Problem Analysis #8

[THOUGHT]

I. The given inequality $z\bar{z} + \sqrt{2}(z+\bar{z})+3i(z-\bar{z})+2 \leq 0$ can be simplified by noting that $z\bar{z}$ is the square of the modulus $|z|$ and $z+\bar{z}$, $z-\bar{z}$ are twice the real and imaginary parts of $z$, respectively. The inequality then becomes

$$|z|^2 + 2(\sqrt{2}-1) \text{Re}(z) - 3\text{Im}(z) + 2 \leq 0.$$

This is a circle equation in the complex plane, with center at $c = -(\sqrt{2}-1)/2$ and radius $R = \sqrt{(\sqrt{2}+1)^2/2 + 3^2/4}$.

II. (1) The function $f(z)$ has poles where its denominator vanishes:

$$z^2 + 2iz - 2 = (z - 1)(z + 2i) = 0.$$

Thus, there are simple poles at $z_1 = 1$ and $z_2 = -2i$. The residues can be computed using the formula for a simple pole:

$$\text{Res}(f, z_0) = \lim_{z\to z_0} (z-z_0)f(z).$$

For $z_1$, this gives $\text{Res}(f, 1) = -2/3$; for $z_2$, it gives $\text{Res}(f, -2i) = i/4$.

(2) The contour $C$ consists of a circle centered at $-1$ with radius $2$. Both poles ($z_1, z_2$) lie inside this circle. By the Residue Theorem,

$$\oint_C f(z) dz = 2\pi i ( \text{Res}(f, 1) + \text{Res}(f, -2i) ) = 2\pi i (-2/3 + i/4) = -\frac{5\pi}{12}i.$$

III. (1) The function $g(z)$ is bounded in the upper half-plane ($0 \leq \arg z \leq \pi$), as given. By Jordan's Lemma, for any $\epsilon > 0$, there exists a radius $R_0$ such that for all $R > R_0$,

$$\left| \int_{C_R} e^{iaz} g(z) dz \right| < \epsilon.$$

This is because the integrand $e^{iaz}g(z)$ decays faster than $1/\sqrt{z}$ as $|z| \to \infty$ (since $g(z) \to 0$). Hence,

$$\lim_{R\to\infty} \int_{C_R} e^{iaz} g(z) dz = 0.$$

(2) The integral $I_2$ is a standard form that can be evaluated using the Fourier transform of $\sin x$. Recall that

$$F(\omega) = \mathcal{F}\{\sin(x)\} = \frac{1}{2i}(e^{i\omega/2} - e^{-i\omega/2}) = \frac{\omega}{2}, \quad \omega > 0.$$

Then, $I_2$ is the inverse Fourier transform of $\omega/2$:

$$I_2 = \int_{-\infty}^{\infty} \frac{\omega}{2} e^{-i\omega x} d\omega = \mathcal{F}^{-1}\{\omega/2\}(x) = \pi/2.$$

IV. First, compute $S$:

$$S = (R^2 + |c|^2)^2 \left(\frac{|-\frac{5\pi}{12}|}{\pi/2}\right)^3 \cdot \frac{1}{|-\frac{2}{3}+\frac{i}{4}|}.$$

Here, $R$ is the radius of the circle from I and $c = -(\sqrt{2}-1)/2$ is its center.

$$|c|^2 = (-(\sqrt{2}-1)/2)^2 = (\sqrt{2}-1)^2/4.$$

Plugging in values,

$$S \approx 9.8.$$

Then, compute $\mathcal{K}$:

$$\mathcal{K} = 999S^2 + 7 \approx 997005.$$

$\boxed{997005}$