

Therefore, z_0 ≈ (1.0905 * 0.98078, 1.0905 * 0.19509 - 1) ≈ (1.070, 0.213 - 1) ≈ (1.070, -0.787).

Similarly, z_1: angle 9π/16 ≈ 1.767 radians. cos(9π/16) ≈ cos(1.767) ≈ -0.19509, sin(9π/16) ≈ 0.98078. Therefore, z_1 ≈ (1.0905 * (-0.19509), 1.0905 * 0.98078 - 1) ≈ (-0.213, 1.070 - 1) ≈ (-0.213, 0.070).

Similarly, z_2: angle 17π/16 ≈ 3.334 radians. cos(17π/16) ≈ cos(π + π/16) = -cos(π/16) ≈ -0.98078, sin(17π/16) ≈ sin(π + π/16) = -sin(π/16) ≈ -0.19509. Therefore, z_2 ≈ (1.0905 * (-0.98078), 1.0905 * (-0.19509) - 1) ≈ (-1.070, -0.213 - 1) ≈ (-1.070, -1.213).

Similarly, z_3: angle 25π/16 ≈ 4.909 radians. cos(25π/16) = cos(2π - 7π/16) = cos(7π/16) ≈ 0.19509, sin(25π/16) = sin(2π - 7π/16) = -sin(7π/16) ≈ -0.98078. Therefore, z_3 ≈ (1.0905 * 0.19509, 1.0905 * (-0.98078) - 1) ≈ (0.213, -1.070 - 1) ≈ (0.213, -2.070).

Wait, hold on. Wait, for z_3, angle is 25π/16. Let me check:

25π/16 = π + 9π/16? Wait, π is 16π/16, so 25π/16 = π + 9π/16. Therefore, cos(25π/16) = cos(π + 9π/16) = -cos(9π/16) ≈ -(-0.19509) = 0.19509? Wait, no. Wait, cos(π + θ) = -cos θ. So cos(25π/16) = cos(π + 9π/16) = -cos(9π/16) ≈ -(-0.19509) = 0.19509? Wait, earlier I said cos(9π/16) ≈ -0.19509? Wait, let me check:

Wait, 9π/16 is 101.25 degrees. cos(101.25 degrees) is cos(90 + 11.25) = -sin(11.25) ≈ -0.19509. Therefore, cos(9π/16) ≈ -0.19509. Therefore, cos(25π/16) = cos(π + 9π/16) = -cos(9π/16) ≈ -(-0.19509) = 0.19509. Similarly, sin(25π/16) = sin(π + 9π/16) = -sin(9π/16) ≈ -0.98078. Therefore, z_3 ≈ (1.0905 * 0.19509, 1.0905 * (-0.98078) - 1) ≈ (0.213, -1.070 - 1) ≈ (0.213, -2.070). Wait, but earlier I thought z_3 was (0.213, -2.070). Wait, but then z_0 is (1.070, -0.787), z_1 is (-0.213, 0.070), z_2 is (-1.070, -1.213), z_3 is (0.213, -2.070). Let me plot these points approximately:

z_0: (1.07, -0.787)

z_1: (-0.213, 0.07)

z_2: (-1.07, -1.213)

z_3: (0.213, -2.07)

Wait, connecting these points, is this a square? Let me check distances between consecutive points.

First, distance between z_0 and z_1:

sqrt[(1.07 - (-0.213))² + (-0.787 - 0.07)²] = sqrt[(1.283)² + (-0.857)²] ≈ sqrt[1.646 + 0.734] ≈ sqrt[2.38] ≈ 1.543.

Distance between z_1 and z_2:

sqrt[(-0.213 - (-1.07))² + (0.07 - (-1.213))²] = sqrt[(0.857)² + (1.283)²] ≈ same as before ≈ 1.543.

Distance between z_2 and z_3:

sqrt[(-1.07 - 0.213)² + (-1.213 - (-2.07))²] = sqrt[(-1.283)² + (0.857)²] ≈ same ≈ 1.543.

Distance between z_3 and z_0:

sqrt[(0.213 - 1.07)² + (-2.07 - (-0.787))²] = sqrt[(-0.857)² + (-1.283)²] ≈ same ≈ 1.543.

So all sides are equal. Now check the diagonals. Distance between z_0 and z_2:

sqrt[(1.07 - (-1.07))² + (-0.787 - (-1.213))²] = sqrt[(2.14)² + (0.426)²] ≈ sqrt[4.5796 + 0.181] ≈ sqrt[4.76] ≈ 2.18.

Distance between z_1 and z_3:

sqrt[(-0.213 - 0.213)² + (0.07 - (-2.07))²] = sqrt[(-0.426)² + (2.14)²] ≈ same as above ≈ 2.18.

Now, for a square, the diagonals should be equal to side * sqrt(2). Let me check: side ≈ 1.543, so diagonal ≈ 1.543 * 1.414 ≈ 2.18. Which matches the computed diagonal. Therefore, yes, it's a square. Therefore, the area is (side)^2 ≈ (1.543)^2 ≈ 2.38. Which is approximately 2^{5/4} ≈ 2.378. Therefore, correct. Therefore, the area is indeed 2^{5/4}.

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}". 2^{5/4} is already in that form. However, maybe the problem expects the answer in another form? For example, maybe I need to rationalize or something? Wait, 2^{5/4} is equal to sqrt(2^{5/2})? Wait, no. Wait, 2^{5/4} is the same as (2^{1/4})^5 or something else? 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, answer is 5 + 4 + 2 = 11. Therefore, I think the answer is 11.

But just to be thorough, let me check if there's another approach. For example, maybe the original equation can be transformed into something else. Wait, we had (z + i)^4 = 1 + i. Therefore, the solutions are z = -i + (1 + i)^{1/4}. But since we already did that, and found the area. Alternatively, maybe there's a mistake in assuming that the polygon is a square? But with the coordinates checked, it's a square. Therefore, I think the answer is 11.

But just to check, maybe the problem is not a square? Wait, the equation is quartic, so four roots, which form a convex polygon. Since they are four points, the only convex polygon is a quadrilateral. If they are symmetrically placed, it's a square. Therefore, yes. Therefore, answer is 11. 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 resembles the binomial expansion of $(z + i)^4$, which is:

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

Comparing this with the given equation, we see that:

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

Thus, the equation becomes:

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

---

### Step 1: Solve the Equation

Let $ w = z + i $, so the equation becomes:

$$
w^4 = 1 + i
$$

We now find the fourth roots of $ 1 + i $. First, express $ 1 + i $ in polar form:

- Modulus: $ |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2} $
- Argument: $ \theta = \frac{\pi}{4} $

So:

$$
1 + i = \sqrt{2} \left( \cos\frac{\pi}{4} + i \sin\frac{\pi}{4} \right)
$$

The fourth roots of $ 1 + i $ are:

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

This gives four complex numbers $ w_k $, each lying on a circle of radius $ r = \left( \sqrt{2} \right)^{1/4} = 2^{1/8} $, and spaced at angles $ \frac{\pi}{16}, \frac{9\pi}{16}, \frac{17\pi}{16}, \frac{25\pi}{16} $, which are equally spaced every $ \frac{\pi}{2} $ radians. Therefore, the four roots form a square centered at the origin.

