<Question>
The solutions of the equation $z^4+4z^3i-6z^2-4zi-i=0$ are the vertices of a convex polygon in the complex plane.  The area of this polygon can be expressed in the form $p^{a/b},$ where $a,$ $b,$ $p$ are positive integers, $p$ is prime, and $a$ and $b$ are relatively prime.  Find $a + b + p.$
</Question>

<think>
<Plan_1>
Okay, so I need to solve this complex equation: z^4 + 4z^3i - 6z^2 - 4zi - i = 0. The solutions are the vertices of a convex polygon, and I need to find the area of that polygon, which can be expressed in the form p^(a/b) where p is a prime, and a and b are coprime. Then find a + b + p. Hmm, right. Let me start by trying to figure out what kind of equation this is and how to solve it.

<prompt_1.1> Identify the type of equation and plan to solve it by recognizing a binomial expansion pattern. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, it's a quartic equation in z, which can be tough, but maybe there's a pattern or substitution that can simplify it. The coefficients involve i, which is the imaginary unit. Let me write the equation again:

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

Hmm. The coefficients have a mix of real and imaginary terms. Maybe this can be related to a binomial expansion? Sometimes, complex equations can be factored or transformed into a known form. Let me think.

Wait, the coefficients 4i, -6, -4i, -i... Maybe this is related to expanding (z + i)^4? Let me check the expansion of (z + i)^4. Let's compute that:

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

Compute each term:

- z^4 is z^4.
- 4i z^3 is 4i z^3.
- 4i^2 z^2: i^2 is -1, so 4*(-1) = -4, so term is -4 z^2.
- 4i^3 z: i^3 is -i, so 4*(-i) = -4i, so term is -4i z.
- i^4 is (i^2)^2 = (-1)^2 = 1.

So, (z + i)^4 = z^4 + 4i z^3 - 4 z^2 -4i z + 1.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Comparing this to the given equation: z^4 +4i z^3 -6 z^2 -4i z -i =0. The given equation has -6 z^2 instead of -4 z^2, and the constant term is -i instead of 1. So, there's a discrepancy here.

Hmm. So maybe the given equation is (z + i)^4 but with some modifications? Let's see.

<prompt_2.1> Compare the given equation with the expansion of (z + i)^4 and identify the discrepancy. </prompt_2.1>

Maybe the original equation is (z + i)^4 - 2 z^2 - 2i z -i -1 =0?

Wait, perhaps not. Let me think differently. Let's suppose that the equation can be written as (z + i)^4 = something.

<prompt_2.2> Consider the possibility of the equation being related to (z + i)^4 and rewrite the equation accordingly. </prompt_2.2>
-----

Wait, but let me confirm that all roots are distinct. The equation (z + i)^4 = 2 z^2 + 2i z + i +1. Wait, but when we set (z + i)^4 equal to that expression, does that introduce any multiplicities? Or is the original equation equivalent to (z + i)^4 = 2 z^2 + 2i z + i +1, which is a quartic equation, so degree 4, so four roots. Since we started with a quartic equation, and after manipulation, we have four roots. Unless there's a factorization that causes multiple roots, but given that we have a quartic equation transformed into (z + i)^4 equals another quadratic expression, unless that expression is a perfect square or something, but 2 z^2 + 2i z + i +1 is not a perfect square.

Wait, but let's suppose that maybe when we moved terms around, we squared something or did operations that might introduce extraneous solutions? But no, because we just added (z + i)^4 to one side and rearranged. So the equation is equivalent.
Therefore, the four roots are indeed the four fourth roots of (2 z^2 + 2i z + i +1). Wait, but how does that help?

Alternatively, perhaps we can parametrize the roots. Let me think.

Alternatively, maybe since (z + i)^4 = something, but not sure. Alternatively, perhaps if we let w = z + i, then z = w - i. Substitute into the original equation.

<prompt_2.3> Substitute w = z + i and z = w - i into the original equation and simplify. </prompt_2.3>
</Plan_2>

<Execution_2>
<execution_2.1>
If I subtract 2 z^2 - 2i z - i - 1 from (z + i)^4, I get:

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

Combine like terms:

z^4 + 4i z^3 -6 z^2 -6i z -i =0.
</execution_2.1>

<execution_2.2>
Let's see:

If we take (z + i)^4, as above, that's z^4 + 4i z^3 -4 z^2 -4i z +1. The given equation is z^4 +4i z^3 -6 z^2 -4i z -i =0. Let's subtract (z + i)^4 from the given equation:

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

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

Adding these together:

(z^4 - z^4) + (4i z^3 -4i z^3) + (-6 z^2 +4 z^2) + (-4i z +4i z) + (-i -1) =0

Which simplifies to:

0 + 0 -2 z^2 +0 -i -1 =0

So, that gives -2 z^2 -i -1 =0 → -2 z^2 =1 +i → z^2 = -(1 +i)/2
</execution_2.2>

<execution_2.3>
Let me try that substitution:

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

Substitute z = w - i:

(w - i)^4 +4i (w - i)^3 -6 (w - i)^2 -4i (w - i) -i =0.

