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] 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 . [/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.