---

### Step 2: Translate the Roots

Recall that $ z = w - i $, so the solutions $ z $ are the fourth roots of $ 1 + i $, shifted down by $ i $ (i.e., translated vertically by 1 unit in the imaginary direction). This translation does **not** affect the shape or area of the polygon formed by the roots.

Thus, the four solutions $ z $ form a square, and the area of this square is the same as the area of the square formed by the roots $ w $, which are centered at the origin.

---

### Step 3: Compute the Area of the Square

For a square inscribed in a circle of radius $ r $, the diagonal of the square is $ 2r $, and the area is:

$$
\text{Area} = \frac{(\text{diagonal})^2}{2} = \frac{(2r)^2}{2} = 2r^2
$$

Here, $ r = 2^{1/8} $, so:

$$
\text{Area} = 2 \cdot (2^{1/8})^2 = 2 \cdot 2^{1/4} = 2^{1 + 1/4} = 2^{5/4}
$$

---

### Step 4: Express in the Required Form

We are told the area can be expressed in the form $ p^{a/b} $, where $ p $ is prime, and $ a $ and $ b $ are coprime positive integers.

We have:

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

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 with alternating signs. Wait, 1, 4, 6, 4, 1... but here it's 1, 4i, -6, -4i, -i. Maybe there's a connection with (z + i)^4? Let me check.

Let me compute $ (z + i)^4 $. Using binomial theorem:

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

Calculating each term:

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

So altogether:

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

Wait, comparing this with our original equation:

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

So if I subtract i from $ (z + i)^4 $, I get:

$ (z + i)^4 - i - 1 = 0 $? Wait, no. Let me check:

Original equation is $ (z + i)^4 - 1 - i = 0 $? Wait:

Original equation is $ (z + i)^4 - 1 - i = 0 $? Let me check:

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

But $ (z + i)^4 = z^4 + 4i z^3 - 6 z^2 - 4i z + 1 $. Therefore, if I subtract 1 and then subtract i, I get:

$ (z + i)^4 - 1 - i = z^4 + 4i z^3 - 6 z^2 - 4i z + 1 - 1 - i = z^4 + 4i z^3 - 6 z^2 - 4i z - i $, which is exactly our original equation. Therefore, the equation can be rewritten as:

$ (z + i)^4 - 1 - i = 0 $, so $ (z + i)^4 = 1 + i $.

Therefore, the solutions are the fourth roots of $ 1 + i $, shifted by -i. So if I let $ w = z + i $, then $ w^4 = 1 + i $, so solving for w gives the fourth roots of $ 1 + i $, and then z = w - i.

So first, I need to find the fourth roots of $ 1 + i $. Let me recall that to find the nth roots of a complex number, we can express the number in polar form, then take the nth root of the modulus and divide the angle by n, adding multiples of 2π/n.

So first, let me write $ 1 + i $ in polar form. The modulus is $ \sqrt{1^2 + 1^2} = \sqrt{2} $. The argument is $ \pi/4 $, since it's in the first quadrant. Therefore, $ 1 + i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) $.

Therefore, the fourth roots of $ 1 + i $ will have modulus $ (\sqrt{2})^{1/4} = 2^{1/8} $, and arguments $ \frac{\pi/4 + 2\pi k}{4} $ for k = 0, 1, 2, 3.

So the four roots are:

For k = 0: $ 2^{1/8} \left( \cos \frac{\pi}{16} + i \sin \frac{\pi}{16} \right) $

For k = 1: $ 2^{1/8} \left( \cos \frac{9\pi}{16} + i \sin \frac{9\pi}{16} \right) $

For k = 2: $ 2^{1/8} \left( \cos \frac{17\pi}{16} + i \sin \frac{17\pi}{16} \right) $

For k = 3: $ 2^{1/8} \left( \cos \frac{25\pi}{16} + i \sin \frac{25\pi}{16} \right) $

But then, since z = w - i, each root w is shifted by -i. Therefore, the solutions z are:

z = w - i, so each of these roots minus i.

Therefore, the four solutions are:

z_k = 2^{1/8} [cos( (π/4 + 2πk)/4 ) + i sin( (π/4 + 2πk)/4 ) ] - i, for k = 0, 1, 2, 3.

Alternatively, maybe I can write them more neatly. Let me compute each root.

Alternatively, maybe there's a better way to think about this. Since the original equation is (z + i)^4 = 1 + i, so the solutions are z = -i + w, where w^4 = 1 + i. Therefore, the four solutions are located at the fourth roots of 1 + i, shifted down by i.

But maybe I can think geometrically. The fourth roots of 1 + i are equally spaced around a circle of radius 2^{1/8}, starting at angle π/16, then π/16 + π/2, etc. Then shifting them down by i would translate all points down by 1 unit in the imaginary axis.

But perhaps I need to find the coordinates of each z_k and then figure out the polygon they form. Since they are the fourth roots of 1 + i shifted by -i, maybe the polygon is a square? Because the fourth roots of a complex number are equally spaced, so if you shift them, they might still form a square, but rotated or translated.

But since the original roots are symmetrically placed, shifting them by -i would translate the entire figure. However, the polygon's shape might still be a square, but its position is shifted. However, the area might be preserved? Wait, translation doesn't change the area. So if the original roots form a square, then shifting them would still form a square with the same area. Therefore, maybe I can compute the area of the square formed by the fourth roots of 1 + i, and that would be the same as the area of the polygon formed by the z's.

But let me check. Let me think: if the roots w are the fourth roots of 1 + i, then they form a square centered at the origin, with radius 2^{1/8}. Then shifting them by -i would translate the square down by 1 unit in the imaginary axis. However, the area of the square would remain the same, since translation doesn't affect area. Therefore, the area of the polygon (which is a square) is the same as the area of the square formed by the roots w.

Therefore, maybe I can compute the area of the square formed by the fourth roots of 1 + i. Let me recall that for a square centered at the origin with vertices on a circle of radius r, the area can be calculated. Wait, but the square is not necessarily centered at the origin. Wait, the fourth roots of 1 + i are equally spaced around a circle, so they form a square. However, the center of the square is at the origin? Wait, if you take the fourth roots of a complex number, they are symmetrically placed around the origin. So yes, the square is centered at the origin. Therefore, if we shift all points by -i, the square is translated down, but the area remains the same. Therefore, the area of the polygon is the same as the area of the square formed by the fourth roots of 1 + i.

Therefore, I need to compute the area of that square. How?

Well, for a square centered at the origin with vertices on a circle of radius r, the distance between adjacent vertices is the side length. However, the area can be calculated if we know the side length or the radius.

Alternatively, since the square is inscribed in a circle of radius r, the diagonal of the square is equal to 2r. For a square, diagonal d = side length * sqrt(2). Therefore, side length s = d / sqrt(2) = 2r / sqrt(2) = sqrt(2) r. Then area is s^2 = 2 r^2.

Wait, let me check. If the diagonal is 2r, then area is (diagonal^2)/2 = (4r^2)/2 = 2r^2. Yes, that's correct. So area is 2r^2. Therefore, if the square is inscribed in a circle of radius r, then area is 2r^2.

In our case, the radius r is 2^{1/8}, so area would be 2*(2^{1/8})^2 = 2*(2^{1/4}) = 2^{1 + 1/4} = 2^{5/4}.