This will expand, but maybe it's manageable. Let me compute each term step by step.

First term: (w - i)^4. Let's expand that using binomial theorem:

= w^4 - 4i w^3 + ( -6 ) w^2 + 4i^3 w + i^4

Wait, compute each term:

(w - i)^4 = w^4 + 4 w^3 (-i) + 6 w^2 (-i)^2 + 4 w (-i)^3 + (-i)^4

Compute each coefficient:

- 4i w^3: correct.

- 6 w^2 (-i)^2: (-i)^2 = (-1)^2 i^2 = 1*(-1) = -1, so 6*(-1) = -6. So term is -6 w^2.

- 4 w (-i)^3: (-i)^3 = (-i)*(-i)^2 = (-i)*(-1) = i. So 4 w*(i) = 4i w.

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

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

Second term: 4i (w - i)^3. Let's compute (w - i)^3 first:

(w - i)^3 = w^3 + 3 w^2 (-i) + 3 w (-i)^2 + (-i)^3

Compute each term:

- 3 w^2 (-i) = -3i w^2.

- 3 w (-i)^2: (-i)^2 = -1, so 3w*(-1) = -3w.

- (-i)^3 = - (-i) = i. So term is i.

Thus, (w - i)^3 = w^3 -3i w^2 -3 w +i.

Multiply by 4i:

4i*(w^3 -3i w^2 -3 w +i) = 4i w^3 -12i^2 w^2 -12i w +4i^2

Simplify:

- 4i w^3.

- -12i^2 w^2: i^2 = -1, so -12*(-1) = 12, so term is +12 w^2.

- -12i w.

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

Third term: -6 (w - i)^2. Compute (w - i)^2:

(w - i)^2 = w^2 - 2i w + (-i)^2 = w^2 -2i w -1.

Multiply by -6: -6 w^2 +12i w +6.

Fourth term: -4i (w - i). Compute:

-4i w +4i^2 = -4i w -4.

Fifth term: -i.

Now, sum all terms together:

First term: w^4 -4i w^3 -6 w^2 +4i w +1

Second term: +4i w^3 +12 w^2 -12i w -4

Third term: -6 w^2 +12i w +6

Fourth term: -4i w -4

Fifth term: -i

Combine all terms:

Start with w^4.

Then w^3 terms: -4i w^3 +4i w^3 = 0.

w^2 terms: -6 w^2 +12 w^2 -6 w^2 = 0.

w terms: 4i w -12i w +12i w -4i w = 0.

Constant terms: 1 -4 +6 -4 -i = (1 -4) + (6 -4) -i = (-3) + 2 -i = -1 -i.

So overall, after substitution, the equation becomes:

w^4 -1 -i =0.

Therefore, w^4 = 1 + i.

So, w is a fourth root of (1 + i).
</execution_2.3>
</Execution_2>

<Plan_3>
Based on execution_2.3:
Wow, that's a significant simplification! So, by substituting w = z + i, the equation becomes w^4 = 1 + i. Therefore, the solutions for w are the fourth roots of the complex number 1 + i. Then, once we find w, we can get z by z = w - i.

Therefore, the solutions z are the fourth roots of (1 + i) shifted down by i.

So, first, let's find the fourth roots of 1 + i.

To find the fourth roots of 1 + i, we can represent 1 + i in polar form. Let's do that.

<prompt_3.1> Find the fourth roots of the complex number 1 + i. </prompt_3.1>
-----

Therefore, the polygon is a square rotated by 45 degrees, centered at the origin, with vertices at these four points. Wait, but actually, since we have four roots, which are fourth roots of 1 + i shifted by -i, so their positions in the complex plane would form a square?

<prompt_3.2> Analyze the geometric configuration of the roots to determine the area of the polygon. </prompt_3.2>
-----

But let me just confirm once again. The roots are fourth roots of 1 + i, each multiplied by e^{-i\pi/8} and then shifted by -i. Wait, but does the shifting by -i affect the regularity of the polygon? Since all roots are shifted by the same amount, the polygon is just translated downward by i. However, translation doesn't affect the area. So if the original roots (without the -i) would form a square, then shifting all of them by -i would just move the square but not change its area.

But actually, wait, the fourth roots of a complex number are equally spaced around a circle. So, the fourth roots of 1 + i would lie on a circle of radius r = |1 + i|^(1/4). Let's compute |1 + i|.

