[THOUGHT] Let me solve this problem step by step.

Part I: Finding the region Given: $z\bar{z} + \sqrt{2}(z + \bar{z}) + 3i(z - \bar{z}) + 2 \leq 0$

Let $z = x + iy$, then:

• $z\bar{z} = x^2 + y^2$
• $z + \bar{z} = 2x$
• $z - \bar{z} = 2iy$

Substituting: $x^2 + y^2 + \sqrt{2}(2x) + 3i(2iy) + 2 \leq 0$ $x^2 + y^2 + 2\sqrt{2}x - 6y + 2 \leq 0$

Completing the square: $(x + \sqrt{2})^2 - 2 + (y - 3)^2 - 9 + 2 \leq 0$ $(x + \sqrt{2})^2 + (y - 3)^2 \leq 9$

So the center is $c = -\sqrt{2} + 3i$ and radius $R = 3$.

Part II: Finding poles and residues of $f(z) = \frac{z^2 - 2}{(z^2 + 2i)z^2}$

Poles:

• $z = 0$ (from $z^2$) - order 2
• $z^2 + 2i = 0 \Rightarrow z^2 = -2i = 2e^{-i\pi/2}$
  • $z = \sqrt{2}e^{-i\pi/4} = 1 - i$
  • $z = \sqrt{2}e^{i3\pi/4} = -1 + i$

For the circle $|z + 1| = 2$:

• Center at $-1$, radius $2$
• Check which poles are inside

Let me compute this with Python. [/THOUGHT] [PYTHON] import numpy as np from sympy import *

Part I

Center c = -sqrt(2) + 3i, radius R = 3

R = 3 c = -sqrt(2) + 3*I c_abs = Abs(c) print(f"Part I: R = {R}, c = {c}, |c| = {simplify(c_abs)}") print(f"|c|^2 = {simplify(c_abs**2)}")

Part II

z = symbols('z') f = (z2 - 2) / ((z2 + 2*I) * z**2)

Find poles

z = 0 is a pole of order 2

z^2 + 2i = 0 => z = sqrt(-2i)

-2i = 2exp(-ipi/2)

sqrt(-2i) = sqrt(2)exp(-ipi/4) = sqrt(2)(cos(-pi/4) + isin(-pi/4)) = 1 - i

and sqrt(2)exp(i3pi/4) = sqrt(2)(cos(3pi/4) + isin(3*pi/4)) = -1 + i

pole1 = 0 pole2 = 1 - I # sqrt(-2i) pole3 = -1 + I # -sqrt(-2i)

print(f"\nPoles: z = {pole1}, z = {pole2}, z = {pole3}")

Verify poles

print(f"Check pole2: (1-i)^2 + 2i = {expand((1-I)**2 + 2I)}") print(f"Check pole3: (-1+i)^2 + 2i = {expand((-1+I)**2 + 2I)}")

Check which poles are inside |z + 1| = 2

Distance from pole to -1

dist1 = Abs(pole1 - (-1)) dist2 = Abs(pole2 - (-1)) dist3 = Abs(pole3 - (-1))

print(f"\nDistance from poles to -1:") print(f"|0 - (-1)| = {dist1}") print(f"|(1-i) - (-1)| = {simplify(dist2)}") print(f"|(-1+i) - (-1)| = {simplify(dist3)}")

Poles inside: those with distance < 2

print(f"\nPoles inside |z+1| = 2:") print(f"z = 0: {dist1} < 2 ? {dist1 < 2}") print(f"z = 1-i: {simplify(dist2)} < 2 ? {simplify(dist2) < 2}") print(f"z = -1+i: {simplify(dist3)} < 2 ? {simplify(dist3) < 2}") [/PYTHON]

[THOUGHT] Let me continue computing the residues and integrals. [/THOUGHT] [PYTHON] from sympy import *

z = symbols('z') f = (z2 - 2) / ((z2 + 2*I) * z**2)