Wait, but hold on. Wait, the square is formed by the fourth roots of 1 + i. But is the square inscribed in a circle of radius |1 + i|^{1/4}? Wait, yes. Because each root w has modulus |1 + i|^{1/4} = (sqrt(2))^{1/4} = 2^{1/8}, as we had earlier. Therefore, the square is inscribed in a circle of radius 2^{1/8}, so area is 2*(2^{1/8})^2 = 2*2^{1/4} = 2^{1 + 1/4} = 2^{5/4}.

But wait, the problem says the polygon is convex, which a square is, so that seems okay. Therefore, the area would be 2^{5/4}, which is p^{a/b} where p is 2, a = 5, b = 4. Then a + b + p = 5 + 4 + 2 = 11. But wait, before I get too confident, let me check if my reasoning is correct.

Wait, but hold on. Is the square actually inscribed in a circle of radius 2^{1/8}? Let me think again. The fourth roots of 1 + i are equally spaced around a circle of radius |1 + i|^{1/4} = (sqrt(2))^{1/4} = 2^{1/8}. Therefore, yes, they lie on a circle of radius 2^{1/8}, so the square is inscribed in that circle. Therefore, the area is 2*(radius)^2 = 2*(2^{1/8})^2 = 2*2^{1/4} = 2^{5/4}. Therefore, the area is 2^{5/4}, so p = 2, a = 5, b = 4. Then a + b + p = 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 made a mistake here?

Wait, but let me check again. Maybe the square isn't inscribed in a circle of radius 2^{1/8}? Wait, the fourth roots of 1 + i are points on the circle of radius |1 + i|^{1/4} = 2^{1/8}, so yes. Therefore, the square is inscribed in that circle. Therefore, the area is 2*(radius)^2. Wait, but let me think again. For a square inscribed in a circle of radius r, the diagonal is 2r. Therefore, the side length is 2r / sqrt(2) = sqrt(2) r. Therefore, area is (sqrt(2) r)^2 = 2 r^2. So yes, area is 2r^2. Therefore, with r = 2^{1/8}, area is 2*(2^{1/8})^2 = 2*2^{1/4} = 2^{1 + 1/4} = 2^{5/4}. Therefore, 2^{5/4} is the area. Therefore, p = 2, a = 5, b = 4. Then a + b + p = 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 the polygon is indeed a square? Because if the roots are shifted, maybe the polygon is not a square? Wait, but shifting doesn't change the shape, just the position. So if the original roots form a square, then shifting them would still form a square. Therefore, the area is the same. Therefore, the answer is 11? But I need to check again.

Wait, but maybe my assumption that the square is inscribed in a circle of radius 2^{1/8} is wrong? Let me think again. Let me take specific roots. Let me compute the four roots of w^4 = 1 + i. Let me compute them numerically.

First, modulus of 1 + i is sqrt(2), so modulus of each root is (sqrt(2))^{1/4} = 2^{1/8} ≈ 1.0905. The arguments are π/16, π/16 + π/2, π/16 + π, π/16 + 3π/2. So angles are π/16, 9π/16, 17π/16, 25π/16. So plotting these points on the complex plane, they are equally spaced every π/2 radians, so they form a square. Therefore, the square is centered at the origin, with vertices on the circle of radius 2^{1/8}. Therefore, the area is 2*(2^{1/8})^2 = 2^{5/4} as before. Therefore, the area is 2^{5/4}, so answer is 5 + 4 + 2 = 11. But wait, let me check if the problem says "the solutions are the vertices of a convex polygon". Since it's a square, which is convex, so that's okay.

But wait, maybe I made a mistake in assuming that the square's area is 2r^2. Let me check with coordinates. Let me take two adjacent roots and compute the distance between them, then compute the area.

Let me take two adjacent roots. Let me compute the coordinates of two adjacent roots. Let me take w_0 = 2^{1/8} [cos(π/16) + i sin(π/16)] and w_1 = 2^{1/8} [cos(9π/16) + i sin(9π/16)]. The distance between w_0 and w_1 is the side length of the square. Let me compute that.

The distance squared between w_0 and w_1 is |w_0 - w_1|^2. Let me compute that. Since they are two adjacent vertices of a square inscribed in a circle of radius r, the angle between them is π/2. Therefore, the distance between them is sqrt(2) r sin(π/4) = sqrt(2) * r * sqrt(2)/2 = r. Wait, no. Wait, the chord length between two points on a circle separated by angle θ is 2r sin(θ/2). Here, θ is π/2, so chord length is 2r sin(π/4) = 2r*(√2/2) = r√2. Therefore, the side length is r√2, so area is (r√2)^2 = 2r^2. Which matches the previous result. Therefore, area is 2r^2. Therefore, with r = 2^{1/8}, area is 2*(2^{1/8})^2 = 2*2^{1/4} = 2^{5/4}. Therefore, that seems correct.

But let me check with coordinates. Let me compute coordinates of w_0 and w_1. Let me take r = 2^{1/8} ≈ 1.0905. Let me compute cos(π/16) and sin(π/16). π/16 is 11.25 degrees. cos(π/16) ≈ 0.98078, sin(π/16) ≈ 0.19509. Therefore, w_0 ≈ 1.0905*(0.98078 + 0.19509i) ≈ 1.070 + 0.213i. Similarly, w_1 is at angle 9π/16, which is 101.25 degrees. cos(9π/16) ≈ -0.19509, sin(9π/16) ≈ 0.98078. Therefore, w_1 ≈ 1.0905*(-0.19509 + 0.98078i) ≈ -0.213 + 1.070i. Then the distance between w_0 and w_1 is sqrt[ (1.070 - (-0.213))^2 + (0.213 - 1.070)^2 ] = sqrt[ (1.283)^2 + (-0.857)^2 ] ≈ sqrt[1.646 + 0.734] ≈ sqrt[2.38] ≈ 1.543. Now, if r = 2^{1/8} ≈ 1.0905, then r√2 ≈ 1.0905 * 1.414 ≈ 1.543, which matches. Therefore, the side length is indeed r√2, so area is 2r^2. Therefore, 2*(2^{1/8})^2 = 2^{5/4} ≈ 2.378. Therefore, seems correct.

But then, the problem says the solutions are the vertices of a convex polygon. So if the original roots form a square, then shifting them by -i would translate the square down by 1 unit in the imaginary axis. However, the area remains the same. Therefore, the area is 2^{5/4}, so answer is 5 + 4 + 2 = 11. However, I need to check if the problem is expecting something else. Wait, maybe I made a mistake in assuming that the polygon is a square? Let me think again.

Wait, the original equation is (z + i)^4 = 1 + i. Therefore, the solutions are z = -i + w, where w^4 = 1 + i. Therefore, the four solutions are the four fourth roots of 1 + i shifted down by i. Therefore, if the original roots w form a square centered at the origin, then shifting them down by i would translate the square down. However, the square is not necessarily centered at the origin anymore. Wait, but does that affect the area? No, translation doesn't change the area. Therefore, the area is still 2^{5/4}. Therefore, answer is 11. But maybe I need to check if the polygon is actually a square? Let me think.

Alternatively, maybe the roots are not forming a square? Wait, if you take the fourth roots of a complex number, they are equally spaced around a circle, so they form a square. Therefore, yes, they form a square. Therefore, shifting them doesn't change the shape, just the position. Therefore, the area is same. Therefore, answer is 11. However, I need to check if the problem is perhaps expecting a different answer. Maybe I made a mistake in the modulus?