<prompt_3.3> Verify the geometric configuration of the roots and confirm the area of the polygon. </prompt_3.3>
</Plan_3>

<Execution_3>
<execution_3.1>
1 + i is a complex number. In the complex plane, 1 + i is located at (1,1). Its modulus (magnitude) is sqrt(1^2 + 1^2) = sqrt(2). Its argument (angle) is arctan(1/1) = π/4 radians. Therefore, 1 + i can be written in polar form as sqrt(2) e^{iπ/4}.

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

So, the fourth roots are:

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

Therefore, the four roots w are located at angles π/16, π/16 + π/2, π/16 + π, π/16 + 3π/2, all with radius 2^{1/8}.
</execution_3.1>

<execution_3.2>
Wait, actually, no. The fourth roots of a complex number are equally spaced around a circle, so they form the vertices of a square (rotated appropriately) inscribed in that circle. Then, shifting all of them by the same complex number (-i) just translates the square in the complex plane. So the shape remains a square, and the area is the same as the original square formed by the roots of 1 + i.

Therefore, the area should be the same as the area of the square formed by the fourth roots of 1 + i.
</execution_3.2>

<execution_3.3>
|1 + i| = sqrt(1^2 + 1^2) = sqrt(2). Therefore, the modulus of each fourth root is (sqrt(2))^(1/4) = (2^(1/2))^(1/4) = 2^(1/8). So radius is 2^(1/8).

But when we shift each root by -i, it's a translation in the complex plane. However, translating all points by the same vector doesn't change the shape or the area of the polygon. It just moves it somewhere else. So the area remains the same as the original square.

Therefore, the area of the polygon is 2, which is 2^(1/1), so a = 2, b =1, p=2. Therefore, a + b + p = 5.
</execution_3.3>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Let me verify that.

<prompt_4.1> Verify the fourth roots of 1 + i and their properties. </prompt_4.1>
-----

Based on execution_3.3:
Wait, but let me check again. Let me compute the side length.

<prompt_4.2> Recheck the geometric configuration and area calculation of the polygon. </prompt_4.2>
-----

Wait, but let me check if the fourth roots of 1 + i form a square.

<prompt_4.3> Verify the geometric configuration of the roots and finalize the area calculation. </prompt_4.3>
</Plan_4>

<Execution_4>
<execution_4.1>
The modulus of each fourth root is 2^{1/8}, so the roots lie on a circle of radius 2^{1/8}. The angles start at π/16, then each subsequent root is separated by π/2 (90 degrees). So, the four roots are spaced at 90-degree intervals starting from π/16. Therefore, their arguments are π/16, π/16 + π/2 = 9π/16, π/16 + π = 17π/16, and π/16 + 3π/2 = 25π/16. Each of these roots is a vertex of a square rotated by π/16 radians (or 11.25 degrees) relative to the standard position.

But when we take these roots and translate them by -i (which is a downward shift in the complex plane), the figure is just translated downward, not rotated or scaled. Therefore, the distances between the points remain the same, so the area remains the same as the original square.
</execution_4.1>

<execution_4.2>
The fourth roots of 1 + i are at radius 2^(1/8). The angle between consecutive roots is π/2 (since 2π/4 = π/2). So, the distance between two adjacent roots (side length of the square) can be found using the chord length formula: 2r sin(θ/2), where θ is the central angle. Here, θ = π/2, so chord length is 2 * 2^(1/8) * sin(π/4) = 2 * 2^(1/8) * sqrt(2)/2 = 2^(1/8) * sqrt(2) = 2^(1/8) * 2^(1/2) = 2^(3/8). Then, the area of the square is (side length)^2 = (2^(3/8))^2 = 2^(3/4). Wait, but 2^(3/4) is the same as (2^(1/4))^3, but maybe we can write it as 2^{3/4}. However, the problem states that the area can be expressed as p^{a/b}. So 2^{3/4} is already in that form with p=2, a=3, b=4. Since 2 is prime and 3 and 4 are coprime.
</execution_4.2>

<execution_4.3>
Suppose the fourth roots of a complex number are equally spaced around a circle. For four roots, the angles between them are π/2 apart. Therefore, they form a square. So yes, the original roots (w) form a square of side length sqrt(2) * r, where r is the radius. Wait, no. Wait, the distance between two adjacent roots on the circle of radius r is 2r sin(π/4) = sqrt(2) r. But the area of the square would be (sqrt(2) r)^2 = 2 r^2. Wait, but if the roots form a square, the area can also be computed as (diagonal)^2 / 2. The diagonal of the square is 2r (distance between two opposite vertices). So area is (2r)^2 / 2 = 4r^2 / 2 = 2r^2. So same as before.

