[THOUGHT] We need to compute several quantities from parts I, II, III and then compute S and K. Let's break it down.

Part I: Find the region for z satisfying z\bar{z} + sqrt(2)(z + \bar{z}) + 3i(z - \bar{z}) + 2 ≤ 0.

Let z = x + iy, where x, y real. Then: z\bar{z} = x^2 + y^2. z + \bar{z} = 2x. z - \bar{z} = 2iy. So 3i(z - \bar{z}) = 3i * (2iy) = 3i * 2i y = 6 i^2 y = -6y.

Thus inequality becomes: x^2 + y^2 + sqrt(2)*(2x) + (-6y) + 2 ≤ 0 => x^2 + y^2 + 2√2 x - 6y + 2 ≤ 0.

Complete squares: x^2 + 2√2 x = (x + √2)^2 - 2. y^2 - 6y = (y - 3)^2 - 9.

So inequality: (x + √2)^2 - 2 + (y - 3)^2 - 9 + 2 ≤ 0 => (x + √2)^2 + (y - 3)^2 - 9 ≤ 0 => (x + √2)^2 + (y - 3)^2 ≤ 9.

Thus center c = -√2 + 3i, radius R = 3.

So R = 3, |c| = sqrt((√2)^2 + 3^2) = sqrt(2 + 9) = sqrt(11).

Part II: f(z) = (z^2 - 2) / ((z^2 + 2i) z^2).

Find poles: denominator = (z^2 + 2i) z^2. So poles at z=0 (order 2) and z^2 = -2i => z = ±√(-2i). Compute -2i = 2 e^{-iπ/2} (since -i = e^{-iπ/2}). Actually careful: -2i = 2 * e^{-iπ/2}. Then sqrt: z = ±√2 e^{-iπ/4} and z = ±√2 e^{i( -π/4 + π)}? Actually square roots: z = ± (√2) e^{-iπ/4} and z = ± (√2) e^{i(3π/4)}? Let's compute properly.

We need z^2 = -2i. Write -2i in polar: magnitude 2, argument -π/2 (or 3π/2). Then square roots: magnitude √2, arguments (-π/2)/2 = -π/4 and (-π/2 + 2π)/2 = (3π/2)/2 = 3π/4. So poles: z = √2 e^{-iπ/4} and z = √2 e^{i3π/4}. But note e^{i3π/4} = e^{-iπ/4} * e^{iπ}? Actually e^{i3π/4} = e^{iπ} e^{-iπ/4}? No, e^{iπ} e^{-iπ/4} = e^{i3π/4} indeed. But anyway they are distinct.

So poles: z1 = 0 (order 2), z2 = √2 e^{-iπ/4}, z3 = √2 e^{i3π/4}.

Now we need residues at these poles for later? Actually for part II(2), we need R_in = sum of residues of f(z) over all poles lying inside the circle |z+1| = 2. So we need to determine which poles lie inside that circle.

Circle: center at -1, radius 2.

We need to compute distances from each pole to center -1.

Let's compute numerically.

Also need residues for II(2) maybe not needed because we use residue theorem to compute I1 directly? Actually part II(2) asks: "By applying the residue theorem, find the value of the integral I1 = ∮_C f(z) dz, where the integration path C is the circle in the counterclockwise direction given by |z+1| = 2." So I1 = 2πi * (sum of residues inside C). That sum is defined as R_in? Wait: In the problem statement: "In II(2), let R_in be the sum of the residues of f(z) over all poles lying inside the circle |z+1| = 2." So R_in is that sum. Then I1 = 2πi * R_in.

But careful: The residue theorem says ∮ f(z) dz = 2πi * sum of residues inside. So I1 = 2πi * R_in.

Thus |I1| = 2π * |R_in| (since R_in is complex, but we take absolute value of I1? Actually |I1| = |2πi * R_in| = 2π |R_in|.

Now I2: ∫_0^∞ sin x / x dx = π/2.

Thus |I1| / I2 = (2π |R_in|) / (π/2) = (2π |R_in|) * (2/π) = 4 |R_in|.

Thus (|I1| / I2)^3 = (4 |R_in|)^3 = 64 |R_in|^3.

Now S = (R^2 + |c|^2)^2 * (|I1|/I2)^3 * (1/|R_in|).

But careful: S = (R^2 + |c|^2)^2 * (|I1|/I2)^3 * (1/|R_in|).