Wait, let me check modulus again. The modulus of 1 + i is sqrt(2). Therefore, modulus of each root is (sqrt(2))^{1/4} = 2^{1/8}. Therefore, radius is 2^{1/8}. Therefore, area is 2*(2^{1/8})^2 = 2^{1 + 1/4} = 2^{5/4}. Therefore, that's correct. Therefore, 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}". 2^{5/4} is already in that form, with p=2, a=5, b=4. Since 5 and 4 are coprime, and 2 is prime. Therefore, answer is 5 + 4 + 2 = 11. Therefore, maybe the answer is 11. However, I need to check again if the polygon is indeed a square. Let me think of another approach.

Alternatively, maybe I can find the coordinates of the four solutions z and compute the area using coordinates. Let me try that. Let me compute each z.

First, let me recall that w^4 = 1 + i, so w = 2^{1/8} e^{i(π/16 + kπ/2)}, for k = 0, 1, 2, 3. Therefore, z = w - i. Therefore, each z is:

z_k = 2^{1/8} e^{i(π/16 + kπ/2)} - i.

Let me compute coordinates for each z_k.

Let me compute for k = 0:

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

Similarly for k = 1:

z_1 = 2^{1/8} [cos(9π/16) + i sin(9π/16)] - i.

k = 2:

z_2 = 2^{1/8} [cos(17π/16) + i sin(17π/16)] - i.

k = 3:

z_3 = 2^{1/8} [cos(25π/16) + i sin(25π/16)] - i.

Alternatively, since cos(17π/16) = cos(π + π/16) = -cos(π/16), and sin(17π/16) = sin(π + π/16) = -sin(π/16). Similarly, cos(25π/16) = cos(2π - 7π/16) = cos(7π/16), and sin(25π/16) = sin(2π - 7π/16) = -sin(7π/16). Wait, maybe I need to compute each coordinate.

Alternatively, maybe I can think of the four points z_0, z_1, z_2, z_3. Since they are the fourth roots of 1 + i shifted down by i, maybe they form a square. However, if I shift a square down by i, it's still a square. Therefore, the area is same as before. Therefore, 2^{5/4} is correct. Therefore, answer is 11. However, maybe I need to check with coordinates.

Alternatively, maybe I can compute the coordinates of the four points and use the shoelace formula to compute the area. Let me try that. Let me denote the four points as z_0, z_1, z_2, z_3. Let me compute their coordinates.

First, let me compute the coordinates of w_k, then subtract i.

Let me compute w_0:

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

Similarly, w_1 = 2^{1/8} [cos(9π/16) + i sin(9π/16)].

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

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

Therefore, z_k = w_k - i. Therefore, coordinates:

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

For z_1: x = 2^{1/8} cos(9π/16), y = 2^{1/8} sin(9π/16) - 1.

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

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

Now, to compute the area using shoelace formula, I need coordinates of all four points. However, since the points are symmetric, maybe I can find the coordinates more easily.

Alternatively, since the original square is centered at the origin, and we are shifting it down by 1 unit in the imaginary axis, the center of the square is now at (0, -1). However, the area remains the same. Therefore, regardless of the position, the area is 2^{5/4}. Therefore, answer is 11. However, maybe I need to check with coordinates.

Alternatively, maybe I can think of the square as having vertices at points (a, b), (-b, a), (-a, -b), (b, -a), or something like that. Wait, if the original square is centered at the origin, with vertices on the circle of radius r, then coordinates can be written as (r cos θ, r sin θ), where θ = π/16, 9π/16, 17π/16, 25π/16. Then shifting down by i (i.e., subtracting 1 from the imaginary part) would make the coordinates:

For each vertex (x, y), new coordinates are (x, y - 1). Therefore, the square is translated down by 1 unit. However, the area remains the same. Therefore, the area is still 2r^2 = 2*(2^{1/8})^2 = 2^{5/4}.

Alternatively, maybe I can compute the distance between two adjacent points and then compute the area. Let me take two adjacent points, say z_0 and z_1. The distance between them is the same as the distance between w_0 and w_1, which we already calculated as r√2. Therefore, the side length is r√2, so area is (r√2)^2 = 2r^2. Therefore, same result. Therefore, answer is 2^{5/4}.

But let me check if the problem says "the area of this polygon can be expressed in the form p^{a/b}". 2^{5/4} is already in that form. However, maybe the problem expects a different form? For example, maybe I made a mistake in the modulus?

Wait, let me check modulus again. The original equation is (z + i)^4 = 1 + i. Therefore, modulus of both sides: |z + i|^4 = |1 + i| = sqrt(2). Therefore, |z + i| = (sqrt(2))^{1/4} = 2^{1/8}. Therefore, the distance from each solution z to -i is 2^{1/8}. Therefore, the four solutions lie on a circle of radius 2^{1/8} centered at -i. Therefore, the polygon is a square inscribed in a circle of radius 2^{1/8}, centered at -i. Therefore, the area of the square is still 2*(2^{1/8})^2 = 2^{5/4}. Therefore, same result. Therefore, answer is 11.

But maybe I need to check with coordinates. Let me try to compute coordinates of z_0 and z_1 and compute the area.

Let me take specific values. Let me compute coordinates numerically.

First, compute 2^{1/8} ≈ e^{(ln 2)/8} ≈ e^{0.0866} ≈ 1.0905.

Compute π/16 ≈ 0.19635 radians.

cos(π/16) ≈ 0.98078, sin(π/16) ≈ 0.19509.

Therefore, z_0 ≈ (1.0905 * 0.98078, 1.0905 * 0.19509 - 1) ≈ (1.070, 0.213 - 1) ≈ (1.070, -0.787).

Similarly, z_1: angle 9π/16 ≈ 1.767 radians. cos(9π/16) ≈ cos(1.767) ≈ -0.19509, sin(9π/16) ≈ 0.98078. Therefore, z_1 ≈ (1.0905 * (-0.19509), 1.0905 * 0.98078 - 1) ≈ (-0.213, 1.070 - 1) ≈ (-0.213, 0.070).

Similarly, z_2: angle 17π/16 ≈ 3.334 radians. cos(17π/16) ≈ cos(π + π/16) = -cos(π/16) ≈ -0.98078, sin(17π/16) ≈ sin(π + π/16) = -sin(π/16) ≈ -0.19509. Therefore, z_2 ≈ (1.0905 * (-0.98078), 1.0905 * (-0.19509) - 1) ≈ (-1.070, -0.213 - 1) ≈ (-1.070, -1.213).

Similarly, z_3: angle 25π/16 ≈ 4.909 radians. cos(25π/16) = cos(2π - 7π/16) = cos(7π/16) ≈ 0.19509, sin(25π/16) = sin(2π - 7π/16) = -sin(7π/16) ≈ -0.98078. Therefore, z_3 ≈ (1.0905 * 0.19509, 1.0905 * (-0.98078) - 1) ≈ (0.213, -1.070 - 1) ≈ (0.213, -2.070).