But in our case, the original roots (w) are on a circle of radius 2^{1/8}, so the area would be 2*(2^{1/8})^2 = 2*2^{1/4} = 2^{1 + 1/4} = 2^{5/4}. But wait, hold on. Wait, let me clarify.

Wait, if the original roots (w) form a square inscribed in a circle of radius R, then the side length s of the square is R*sqrt(2). Because the diagonal of the square is 2R (the diameter), so s*sqrt(2) = 2R → s = 2R / sqrt(2) = sqrt(2) R. Then area is s^2 = 2 R^2. Alternatively, if the square is rotated, the distance between adjacent points is 2R sin(π/4) = sqrt(2) R. Wait, maybe I need to think differently.

Wait, actually, for a square inscribed in a circle of radius R, the side length is sqrt(2) R. Wait, no. Let's imagine a square with vertices on a circle of radius R. The diagonal of the square is equal to the diameter of the circle, which is 2R. For a square, diagonal d = s*sqrt(2), so s = d / sqrt(2) = 2R / sqrt(2) = sqrt(2) R. Therefore, the side length is sqrt(2) R. Then area is s^2 = (sqrt(2) R)^2 = 2 R^2.

But in our case, R is 2^{1/8}. Therefore, the area would be 2*(2^{1/8})^2 = 2*2^{1/4} = 2^{1 + 1/4} = 2^{5/4}. But wait, but the roots are shifted by -i, which is a translation, so the area remains the same. Therefore, the area of the polygon is 2^{5/4}.

But wait, let me confirm once again. Let's compute the coordinates of the roots.

Original roots w_k are 2^{1/8} e^{i (π/16 + π k / 2)} for k=0,1,2,3. Then z_k = w_k - i. So each w_k is a point on the complex plane, and then we subtract i, which is equivalent to shifting down by 1 unit.

But the shape of the polygon is just the original square (from the w_k) shifted down by i. The area is preserved under translation. Therefore, the area is indeed 2^{5/4}.

But let me verify the area calculation again.

Compute R = 2^{1/8}. Then area is 2 R^2 = 2*(2^{1/8})^2 = 2*2^{1/4} = 2^{1 + 1/4} = 2^{5/4}. So that's correct.

But the problem states that the area can be expressed as p^{a/b}, where p is prime, and a and b are coprime. Here, 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, a + b + p = 5 + 4 + 2 = 11.
</execution_4.3>
</Execution_4>

<Plan_5>
Based on execution_4.3:
But let me check if my initial substitution was correct. Let me recap:

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

Substitution: w = z + i ⇒ z = w - i.

Substituting into the equation, after expanding all terms, we get w^4 -1 -i =0 ⇒ w^4 =1 +i.

Therefore, the roots w are the fourth roots of 1 +i, which as we found, form a square of area 2^{5/4}.

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

But let me check with another approach to ensure that there isn't a miscalculation.

Alternative approach: Let's compute the area directly from the coordinates.

<prompt_5.1> Explore an alternative approach to verify the area calculation of the polygon. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
Suppose we find the coordinates of the roots z_k = w_k -i. Let's compute them:

Each w_k is 2^{1/8} e^{i (π/16 + π k /2)}. Then z_k = w_k -i = 2^{1/8} e^{i (π/16 + π k /2)} - i.

But perhaps this is more complicated. Alternatively, since the original roots w_k form a square of area 2^{5/4}, then shifting all points by -i (translation) doesn't change the area, so the area remains 2^{5/4}. Therefore, the answer is 2^{5/4}, leading to a + b + p = 11.

But to be thorough, let's compute the area using coordinates.

First, find the coordinates of each z_k.

Let me compute each w_k:

w_0: angle π/16, modulus 2^{1/8}.

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

Then z_0 = w_0 - i = 2^{1/8} cos(π/16) + i (2^{1/8} sin(π/16) -1).

Similarly for other roots:

w_1: angle π/16 + π/2 = 9π/16.

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

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

Similarly:

w_2: angle 17π/16.

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

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

w_3: angle 25π/16.

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

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

These points z_0, z_1, z_2, z_3 form a square? Let's check the distances between consecutive points.

But calculating all distances would be tedious. Alternatively, since the original w_k form a square, shifting all points by -i (which is a translation) preserves the shape and area. Therefore, the area remains the same as the original square formed by w_k, which we found to be 2^{5/4}.
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
**Final Answer**
\boxed{11}
</Plan_6>

<Execution_6>
No prompt to execute.
</Execution_6>

<Final_answer>
Therefore, the final answer, following the instructions, is encapsulated as requested. The area of the polygon can be expressed as $2^{5/4}$, leading to $a = 5$, $b = 4$, and $p = 2$.
Thus, $a + b + p = 5 + 4 + 2 = \boxed{11}$.
</Final_answer>
</think>