We have R = 3, |c|^2 = 11, so R^2 + |c|^2 = 9 + 11 = 20. So (R^2 + |c|^2)^2 = 400.

(|I1|/I2)^3 = 64 |R_in|^3.

Thus S = 400 * 64 |R_in|^3 * (1/|R_in|) = 400 * 64 * |R_in|^2.

So S = 25600 * |R_in|^2.

Then K = 999 S^2 + 7.

Thus we need R_in: sum of residues of f(z) inside circle |z+1|=2.

We need to compute residues at poles inside that circle.

First, find which poles are inside.

Poles: z=0, z = √2 e^{-iπ/4}, z = √2 e^{i3π/4}.

Compute coordinates:

z0 = 0.

z2 = √2 e^{-iπ/4} = √2*(cos(-π/4) + i sin(-π/4)) = √2*(√2/2 - i√2/2) = 1 - i.

z3 = √2 e^{i3π/4} = √2*(cos(3π/4) + i sin(3π/4)) = √2*(-√2/2 + i√2/2) = -1 + i.

Check: e^{i3π/4} = cos(135°) + i sin(135°) = -√2/2 + i√2/2. Multiply by √2 gives -1 + i. Yes.

So poles: 0, 1 - i, -1 + i.

Now center at -1 (real -1, imag 0). Radius 2.

Compute distances:

For z=0: distance to -1: |0 - (-1)| = 1. So inside.

For z=1 - i: distance = |(1 - i) - (-1)| = |2 - i| = sqrt(4 + 1) = √5 ≈ 2.236 > 2. So outside.

For z=-1 + i: distance = |(-1 + i) - (-1)| = |i| = 1. So inside.

Thus poles inside: z=0 (order 2) and z=-1 + i (order 1). Wait, check: z=-1 + i is exactly the pole from z^2 = -2i? Yes, because -1 + i = √2 e^{i3π/4}. So poles inside: z=0 and z=-1 + i.

Now compute residues.

First, residue at z=0 (double pole). f(z) = (z^2 - 2) / ((z^2 + 2i) z^2). Write f(z) = (z^2 - 2) / (z^2 (z^2 + 2i)). For double pole at 0, we can compute residue as limit: Res(f,0) = limit as z->0 of d/dz [ z^2 f(z) ].

Compute z^2 f(z) = (z^2 - 2) / (z^2 + 2i). Then derivative w.r.t z: d/dz [(z^2 - 2)/(z^2 + 2i)].

Let numerator N = z^2 - 2, denominator D = z^2 + 2i.

Derivative: (N' D - N D') / D^2.

N' = 2z, D' = 2z.

So derivative = (2z*(z^2 + 2i) - (z^2 - 2)*2z) / (z^2 + 2i)^2 = [2z(z^2 + 2i - z^2 + 2)] / (z^2 + 2i)^2 = [2z(2i + 2)] / (z^2 + 2i)^2 = 4z(1 + i) / (z^2 + 2i)^2.

Now take limit as z->0: derivative -> 0. So Res(f,0) = 0? Wait, careful: For a double pole, residue is coefficient of 1/z in Laurent series. The formula Res(f,0) = limit_{z->0} d/dz [ z^2 f(z) ]. That limit is 0. So residue at 0 is 0.

Check: maybe I made mistake. Actually derivative at 0: plug z=0: numerator 4z(1+i)=0, denominator (0+2i)^2 = -4. So derivative = 0. Yes.

Thus residue at 0 is 0.

Now residue at z = -1 + i (simple pole). For simple pole, Res(f, z0) = limit_{z->z0} (z - z0) f(z). Since denominator factor (z^2 + 2i) has simple zero at z0 (because z0^2 = -2i, and derivative 2z0 ≠ 0). So we can write f(z) = (z^2 - 2) / (z^2 * (z^2 + 2i)). At z0, factor (z^2 + 2i) = (z - z0)(z + z0) * something? Actually since z0^2 = -2i, then z^2 + 2i = (z - z0)(z + z0). But careful: z^2 + 2i = (z - z0)(z + z0). Yes because product of roots: z0 * (-z0) = -z0^2 = 2i? Wait: if z0^2 = -2i, then (-z0)^2 = z0^2 = -2i, so both z0 and -z0 are roots of z^2 + 2i? Actually if z0^2 = -2i, then (-z0)^2 = z0^2 = -2i, so indeed -z0 is also a root. So z^2 + 2i = (z - z0)(z + z0). Good.