Wait, hold on. Wait, for z_3, angle is 25π/16. Let me check:

25π/16 = π + 9π/16? Wait, π is 16π/16, so 25π/16 = π + 9π/16. Therefore, cos(25π/16) = cos(π + 9π/16) = -cos(9π/16) ≈ -(-0.19509) = 0.19509? Wait, no. Wait, cos(π + θ) = -cos θ. So cos(25π/16) = cos(π + 9π/16) = -cos(9π/16) ≈ -(-0.19509) = 0.19509? Wait, earlier I said cos(9π/16) ≈ -0.19509? Wait, let me check:

Wait, 9π/16 is 101.25 degrees. cos(101.25 degrees) is cos(90 + 11.25) = -sin(11.25) ≈ -0.19509. Therefore, cos(9π/16) ≈ -0.19509. Therefore, cos(25π/16) = cos(π + 9π/16) = -cos(9π/16) ≈ -(-0.19509) = 0.19509. Similarly, sin(25π/16) = sin(π + 9π/16) = -sin(9π/16) ≈ -0.98078. Therefore, z_3 ≈ (1.0905 * 0.19509, 1.0905 * (-0.98078) - 1) ≈ (0.213, -1.070 - 1) ≈ (0.213, -2.070). Wait, but earlier I thought z_3 was (0.213, -2.070). Wait, but then z_0 is (1.070, -0.787), z_1 is (-0.213, 0.070), z_2 is (-1.070, -1.213), z_3 is (0.213, -2.070). Let me plot these points approximately:

z_0: (1.07, -0.787)

z_1: (-0.213, 0.07)

z_2: (-1.07, -1.213)

z_3: (0.213, -2.07)

Wait, connecting these points, is this a square? Let me check distances between consecutive points.

First, distance between z_0 and z_1:

sqrt[(1.07 - (-0.213))² + (-0.787 - 0.07)²] = sqrt[(1.283)² + (-0.857)²] ≈ sqrt[1.646 + 0.734] ≈ sqrt[2.38] ≈ 1.543.

Distance between z_1 and z_2:

sqrt[(-0.213 - (-1.07))² + (0.07 - (-1.213))²] = sqrt[(0.857)² + (1.283)²] ≈ same as before ≈ 1.543.

Distance between z_2 and z_3:

sqrt[(-1.07 - 0.213)² + (-1.213 - (-2.07))²] = sqrt[(-1.283)² + (0.857)²] ≈ same ≈ 1.543.

Distance between z_3 and z_0:

sqrt[(0.213 - 1.07)² + (-2.07 - (-0.787))²] = sqrt[(-0.857)² + (-1.283)²] ≈ same ≈ 1.543.

So all sides are equal. Now check the diagonals. Distance between z_0 and z_2:

sqrt[(1.07 - (-1.07))² + (-0.787 - (-1.213))²] = sqrt[(2.14)² + (0.426)²] ≈ sqrt[4.5796 + 0.181] ≈ sqrt[4.76] ≈ 2.18.

Distance between z_1 and z_3:

sqrt[(-0.213 - 0.213)² + (0.07 - (-2.07))²] = sqrt[(-0.426)² + (2.14)²] ≈ same as above ≈ 2.18.

Now, for a square, the diagonals should be equal to side * sqrt(2). Let me check: side ≈ 1.543, so diagonal ≈ 1.543 * 1.414 ≈ 2.18. Which matches the computed diagonal. Therefore, yes, it's a square. Therefore, the area is (side)^2 ≈ (1.543)^2 ≈ 2.38. Which is approximately 2^{5/4} ≈ 2.378. Therefore, correct. Therefore, the area is indeed 2^{5/4}.

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}". 2^{5/4} is already in that form. However, maybe the problem expects the answer in another form? For example, maybe I need to rationalize or something? Wait, 2^{5/4} is equal to sqrt(2^{5/2})? Wait, no. Wait, 2^{5/4} is the same as (2^{1/4})^5 or something else? 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, answer is 5 + 4 + 2 = 11. Therefore, I think the answer is 11.

But just to be thorough, let me check if there's another approach. For example, maybe the original equation can be transformed into something else. Wait, we had (z + i)^4 = 1 + i. Therefore, the solutions are z = -i + (1 + i)^{1/4}. But since we already did that, and found the area. Alternatively, maybe there's a mistake in assuming that the polygon is a square? But with the coordinates checked, it's a square. Therefore, I think the answer is 11.

But just to check, maybe the problem is not a square? Wait, the equation is quartic, so four roots, which form a convex polygon. Since they are four points, the only convex polygon is a quadrilateral. If they are symmetrically placed, it's a square. Therefore, yes. Therefore, answer is 11. 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 resembles the binomial expansion of $(z + i)^4$, which is:

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

Comparing this with the given equation, we see that:

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

Thus, the equation becomes:

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

---

### Step 1: Solve the Equation

Let $ w = z + i $, so the equation becomes:

$$
w^4 = 1 + i
$$

We now find the fourth roots of $ 1 + i $. First, express $ 1 + i $ in polar form:

- Modulus: $ |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2} $
- Argument: $ \theta = \frac{\pi}{4} $

So:

$$
1 + i = \sqrt{2} \left( \cos\frac{\pi}{4} + i \sin\frac{\pi}{4} \right)
$$

The fourth roots of $ 1 + i $ are:

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

This gives four complex numbers $ w_k $, each lying on a circle of radius $ r = \left( \sqrt{2} \right)^{1/4} = 2^{1/8} $, and spaced at angles $ \frac{\pi}{16}, \frac{9\pi}{16}, \frac{17\pi}{16}, \frac{25\pi}{16} $, which are equally spaced every $ \frac{\pi}{2} $ radians. Therefore, the four roots form a square centered at the origin.

---

### Step 2: Translate the Roots

Recall that $ z = w - i $, so the solutions $ z $ are the fourth roots of $ 1 + i $, shifted down by $ i $ (i.e., translated vertically by 1 unit in the imaginary direction). This translation does **not** affect the shape or area of the polygon formed by the roots.

Thus, the four solutions $ z $ form a square, and the area of this square is the same as the area of the square formed by the roots $ w $, which are centered at the origin.

---

### Step 3: Compute the Area of the Square

For a square inscribed in a circle of radius $ r $, the diagonal of the square is $ 2r $, and the area is:

$$
\text{Area} = \frac{(\text{diagonal})^2}{2} = \frac{(2r)^2}{2} = 2r^2
$$

Here, $ r = 2^{1/8} $, so:

$$
\text{Area} = 2 \cdot (2^{1/8})^2 = 2 \cdot 2^{1/4} = 2^{1 + 1/4} = 2^{5/4}
$$

---

### Step 4: Express in the Required Form

We are told the area can be expressed in the form $ p^{a/b} $, where $ p $ is prime, and $ a $ and $ b $ are coprime positive integers.

We have:

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

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. Let me multiply out $ (x + 5)(x + 2) $.

So, $ (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 $ with a remainder of 4. Therefore, the original expression simplifies to:

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

So, $ \frac{(x + 5)(x + 2)}{x + 1} = x + 6 + \frac{4}{x + 1} $.

Okay, so now the function is expressed as $ x + 6 + \frac{4}{x + 1} $. Maybe this is easier to work with. Now, I need to find the minimum value of this expression for $ x > 0 $.

Let me denote $ f(x) = x + 6 + \frac{4}{x + 1} $. To find the minimum, I can take the derivative and set it equal to zero. Let me recall how to take derivatives of such functions.