Residues

At z = 0 (pole of order 2)

Res = lim_{z->0} d/dz [z^2 * f(z)] = lim d/dz [(z^2 - 2)/(z^2 + 2i)]

g0 = (z2 - 2)/(z2 + 2*I) dg0 = diff(g0, z) res_0 = limit(dg0, z, 0) print(f"Residue at z = 0: {simplify(res_0)}")

At z = -1 + i (simple pole)

Res = lim_{z -> -1+i} (z - (-1+i)) * f(z)

f(z) = (z^2 - 2) / ((z - (1-i))(z - (-1+i)) * z^2)

At simple pole z0: Res = (z0^2 - 2) / ((z0 - (1-i)) * z0^2 * 2*z0)

Actually z^2 + 2i = (z - (1-i))(z - (-1+i))

So Res at z = -1+i is (z^2 - 2) / (2z * z^2) evaluated at z = -1+i

pole3 = -1 + I numerator_at_pole3 = (pole32 - 2) derivative_denom = 2*z # derivative of z^2 + 2i denom_at_pole3 = derivative_denom.subs(z, pole3) * pole32 res_pole3 = simplify(numerator_at_pole3 / denom_at_pole3) print(f"Residue at z = -1+i: {res_pole3}")

Let me compute more carefully using residue function

res_0_sympy = residue(f, z, 0) res_pole3_sympy = residue(f, z, -1+I) res_pole2_sympy = residue(f, z, 1-I)

print(f"\nUsing sympy residue:") print(f"Residue at z = 0: {simplify(res_0_sympy)}") print(f"Residue at z = -1+i: {simplify(res_pole3_sympy)}") print(f"Residue at z = 1-i: {simplify(res_pole2_sympy)}")

Sum of residues inside |z+1| = 2 (poles: z=0 and z=-1+i)

R_in = res_0_sympy + res_pole3_sympy print(f"\nR_in = {simplify(R_in)}") print(f"|R_in| = {simplify(Abs(R_in))}")

I_1 = 2pii * R_in

I_1 = 2 * pi * I * R_in print(f"\nI_1 = {simplify(I_1)}") print(f"|I_1| = {simplify(Abs(I_1))}")

Part III(2): I_2 = integral from 0 to infinity of sin(x)/x dx = pi/2

I_2 = pi/2 print(f"\nI_2 = {I_2}")

Final calculation

R = 3 c_abs_sq = 11 # |c|^2 = 2 + 9 = 11 R_sq_plus_c_sq = R**2 + c_abs_sq # 9 + 11 = 20

print(f"\nR^2 + |c|^2 = {R_sq_plus_c_sq}") print(f"(R^2 + |c|^2)^2 = {R_sq_plus_c_sq**2}")

Compute S

abs_I1 = simplify(Abs(I_1)) abs_R_in = simplify(Abs(R_in))

print(f"\n|I_1| = {abs_I1}") print(f"|R_in| = {abs_R_in}") print(f"I_2 = {I_2}")

S = (R^2 + |c|^2)^2 * (|I_1|/I_2)^3 * 1/|R_in|

S = (R_sq_plus_c_sq**2) * (abs_I1/I_2)**3 / abs_R_in S_simplified = simplify(S) print(f"\nS = {S_simplified}")

K = 999 * S^2 + 7

K = 999 * S_simplified**2 + 7 K_simplified = simplify(K) print(f"\nK = {K_simplified}") [/PYTHON]

[THOUGHT] The Python calculation gives us:

• R = 3, |c|² = 11, so R² + |c|² = 20
• R_in = -1/2, so |R_in| = 1/2
• I_1 = -iπ, so |I_1| = π
• I_2 = π/2

S = (20)² × (π/(π/2))³ × (1/(1/2)) = 400 × 8 × 2 = 6400

K = 999 × 6400² + 7 = 999 × 40960000 + 7 = 40919040000 + 7 = 40919040007 [/THOUGHT]

\boxed{40919040007}