Thus f(z) = (z^2 - 2) / (z^2 (z - z0)(z + z0)).

Then residue at z0 is limit_{z->z0} (z - z0) f(z) = (z0^2 - 2) / (z0^2 * (z0 + z0)) = (z0^2 - 2) / (z0^2 * 2z0) = (z0^2 - 2) / (2 z0^3).

But z0^2 = -2i, so z0^2 - 2 = -2i - 2 = -2(1 + i). And z0^3 = z0 * z0^2 = z0 * (-2i) = -2i z0.

Thus Res(f, z0) = [-2(1+i)] / [2 * (-2i z0)] = [-2(1+i)] / [-4i z0] = (1+i) / (2i z0).

Simplify: (1+i)/(2i z0) = (1+i)/(2i) * 1/z0.

Now z0 = -1 + i. Compute 1/z0: z0 = -1 + i, magnitude sqrt(2). 1/z0 = conjugate(z0)/|z0|^2 = (-1 - i)/2.

Also (1+i)/(2i) = (1+i)/(2i) * (-i)/(-i)? Actually multiply numerator and denominator by -i: (1+i)(-i) / (2i)(-i) = (-i - i^2) / (2) = (-i + 1)/2 = (1 - i)/2.

Thus (1+i)/(2i) = (1 - i)/2.

Thus Res(f, z0) = ((1 - i)/2) * (1/z0) = ((1 - i)/2) * ((-1 - i)/2) = (1 - i)(-1 - i) / 4.

Compute (1 - i)(-1 - i) = -(1 - i)(1 + i)? Let's compute directly: (1)(-1) + 1*(-i) + (-i)(-1) + (-i)(-i) = -1 - i + i + i^2? Wait carefully:

(1 - i)(-1 - i) = 1*(-1) + 1*(-i) + (-i)(-1) + (-i)(-i) = -1 - i + i + i^2? Actually (-i)*(-i) = i^2 = -1.

So sum: -1 - i + i -1 = -2.

Thus product = -2.

Thus Res(f, z0) = (-2)/4 = -1/2.

So residue at z0 = -1/2.

Thus sum of residues inside: R_in = Res(f,0) + Res(f,z0) = 0 + (-1/2) = -1/2.

Thus |R_in| = 1/2.

Now compute S: S = 25600 * |R_in|^2 = 25600 * (1/4) = 6400.

Then K = 999 S^2 + 7 = 999 * (6400^2) + 7.

Compute 6400^2 = 40960000. Then 999 * 40960000 = 999 * 4096 * 10000? Actually compute directly.

We'll compute with Python to avoid errors.

Also need to verify I1 = 2πi * R_in = 2πi * (-1/2) = -π i. So |I1| = π. I2 = π/2. So |I1|/I2 = π / (π/2) = 2. Indeed earlier we got 4|R_in| = 4*(1/2) = 2. Good.

Now compute S: (R^2+|c|^2)^2 = 20^2 = 400. (|I1|/I2)^3 = 2^3 = 8. Then S = 400 * 8 * (1/|R_in|) = 3200 * (1/(1/2)) = 3200 * 2 = 6400. Yes matches.

Now compute K = 999*6400^2 + 7.

We'll compute with Python.

Also need to ensure region in I is a disk, which we have.

Now compute K.

Let's write Python code to compute K.

We'll compute step by step.

But also need to verify residues correctly. Let's double-check residue at z0 using sympy or manual.

We'll compute with sympy to be safe.

Also check if any pole inside besides those two? Possibly double pole at 0 residue 0, simple pole at -1+i residue -1/2.

Thus R_in = -0.5.

Now compute K.

We'll compute S = (R^2 + |c|^2)^2 * (|I1|/I2)^3 * (1/|R_in|).

But careful: In the problem statement, S = (R^2 + |c|^2)^2 * (|I1|/I2)^3 * (1/|R_in|). Yes.

We computed earlier S = 400 * 8 * (1/(0.5)) = 40082 = 6400.

Now K = 999*S^2 + 7.

Compute S^2 = 6400^2 = 40960000.

999 * 40960000 = ?

Compute: 1000 * 40960000 = 40960000000, subtract 40960000 = 40919040000.

Then +7 = 40919040007.