First, the derivative of $ x $ is 1. The derivative of 6 is 0. The derivative of $ \frac{4}{x + 1} $ is $ -\frac{4}{(x + 1)^2} $. So putting it all together:

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

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

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

Solving for $ x $:

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

Multiply both sides by $ (x + 1)^2 $:

$ (x + 1)^2 = 4 $

Take square roots:

$ x + 1 = \pm 2 $

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

$ x + 1 = 2 $

Therefore, $ x = 2 - 1 = 1 $.

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

First, compute the second derivative $ f''(x) $:

The first derivative was $ f'(x) = 1 - \frac{4}{(x + 1)^2} $. The derivative of that is:

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

Wait, let me check that again. The derivative of $ -4/(x + 1)^2 $ is:

Let me write it as $ -4(x + 1)^{-2} $. The derivative is $ -4 * (-2)(x + 1)^{-3} * 1 = 8/(x + 1)^3 $. Yes, that's correct.

So $ f''(x) = 8/(x + 1)^3 $. Now, since $ x > 0 $, $ x + 1 > 1 $, so $ (x + 1)^3 > 1 $, so $ f''(x) > 0 $. Therefore, the function is concave upward at $ x = 1 $, which means that the critical point is a local minimum. Since the function tends to infinity as $ x $ approaches infinity and as $ x $ approaches 0 from the right, this local minimum is actually the global minimum on the interval $ x > 0 $.

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

Original function: $ \frac{(1 + 5)(1 + 2)}{1 + 1} = \frac{6 * 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 $. Same result. So that seems correct.

But just to be thorough, maybe I should check the behavior of the function as $ x $ approaches 0 and as $ x $ approaches infinity to make sure there isn't a lower value somewhere else.

As $ x $ approaches 0 from the right:

The original function is $ \frac{(0 + 5)(0 + 2)}{0 + 1} = \frac{5 * 2}{1} = 10 $. So approaching 10. At $ x = 1 $, it's 9, which is lower.

As $ x $ approaches infinity:

The original function $ \frac{(x + 5)(x + 2)}{x + 1} $. Let me divide numerator and denominator by x:

Numerator: $ (x + 5)(x + 2)/x = (x^2 + 7x + 10)/x = x + 7 + 10/x $.

Denominator: $ (x + 1)/x = 1 + 1/x $.

So the whole expression is approximately $ (x + 7 + 10/x) / (1 + 1/x) $. As $ x $ approaches infinity, this behaves like $ x / 1 = x $, which goes to infinity. Therefore, the function tends to infinity as $ x $ approaches infinity. Therefore, the minimum must be at $ x = 1 $, which is 9.

Alternatively, using the simplified expression $ x + 6 + 4/(x + 1) $. As $ x $ approaches infinity, this behaves like $ x + 6 $, which goes to infinity. As $ x $ approaches 0, it's $ 0 + 6 + 4/(0 + 1) = 6 + 4 = 10 $, which matches. So yes, the minimum is at x=1 with value 9.

But just to make sure, maybe I can try another method, like substitution. Let me set $ t = x + 1 $, since there is a $ x + 1 $ in the denominator. Since $ x > 0 $, then $ t > 1 $.

Let me express the original function in terms of t. Let me note that $ x = t - 1 $. Then:

Original function:

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

Simplify numerator:

First term: $ t - 1 + 5 = t + 4 $

Second term: $ t - 1 + 2 = t + 1 $

Therefore, numerator is $ (t + 4)(t + 1) $, denominator is t. So:

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

So the function becomes $ t + 5 + \frac{4}{t} $, where $ t > 1 $.

Now, this seems like a simpler expression. To find the minimum of $ t + 5 + \frac{4}{t} $ for $ t > 1 $.

Again, taking derivative with respect to t:

$ f(t) = t + 5 + 4/t $

$ f'(t) = 1 - 4/t^2 $

Set derivative equal to zero:

$ 1 - 4/t^2 = 0 $

Which gives $ t^2 = 4 $, so $ t = 2 $ or $ t = -2 $. But since $ t > 1 $, only $ t = 2 $ is valid.

So critical point at t=2. Check second derivative:

$ f''(t) = 8/t^3 $, which is positive for t > 0, so it's a minimum. Therefore, minimum at t=2.

Compute the value:

$ f(2) = 2 + 5 + 4/2 = 2 + 5 + 2 = 9 $. Same result.

Alternatively, since $ t + 4/t $ is a standard expression. For t > 0, the minimum of $ t + 4/t $ occurs at t = 2 (by AM-GM inequality, since t and 4/t are positive). Wait, but here t > 1, but the minimum of t + 4/t over t > 0 is at t = 2, which is within t > 1. Therefore, even if we consider t > 1, the minimum is still at t=2.

So, using AM-GM inequality: For positive numbers a and b, $ a + b \geq 2\sqrt{ab} $. So if we take a = t and b = 4/t, then $ t + 4/t \geq 2\sqrt{t * 4/t} = 2 * 2 = 4 $. Equality when t = 4/t => t^2 = 4 => t = 2. Therefore, the minimum of t + 4/t is 4, achieved at t=2. Therefore, the function $ t + 5 + 4/t $ has minimum value 4 + 5 = 9, achieved at t=2.

So that's another way to see it. Therefore, the minimum value is indeed 9.

Alternatively, maybe I can try completing the square or other methods, but since we already have two methods giving the same answer, and checking the endpoints, I think it's safe.

But just to make sure, let me try plugging in some values around x=1 to see if the function is indeed minimized there.

For example, take x=0.5 (which is greater than 0). Compute the original function:

$ (0.5 + 5)(0.5 + 2)/(0.5 + 1) = (5.5)(2.5)/1.5 = (13.75)/1.5 ≈ 9.166..., which is about 9.17, which is higher than 9.

Take x=2:

Original function: (2 + 5)(2 + 2)/(2 + 1) = 7 * 4 / 3 ≈ 28/3 ≈ 9.333..., which is higher than 9.

Take x=1. Let me compute again: (1 + 5)(1 + 2)/2 = 6*3/2 = 9.

Take x approaching 1 from the left, say x=0.9:

(0.9 + 5)(0.9 + 2)/(0.9 + 1) = (5.9)(2.9)/1.9 ≈ (17.11)/1.9 ≈ 9.005... which is slightly above 9.

Similarly, x=1.1:

(1.1 + 5)(1.1 + 2)/(1.1 + 1) = (6.1)(3.1)/2.1 ≈ 18.91 / 2.1 ≈ 9.0047... which is also slightly above 9.

So yes, it seems that x=1 gives the minimum value of 9.

Alternatively, maybe I can use substitution with y = x + 1, but that's similar to what I did before.

Alternatively, maybe using calculus of variations or other methods, but I think the methods I've used are sufficient.

Another thought: since the function simplifies to $ x + 6 + \frac{4}{x + 1} $, maybe I can let $ u = x + 1 $, so that $ x = u - 1 $, and then the function becomes $ (u - 1) + 6 + \frac{4}{u} = u + 5 + \frac{4}{u} $, which is the same as before. So again, same result.

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

Wait, but just to make sure there isn't a mistake in my algebra when I expanded the original expression. Let me check again:

Original numerator: (x + 5)(x + 2) = x^2 + 2x + 5x + 10 = x^2 + 7x + 10. Correct. Then dividing by x + 1.

Polynomial division: x^2 + 7x + 10 divided by x + 1.

