Problem Analysis #8

[THOUGHT] The region in I is the disk |z - (-√2 + 3i)| ≤ 3, so R=3, c=-√2 + 3i, |c|^2 = 2 + 9 = 11, R^2 + |c|^2 = 9 + 11 = 20.

Poles of f(z): z=0 (order 2, res=0), z=1-i (res=1/2), z=-1+i (res=-1/2).

Inside C: |z+1|=2, poles 0 and -1+i, sum res R_in = 0 + (-1/2) = -1/2, |R_in|=1/2.

I1 = 2πi * (-1/2) = -πi, |I1|=π.

I2 = π/2, |I1|/I2 = 2, (|I1|/I2)^3 = 8.

S = 20^2 * 8 * (1 / 0.5) = 400 * 8 * 2 = 6400.

K = 999 * S^2 + 7 = 999 * 6400^2 + 7.

Verify the computation of K with Python. [/THOUGHT] [PYTHON] print(999 * (6400 ** 2) + 7) [/PYTHON]

[THOUGHT] The Python output confirms $\mathcal{K} = 40919040007$. [/THOUGHT] $\boxed{40919040007}$

Explanation

The reasoning follows a step-by-step verification of each component to compute $\mathcal{K}$.

For part I, substitute $z = x + iy$ into the inequality $z\bar{z} + \sqrt{2}(z + \bar{z}) + 3i(z - \bar{z}) + 2 \leq 0$:

• $z\bar{z} = x^2 + y^2$
• $z + \bar{z} = 2x$
• $z - \bar{z} = 2iy$, so $3i(z - \bar{z}) = 3i(2iy) = -6y$
• The inequality simplifies to $x^2 + y^2 + 2\sqrt{2}x - 6y + 2 \leq 0$
• Completing the square: $(x + \sqrt{2})^2 + (y - 3)^2 \leq 9$
• This describes a disk with center $c = -\sqrt{2} + 3i$ and radius $R = 3$
• Compute $|c|^2 = (\sqrt{2})^2 + 3^2 = 2 + 9 = 11$
• Thus, $R^2 + |c|^2 = 9 + 11 = 20$

For part II(1), identify poles of $f(z) = \frac{z^2 - 2}{z^2 (z^2 + 2i)}$:

• Pole at $z = 0$ of order 2; residue is $\lim_{z \to 0} \frac{d}{dz} \left[ \frac{z^2 - 2}{z^2 + 2i} \right] = 0$
• Poles at roots of $z^2 + 2i = 0$, so $z^2 = -2i$; solutions are $z = 1 - i$ and $z = -1 + i$ (simple poles)
• Residue at $z = 1 - i$: $\frac{(1 - i)^2 - 2}{ \frac{d}{dz} [z^2 (z^2 + 2i)] \big|_{z=1-i} } = \frac{-2(1 + i)}{-4i(1 - i)} = \frac{1}{2}$
• Residue at $z = -1 + i$: $\frac{(-1 + i)^2 - 2}{ \frac{d}{dz} [z^2 (z^2 + 2i)] \big|_{z=-1+i} } = \frac{-2(1 + i)}{4(1 + i)} = -\frac{1}{2}$

For part II(2), the contour $C: |z + 1| = 2$ encloses poles at $z = 0$ ($|0 + 1| = 1 < 2$) and $z = -1 + i$ ($|-1 + i + 1| = 1 < 2$), but not $z = 1 - i$ ($|1 - i + 1| = \sqrt{5} > 2$)

• Sum of residues inside $C$: $R_{\mathrm{in}} = 0 + (-\frac{1}{2}) = -\frac{1}{2}$, so $|R_{\mathrm{in}}| = \frac{1}{2}$
• $I_1 = 2\pi i \cdot R_{\mathrm{in}} = 2\pi i \cdot (-\frac{1}{2}) = -\pi i$, so $|I_1| = \pi$

For part III(2), the integral $I_2 = \int_0^\infty \frac{\sin x}{x} \, dx = \frac{\pi}{2}$ (standard result from contour integration)

Compute $S$:

• $\frac{|I_1|}{I_2} = \frac{\pi}{\pi/2} = 2$, so $\left( \frac{|I_1|}{I_2} \right)^3 = 8$
• $S = (20)^2 \cdot 8 \cdot \frac{1}{1/2} = 400 \cdot 8 \cdot 2 = 6400$
• $\mathcal{K} = 999 \cdot S^2 + 7 = 999 \cdot 6400^2 + 7$
• Python verification: print(999 * (6400 ** 2) + 7) yields 40919040007, confirming the exact integer value.