[THOUGHT] I will compute each required quantity step‑by‑step using symbolic Python (SymPy).

1. Region I
   Write (z=x+iy). The inequality becomes
   [ x^{2}+y^{2}+2\sqrt{2},x-6y+2\le0. ]
   Completing squares gives ((x+\sqrt2)^{2}+(y-3)^{2}\le9).
   Hence the region is a disk with centre ((-,\sqrt2,,3)) and radius (R=3).
   So (|c|^{2}= (\sqrt2)^{2}+3^{2}=11).

2. Poles and residues of (f(z)=\frac{z^{2}-2}{(z^{2}+2i)z^{2}})
   Denominator zeros: (z=0) (order 2) and (z^{2}=-2i\Rightarrow z=1-i,,-1+i).
   Residues:

   • At (z=0): (\operatorname{Res}=0).
   • At (z=1-i): (\operatorname{Res}=1/2).
   • At (z=-1+i): (\operatorname{Res}=-1/2).

3. Integral (I_{1}) over (|z+1|=2)
   The circle is centred at (-1) with radius 2. Inside poles are (z=0) (residue 0) and (z=-1+i) (residue (-1/2)).
   By the residue theorem, [ I_{1}=2\pi i\sum\operatorname{Res}=2\pi i\left(-\frac12\right)=-\pi i, ] so (|I_{1}|=\pi).

4. Integral (I_{2}=\displaystyle\int_{0}^{\infty}\frac{\sin x}{x},dx)
   This classic integral equals (\pi/2).

5. Compute (S) and (\mathcal K)
   [ R^{2}+|c|^{2}=9+11=20,\qquad (R^{2}+|c|^{2})^{2}=400. ]
   [ \frac{|I_{1}|}{I_{2}}=\frac{\pi}{\pi/2}=2,\qquad\left(\frac{|I_{1}|}{I_{2}}\right)^{3}=8. ]
   [ |R_{\text{in}}|=\Bigl|!-\frac12\Bigr|=\frac12,\qquad\frac1{|R_{\text{in}}|}=2. ]
   Hence [ S=400\cdot8\cdot2=6400. ]
   Finally, [ \mathcal K=999,S^{2}+7=999\cdot6400^{2}+7=40,919,040,007. ]

I will now verify these calculations with Python code.

[THOUGHT] I will use SymPy to verify the disk parameters, poles, residues, and integrals step-by-step.

1. Disk from I:
   The inequality simplifies to $(x + \sqrt{2})^2 + (y - 3)^2 \leq 9$, confirming radius $R=3$ and center $c=(-\sqrt{2}, 3)$. Thus, $|c|^2 = 11$.

2. Poles and residues of $f(z)$:

   • Denominator: $z^2(z^2 + 2i)$.
   • Poles: $z=0$ (order 2), $z=1-i$, $z=-1+i$.
   • Residues:
     • At $z=0$: $\operatorname{Res}=0$.
     • At $z=1-i$: $\operatorname{Res}=1/2$.
     • At $z=-1+i$: $\operatorname{Res}=-1/2$.

3. Integral $I_1$ over $|z+1|=2$:

   • Circle centered at $-1$ with radius $2$.
   • Poles inside: $z=0$ (residue $0$) and $z=-1+i$ (residue $-1/2$).
   • Sum of residues: $-1/2$.
   • $I_1 = 2\pi i \cdot (-1/2) = -\pi i$, so $|I_1| = \pi$.

4. Integral $I_2 = \int_0^\infty \frac{\sin x}{x} dx$:

   • Known value: $\pi/2$.

5. Compute $S$ and $\mathcal{K}$:

   • $R^2 + |c|^2 = 9 + 11 = 20$.
   • $(R^2 + |c|^2)^2 = 400$.
   • $\frac{|I_1|}{I_2} = \frac{\pi}{\pi/2} = 2$, so $\left(\frac{|I_1|}{I_2}\right)^3 = 8$.
   • $|R_{\text{in}}| = \left|-\frac{1}{2}\right| = \frac{1}{2}$, so $\frac{1}{|R_{\text{in}}|} = 2$.
   • $S = 400 \cdot 8 \cdot 2 = 6400$.
   • $\mathcal{K} = 999 \cdot 6400^2 + 7 = 999 \cdot 40960000 + 7 = 40919040000 + 7 = 40919040007$.

I will now verify these calculations with Python.
[/THOUGHT] [PYTHON] import sympy as sp

Part I: Disk parameters

R = 3 c_x = -sp.sqrt(2) c_y = 3 c_abs_sq = c_x2 + c_y2

Part II: Poles and residues of f(z)

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

Find poles

poles = sp.roots(z2*(z2 + 2*sp.I), z)

Residues

residues = [] for pole in poles: res = sp.residue(f, z, pole) residues.append(res)

Part II(2): Integral I1 over |z+1|=2

Circle centered at -1, radius 2

circle_center = -1 circle_radius = 2

Poles inside the circle

inside_poles = [] for pole in poles: if abs(pole - circle_center) < circle_radius: inside_poles.append(pole) sum_res_inside = sum(residues[i] for i in range(len(poles)) if poles[i] in inside_poles)

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

I2 = sp.integrate(sp.sin(sp.symbols('x'))/sp.symbols('x'), (sp.symbols('x'), 0, sp.oo))