First term x^2 / x = x. Multiply x*(x + 1) = x^2 + x. Subtract from numerator: (x^2 + 7x + 10) - (x^2 + x) = 6x + 10. Then divide 6x by x to get 6. Multiply 6*(x + 1) = 6x + 6. Subtract: (6x + 10) - (6x + 6) = 4. So quotient is x + 6 with remainder 4. Therefore, the expression is x + 6 + 4/(x + 1). Correct.

So that's correct.

Alternatively, maybe I can write the original function as follows:

Let me consider $ \frac{(x + 5)(x + 2)}{x + 1} $. Maybe I can write this as $ \frac{(x + 1 + 4)(x + 1 + 1)}{x + 1} $. Let me check:

x + 1 + 4 = x + 5, and x + 1 + 1 = x + 2. Yes! So that's a good substitution.

Therefore, the expression becomes $ \frac{( (x + 1) + 4 ) ( (x + 1) + 1 ) }{x + 1} $. Let me denote $ y = x + 1 $, so since $ x > 0 $, $ y > 1 $. Then the expression becomes $ \frac{(y + 4)(y + 1)}{y} = \frac{y^2 + 5y + 4}{y} = y + 5 + \frac{4}{y} $. Which is the same as before.

So again, we have $ y + 5 + 4/y $, which is minimized when y = 2 (as per AM-GM), leading to minimum value 2 + 5 + 4/2 = 9.

Therefore, all methods converge to the same answer.

Alternatively, if I didn't think of substitution, maybe I could use calculus as I did before.

But since all methods confirm that the minimum is 9 at x=1, I think that's solid.

Therefore, the minimum value is 9.

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

To find the minimum value of the function
$$
f(x) = \frac{(x + 5)(x + 2)}{x + 1}
$$
for $ x > 0 $, we begin by simplifying the expression.

---

### Step 1: Simplify the Expression

First, expand the numerator:
$$
(x + 5)(x + 2) = x^2 + 7x + 10
$$
So the function becomes:
$$
f(x) = \frac{x^2 + 7x + 10}{x + 1}
$$

Now, perform polynomial division of the numerator by the denominator:

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

So the expression simplifies to:
$$
f(x) = x + 6 + \frac{4}{x + 1}
$$

---

### Step 2: Find the Minimum

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

To find the minimum, take the derivative:
$$
f'(x) = 1 - \frac{4}{(x + 1)^2}
$$

Set the derivative equal to zero:
$$
1 - \frac{4}{(x + 1)^2} = 0 \Rightarrow (x + 1)^2 = 4 \Rightarrow x + 1 = 2 \Rightarrow x = 1
$$

Now, check the second derivative to confirm it's a minimum:
$$
f''(x) = \frac{8}{(x + 1)^3}
$$
Since $ x > 0 $, $ x + 1 > 1 $, so $ f''(x) > 0 $, meaning the function is concave up at $ x = 1 $, and thus has a local (and global) minimum at this point.

---

### Step 3: Evaluate the Function at the Critical Point

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

---

### Step 4: Confirm Behavior at Endpoints

- As $ x \to 0^+ $, the function approaches:
  $$
  \frac{(0 + 5)(0 + 2)}{0 + 1} = \frac{10}{1} = 10
  $$
- As $ x \to \infty $, the function behaves like $ x $, which tends to infinity.

Thus, the function is minimized at $ x = 1 $, and the minimum value is:

$$
\boxed{9}
$$

[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. Let me multiply out $ (x + 5)(x + 2) $.

So, $ (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 $ with a remainder of 4. Therefore, the original expression simplifies to:

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

So, $ \frac{(x + 5)(x + 2)}{x + 1} = x + 6 + \frac{4}{x + 1} $.

Okay, so now the function is expressed as $ x + 6 + \frac{4}{x + 1} $. Maybe this is easier to work with. Now, I need to find the minimum value of this expression for $ x > 0 $.

Let me denote $ f(x) = x + 6 + \frac{4}{x + 1} $. To find the minimum, I can take the derivative and set it equal to zero. Let me recall how to take derivatives of such functions.

First, the derivative of $ x $ is 1. The derivative of 6 is 0. The derivative of $ \frac{4}{x + 1} $ is $ -\frac{4}{(x + 1)^2} $. So putting it all together:

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

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

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

Solving for $ x $:

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

Multiply both sides by $ (x + 1)^2 $:

$ (x + 1)^2 = 4 $

Take square roots:

$ x + 1 = \pm 2 $

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

$ x + 1 = 2 $

Therefore, $ x = 2 - 1 = 1 $.

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

First, compute the second derivative $ f''(x) $:

The first derivative was $ f'(x) = 1 - \frac{4}{(x + 1)^2} $. The derivative of that is:

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

Wait, let me check that again. The derivative of $ -4/(x + 1)^2 $ is:

Let me write it as $ -4(x + 1)^{-2} $. The derivative is $ -4 * (-2)(x + 1)^{-3} * 1 = 8/(x + 1)^3 $. Yes, that's correct.

So $ f''(x) = 8/(x + 1)^3 $. Now, since $ x > 0 $, $ x + 1 > 1 $, so $ (x + 1)^3 > 1 $, so $ f''(x) > 0 $. Therefore, the function is concave upward at $ x = 1 $, which means that the critical point is a local minimum. Since the function tends to infinity as $ x $ approaches infinity and as $ x $ approaches 0 from the right, this local minimum is actually the global minimum on the interval $ x > 0 $.

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

Original function: $ \frac{(1 + 5)(1 + 2)}{1 + 1} = \frac{6 * 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 $. Same result. So that seems correct.

But just to be thorough, maybe I should check the behavior of the function as $ x $ approaches 0 and as $ x $ approaches infinity to make sure there isn't a lower value somewhere else.

As $ x $ approaches 0 from the right:

The original function is $ \frac{(0 + 5)(0 + 2)}{0 + 1} = \frac{5 * 2}{1} = 10 $. So approaching 10. At $ x = 1 $, it's 9, which is lower.

As $ x $ approaches infinity:

The original function $ \frac{(x + 5)(x + 2)}{x + 1} $. Let me divide numerator and denominator by x:

Numerator: $ (x + 5)(x + 2)/x = (x^2 + 7x + 10)/x = x + 7 + 10/x $.

Denominator: $ (x + 1)/x = 1 + 1/x $.

So the whole expression is approximately $ (x + 7 + 10/x) / (1 + 1/x) $. As $ x $ approaches infinity, this behaves like $ x / 1 = x $, which goes to infinity. Therefore, the function tends to infinity as $ x $ approaches infinity. Therefore, the minimum must be at $ x = 1 $, which is 9.

Alternatively, using the simplified expression $ x + 6 + 4/(x + 1) $. As $ x $ approaches infinity, this behaves like $ x + 6 $, which goes to infinity. As $ x $ approaches 0, it's $ 0 + 6 + 4/(0 + 1) = 6 + 4 = 10 $, which matches. So yes, the minimum is at x=1 with value 9.

But just to make sure, maybe I can try another method, like substitution. Let me set $ t = x + 1 $, since there is a $ x + 1 $ in the denominator. Since $ x > 0 $, then $ t > 1 $.

Let me express the original function in terms of t. Let me note that $ x = t - 1 $. Then:

Original function:

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

Simplify numerator:

First term: $ t - 1 + 5 = t + 4 $

Second term: $ t - 1 + 2 = t + 1 $

Therefore, numerator is $ (t + 4)(t + 1) $, denominator is t. So:

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

So the function becomes $ t + 5 + \frac{4}{t} $, where $ t > 1 $.

