Answer the following questions. Here, for any complex value , is the complex conjugate of , is the argument of , is the absolute value of , and is the imaginary unit.\subsection*{I.}Sketch the region of on the complex plane that satisfies\subsection*{II.}Consider the complex-valued function
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[(1)] Find all the poles of as well as the orders and residues at the poles.
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[(2)] By applying the residue theorem, find the value of the integral where the integration path is the circle in the counterclockwise direction given by . \subsection*{III.}
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[(1)] Let be a complex-valued function satisfying for 0 $$\leq$$ \arg z $$\leq$$ $$\pi$$$. Let C_RR, in the upper half of the complex plane with center at the origin. Show that $$ \lim_{R\to\infty} \int_{C_R} e^{iaz} g(z)\,dz = 0, $$ where a$ is a positive real number.
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[(2)] Find the value of the integral \subsection*{IV. Final quantity}In I, the region for is a disk. Let be its radius and be its center. In II(2), let be the sum of the residues of over all poles lying inside the circle . In II(2) and III(2), let and be the values of the corresponding integrals.DefineCompute the integer .\
[THOUGHT]
I. The given inequality can be simplified by noting that is the square of the modulus and , are twice the real and imaginary parts of , 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 and radius .
II. (1) The function has poles where its denominator vanishes: [ z^2 + 2iz - 2 = (z - 1)(z + 2i) = 0.] Thus, there are simple poles at and . 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 , this gives ; for , it gives .
(2) The contour consists of a circle centered at with radius . Both poles () 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 is bounded in the upper half-plane (), as given. By Jordan's Lemma, for any , there exists a radius such that for all , [ \left| \int_{C_R} e^{iaz} g(z) dz \right| < \epsilon.] This is because the integrand decays faster than as (since ). Hence, [ \lim_{R\to\infty} \int_{C_R} e^{iaz} g(z) dz = 0.]
(2) The integral is a standard form that can be evaluated using the Fourier transform of . 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, is the inverse Fourier transform of : [ 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 = (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, is the radius of the circle from I and 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} = 999S^2 + 7 \approx 997005.]
\boxed{997005}