Compute S and K

R2_plus_c_abs_sq = R2 + c_abs_sq S = (R2_plus_c_abs_sq2) * (sp.pi / I2)3 * (1 / abs(sum_res_inside)) K = 999 * S2 + 7

Print results

print("R:", R) print("c_abs_sq:", c_abs_sq) print("Residues:", residues) print("Inside poles:", inside_poles) print("Sum of residues inside:", sum_res_inside) print("I2:", I2) print("S:", S.evalf()) print("K:", K.evalf()) [/PYTHON]

[THOUGHT] The KeyError indicates an issue with pole indexing. I'll fix the code by:

1. Correctly solving for poles using sp.solve
2. Computing residues with sp.residue
3. Checking pole inclusion by distance from circle center (-1,0)
4. Using exact symbolic values for I2 = π/2

The corrected code will:

• Find poles: z=0 (order 2), z=1-i, z=-1+i
• Compute residues: 0, 1/2, -1/2
• Identify poles inside |z+1|=2: z=0 and z=-1+i
• Sum residues inside: -1/2
• Use I2 = π/2
• Compute S = 6400 and K = 40919040007

I'll run the corrected code.

[THOUGHT] The KeyError suggests an issue with pole indexing. I'll fix the code by:

1. Using sp.solve to find poles of the denominator
2. Computing residues directly
3. Checking pole inclusion by distance from circle center (-1,0)
4. Using exact symbolic values for I2 = π/2

The corrected code will:

• Find poles: z=0 (order 2), z=1-i, z=-1+i
• Compute residues: 0, 1/2, -1/2
• Identify poles inside |z+1|=2: z=0 and z=-1+i
• Sum residues inside: -1/2
• Use I2 = π/2
• Compute S = 6400 and K = 40919040007

I'll run the corrected code now.

[THOUGHT] The KeyError indicates an issue with pole indexing. I'll fix the code by:

1. Using sp.solve to find poles of the denominator
2. Computing residues directly
3. Checking pole inclusion by distance from circle center (-1,0)
4. Using exact symbolic values for I2 = π/2

The corrected code will:

• Find poles: z=0 (order 2), z=1-i, z=-1+i
• Compute residues: 0, 1/2, -1/2
• Identify poles inside |z+1|=2: z=0 and z=-1+i
• Sum residues inside: -1/2
• Use I2 = π/2
• Compute S = 6400 and K = 40919040007

I'll run the corrected code now.
[/THOUGHT]
[PYTHON]
import sympy as sp

Part I: Disk parameters

R = 3
c_x = -sp.sqrt(2)
c_y = 3
c_abs_sq = c_x2 + c_y2

Part II: Poles and residues of f(z)

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

Find poles

denominator = sp.denom(f)
poles = sp.solve(denominator, z)
poles = list(set(poles)) # Remove duplicates

Compute residues

residues = []
for pole in poles:
res = sp.residue(f, z, pole)
residues.append(res)

Part II(2): Integral I1 over |z+1|=2

circle_center = -1
circle_radius = 2
inside_poles = []
for pole in poles:
if abs(pole - circle_center) < circle_radius:
inside_poles.append(pole)
sum_res_inside = sum(residues[i] for i in range(len(poles)) if poles[i] in inside_poles)

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

I2 = sp.integrate(sp.sin(sp.symbols('x'))/sp.symbols('x'), (sp.symbols('x'), 0, sp.oo))

Compute S and K

R2_plus_c_abs_sq = R2 + c_abs_sq
S = (R2_plus_c_abs_sq2) * (sp.pi / I2)3 * (1 / abs(sum_res_inside))
K = 999 * S2 + 7

Print results

print("R:", R)
print("c_abs_sq:", c_abs_sq)
print("Residues:", residues)
print("Inside poles:", inside_poles)
print("Sum of residues inside:", sum_res_inside)
print("I2:", I2)
print("S:", S.evalf())
print("K:", K.evalf())
[/PYTHON]
[THOUGHT]
The corrected code confirms:

• Disk: ( R = 3 ), ( |c|^2 = 11 )
• Residues: ( \operatorname{Res}(f, 0) = 0 ), ( \operatorname{Res}(f, 1-i) = 1/2 ), ( \operatorname{Res}(f, -1+i) = -1/2 )
• Poles inside ( |z+1|=2 ): ( z=0 ) and ( z=-1+i ), sum ( = -1/2 )
• ( I_2 = \pi/2 )
• ( S = 6400 ), ( \mathcal{K} = 40919040007 )

Final answer:
[/THOUGHT]
\boxed{40919040007}

[THOUGHT] The Python output confirms the disk parameters and integral values:

• Disk radius ( R = 3 ), center ( c = (-\sqrt{2}, -3) ), so ( |c|^2 = 11 )
• Residues: ( \operatorname{Res}(f, 0) = 0 ), ( \operatorname{Res}(f, 1-i) = 1/2 ), ( \operatorname{Res}(f, -1+i) = -1/2 )
• Poles inside ( |z+1|=2 ): ( z=0 ) and ( z=-1+i ), sum ( = -1/2 )
• ( I_2 = \pi/2 )
• ( S = 6400 )
• ( \mathcal{K} = 999 \cdot 6400^2 + 7 = 40919040007 )

Final answer:
[/THOUGHT]
\boxed{40919040007}