Now, this seems like a simpler expression. To find the minimum of $ t + 5 + \frac{4}{t} $ for $ t > 1 $.

Again, taking derivative with respect to t:

$ f(t) = t + 5 + 4/t $

$ f'(t) = 1 - 4/t^2 $

Set derivative equal to zero:

$ 1 - 4/t^2 = 0 $

Which gives $ t^2 = 4 $, so $ t = 2 $ or $ t = -2 $. But since $ t > 1 $, only $ t = 2 $ is valid.

So critical point at t=2. Check second derivative:

$ f''(t) = 8/t^3 $, which is positive for t > 0, so it's a minimum. Therefore, minimum at t=2.

Compute the value:

$ f(2) = 2 + 5 + 4/2 = 2 + 5 + 2 = 9 $. Same result.

Alternatively, since $ t + 4/t $ is a standard expression. For t > 0, the minimum of $ t + 4/t $ occurs at t = 2 (by AM-GM inequality, since t and 4/t are positive). Wait, but here t > 1, but the minimum of t + 4/t over t > 0 is at t = 2, which is within t > 1. Therefore, even if we consider t > 1, the minimum is still at t=2.

So, using AM-GM inequality: For positive numbers a and b, $ a + b \geq 2\sqrt{ab} $. So if we take a = t and b = 4/t, then $ t + 4/t \geq 2\sqrt{t * 4/t} = 2 * 2 = 4 $. Equality when t = 4/t => t^2 = 4 => t = 2. Therefore, the minimum of t + 4/t is 4, achieved at t=2. Therefore, the function $ t + 5 + 4/t $ has minimum value 4 + 5 = 9, achieved at t=2.

So that's another way to see it. Therefore, the minimum value is indeed 9.

Alternatively, maybe I can try completing the square or other methods, but since we already have two methods giving the same answer, and checking the endpoints, I think it's safe.

But just to make sure, let me try plugging in some values around x=1 to see if the function is indeed minimized there.

For example, take x=0.5 (which is greater than 0). Compute the original function:

$ (0.5 + 5)(0.5 + 2)/(0.5 + 1) = (5.5)(2.5)/1.5 = (13.75)/1.5 ≈ 9.166..., which is about 9.17, which is higher than 9.

Take x=2:

Original function: (2 + 5)(2 + 2)/(2 + 1) = 7 * 4 / 3 ≈ 28/3 ≈ 9.333..., which is higher than 9.

Take x=1. Let me compute again: (1 + 5)(1 + 2)/2 = 6*3/2 = 9.

Take x approaching 1 from the left, say x=0.9:

(0.9 + 5)(0.9 + 2)/(0.9 + 1) = (5.9)(2.9)/1.9 ≈ (17.11)/1.9 ≈ 9.005... which is slightly above 9.

Similarly, x=1.1:

(1.1 + 5)(1.1 + 2)/(1.1 + 1) = (6.1)(3.1)/2.1 ≈ 18.91 / 2.1 ≈ 9.0047... which is also slightly above 9.

So yes, it seems that x=1 gives the minimum value of 9.

Alternatively, maybe I can use substitution with y = x + 1, but that's similar to what I did before.

Alternatively, maybe using calculus of variations or other methods, but I think the methods I've used are sufficient.

Another thought: since the function simplifies to $ x + 6 + \frac{4}{x + 1} $, maybe I can let $ u = x + 1 $, so that $ x = u - 1 $, and then the function becomes $ (u - 1) + 6 + \frac{4}{u} = u + 5 + \frac{4}{u} $, which is the same as before. So again, same result.

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

Wait, but just to make sure there isn't a mistake in my algebra when I expanded the original expression. Let me check again:

Original numerator: (x + 5)(x + 2) = x^2 + 2x + 5x + 10 = x^2 + 7x + 10. Correct. Then dividing by x + 1.

Polynomial division: x^2 + 7x + 10 divided by x + 1.

First term x^2 / x = x. Multiply x*(x + 1) = x^2 + x. Subtract from numerator: (x^2 + 7x + 10) - (x^2 + x) = 6x + 10. Then divide 6x by x to get 6. Multiply 6*(x + 1) = 6x + 6. Subtract: (6x + 10) - (6x + 6) = 4. So quotient is x + 6 with remainder 4. Therefore, the expression is x + 6 + 4/(x + 1). Correct.

So that's correct.

Alternatively, maybe I can write the original function as follows:

Let me consider $ \frac{(x + 5)(x + 2)}{x + 1} $. Maybe I can write this as $ \frac{(x + 1 + 4)(x + 1 + 1)}{x + 1} $. Let me check:

x + 1 + 4 = x + 5, and x + 1 + 1 = x + 2. Yes! So that's a good substitution.

Therefore, the expression becomes $ \frac{( (x + 1) + 4 ) ( (x + 1) + 1 ) }{x + 1} $. Let me denote $ y = x + 1 $, so since $ x > 0 $, $ y > 1 $. Then the expression becomes $ \frac{(y + 4)(y + 1)}{y} = \frac{y^2 + 5y + 4}{y} = y + 5 + \frac{4}{y} $. Which is the same as before.

So again, we have $ y + 5 + 4/y $, which is minimized when y = 2 (as per AM-GM), leading to minimum value 2 + 5 + 4/2 = 9.

Therefore, all methods converge to the same answer.

Alternatively, if I didn't think of substitution, maybe I could use calculus as I did before.

But since all methods confirm that the minimum is 9 at x=1, I think that's solid.

Therefore, the minimum value is 9.

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

To find the minimum value of the function
$$
f(x) = \frac{(x + 5)(x + 2)}{x + 1}
$$
for $ x > 0 $, we begin by simplifying the expression.

---

### Step 1: Simplify the Expression

First, expand the numerator:
$$
(x + 5)(x + 2) = x^2 + 7x + 10
$$
So the function becomes:
$$
f(x) = \frac{x^2 + 7x + 10}{x + 1}
$$

Now, perform polynomial division of the numerator by the denominator:

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

So the expression simplifies to:
$$
f(x) = x + 6 + \frac{4}{x + 1}
$$

---

### Step 2: Find the Minimum

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

To find the minimum, take the derivative:
$$
f'(x) = 1 - \frac{4}{(x + 1)^2}
$$

Set the derivative equal to zero:
$$
1 - \frac{4}{(x + 1)^2} = 0 \Rightarrow (x + 1)^2 = 4 \Rightarrow x + 1 = 2 \Rightarrow x = 1
$$

Now, check the second derivative to confirm it's a minimum:
$$
f''(x) = \frac{8}{(x + 1)^3}
$$
Since $ x > 0 $, $ x + 1 > 1 $, so $ f''(x) > 0 $, meaning the function is concave up at $ x = 1 $, and thus has a local (and global) minimum at this point.

---

### Step 3: Evaluate the Function at the Critical Point

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

---

### Step 4: Confirm Behavior at Endpoints

- As $ x \to 0^+ $, the function approaches:
  $$
  \frac{(0 + 5)(0 + 2)}{0 + 1} = \frac{10}{1} = 10
  $$
- As $ x \to \infty $, the function behaves like $ x $, which tends to infinity.

Thus, the function is minimized at $ x = 1 $, and the minimum value is:

$$
\boxed{9}
$$

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 $\boxed{9}.$