Thus K = 40919040007.

But we need to ensure integer digits without commas: 40919040007.

Now we should verify with Python to avoid arithmetic mistakes.

Let's write Python code.

We'll compute everything step by step.

We'll also compute residues using sympy to confirm.

We'll also verify which poles are inside the circle.

We'll compute distances.

Now produce final answer.

Let's code. </think>

[PYTHON] import sympy as sp import math

Part I: region

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

R = 3 c_center = -sp.sqrt(2) + 3*sp.I c_mod_sq = sp.sqrt(2)2 + 32 # |c|^2 = 11

So R^2 + |c|^2 = 9 + 11 = 20

Part II: poles and residues

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

Find poles

Denominator: z^2 * (z^2 + 2i)

Solve z^2 = 0 -> z=0 (double)

Solve z^2 = -2i

poles = sp.solve(z**2 + 2*sp.I, z) print("Poles from z^2+2i=0:", poles) # [-1 + I, 1 - I]

So poles: 0, -1+I, 1-I

Check which poles inside circle |z+1| = 2 (center -1, radius 2)

center = -1 radius = 2 def distance(zval): return abs(zval - center)

inside_poles = [] for pole in [0, -1+sp.I, 1-sp.I]: d = distance(pole) print(f"Distance from {pole} to center {center}: {d}") if d < radius: inside_poles.append(pole) print("Poles inside:", inside_poles)

Compute residues at poles inside

residues = {}

Residue at z=0 (double pole)

Compute Res(f,0) = limit_{z->0} d/dz [ z^2 f(z) ]

g = z**2 * f # = (z^2 - 2)/(z^2 + 2I) res0 = sp.limit(sp.diff(g, z), z, 0) residues[0] = res0 print("Residue at 0:", res0)

Residue at z = -1 + I (simple pole)

z0 = -1 + sp.I

Compute residue using formula for simple pole: (z - z0) f(z) limit

Or use sp.residue function

res_z0 = sp.residue(f, z, z0) residues[z0] = res_z0 print("Residue at -1+I:", res_z0)

Sum residues inside

R_in = sum(residues.values()) print("Sum of residues inside (R_in):", R_in) print("Absolute value |R_in|:", abs(R_in.evalf()))

Part III(2): I2 = ∫_0^∞ sin x / x dx = π/2

I2 = sp.pi/2 print("I2 =", I2)

Compute I1 = 2πi * R_in

I1 = 2sp.pisp.I * R_in print("I1 =", I1) print("|I1| =", abs(I1.evalf()))

Compute |I1| / I2

ratio = abs(I1.evalf()) / I2.evalf() print("|I1|/I2 =", ratio)

Compute S = (R^2 + |c|^2)^2 * (|I1|/I2)^3 * (1/|R_in|)

R_sq = R**2 c_mod_sq_val = c_mod_sq.evalf() factor1 = (R_sq + c_mod_sq_val)2 factor2 = ratio3 factor3 = 1 / abs(R_in.evalf()) S = factor1 * factor2 * factor3 print("S =", S.evalf())

Compute K = 999*S^2 + 7

K = 999 * S**2 + 7 print("K =", K.evalf())

Convert to integer (should be integer)

K_int = int(round(K)) print("K as integer:", K_int)

Double-check with direct arithmetic

R^2 + |c|^2 = 20

|I1|/I2 = 2 (since |R_in| = 0.5, ratio = 4*|R_in| = 2)

(|I1|/I2)^3 = 8

So S = 400 * 8 * (1/0.5) = 400 * 8 * 2 = 6400

K = 999 * 6400^2 + 7

S_direct = 6400 K_direct = 999 * S_direct**2 + 7 print("Direct K calculation:", K_direct)

Ensure both match

assert abs(K - K_direct) < 1e-10

print("Final integer K:", K_int) [/PYTHON]

[THOUGHT] The Python output shows that ( S = 6400 ), ( |I_1|/I_2 = 2 ), and ( |R_{\text{in}}| = 0.5 ).
Thus ( S = 20^2 \times 2^3 \times (1/0.5) = 400 \times 8 \times 2 = 6400 ).
Then ( \mathcal{K} = 999 \times 6400^2 + 7 = 40919040007 ).
The Python output confirms this integer value.
Therefore, the final answer is ( \boxed{40919040007} ). [/THOUGHT] \boxed{40919040007}