Problem Analysis #8

[THOUGHT] Let's tackle each part of the problem step-by-step.

Part I

We need to sketch the region for ( z ) on the complex plane that satisfies: [ z\bar{z} + \sqrt{2}(z + \bar{z}) + 3i(z - \bar{z}) + 2 \leq 0. ]

Let ( z = x + yi ) where ( x, y ) are real numbers. Then: [ \bar{z} = x - yi, ] [ z + \bar{z} = 2x, ] [ z - \bar{z} = 2yi, ] [ z\bar{z} = x^2 + y^2. ]

Substituting these into the inequality: [ x^2 + y^2 + \sqrt{2}(2x) + 3i(2yi) + 2 \leq 0, ] [ x^2 + y^2 + 2\sqrt{2}x - 6y + 2 \leq 0. ]

This can be rewritten as: [ x^2 + 2\sqrt{2}x + y^2 - 6y + 2 \leq 0. ]

To complete the square: [ x^2 + 2\sqrt{2}x + 2 = (x + \sqrt{2})^2 - 2, ] [ y^2 - 6y + 9 = (y - 3)^2 - 9. ]

So the inequality becomes: [ (x + \sqrt{2})^2 - 2 + (y - 3)^2 - 9 + 2 \leq 0, ] [ (x + \sqrt{2})^2 + (y - 3)^2 \leq 9. ]

This represents a disk centered at ( (-\sqrt{2}, 3) ) with radius ( 3 ).

Part II

(1) Find poles, orders, and residues

The function is: [ f(z) = \frac{z^2 - 2}{(z^2 + 2i)z^2}. ]

Poles:

1. ( z^2 + 2i = 0 ) gives ( z^2 = -2i ). Thus, ( z = \pm \sqrt{-2i} ).
2. ( z^2 = 0 ) gives ( z = 0 ).

Finding ( \sqrt{-2i} ): Let ( z = re^{i\theta} ), then: [ -2i = 2e^{i(3\pi/2)}. ] [ \sqrt{-2i} = \sqrt{2}e^{i(3\pi/4)} = \sqrt{2}\left(\frac{-\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i\right) = -1 + i. ] Thus, the poles are ( z = -1 + i ) and ( z = 1 - i ).

Orders:

• ( z = 0 ): Pole of order 2 (from ( z^2 ) in the denominator).
• ( z = -1 + i ) and ( z = 1 - i ): Pole of order 1 (simple poles).

Residues:

• At ( z = 0 ): [ \text{Res}[f(z), z = 0] = \lim_{z \to 0} \frac{d}{dz}\left(z^2 f(z)\right) = \lim_{z \to 0} \frac{d}{dz}\left(\frac{z^2 - 2}{z^2 + 2i}\right). ] [ = \lim_{z \to 0} \frac{(2z)(z^2 + 2i) - (z^2 - 2)(2z)}{(z^2 + 2i)^2} = \lim_{z \to 0} \frac{4iz}{(z^2 + 2i)^2} = 0. ]

• At ( z = -1 + i ): [ \text{Res}[f(z), z = -1 + i] = \lim_{z \to -1 + i} \frac{z^2 - 2}{(z - (-1 + i))(z^2 + 2i)} = \frac{(-1 + i)^2 - 2}{((-1 + i)^2 + 2i)(2(-1 + i))} = \frac{-2i - 2}{(-2i + 2i)(2(-1 + i))} = \frac{-2i - 2}{4(-1 + i)} = \frac{-1 - i}{-2 + 2i} = \frac{(-1 - i)(-2 - 2i)}{(-2 + 2i)(-2 - 2i)} = \frac{2 + 2i + 2i + 2}{-8} = \frac{4i}{-8} = -\frac{i}{2}. ]

• At ( z = 1 - i ): [ \text{Res}[f(z), z = 1 - i] = \lim_{z \to 1 - i} \frac{z^2 - 2}{(z - (1 - i))(z^2 + 2i)} = \frac{(1 - i)^2 - 2}{((1 - i)^2 + 2i)(2(1 - i))} = \frac{-2i - 2}{(-2i + 2i)(2(1 - i))} = \frac{-2i - 2}{4(1 - i)} = \frac{-1 - i}{2 - 2i} = \frac{(-1 - i)(2 + 2i)}{(2 - 2i)(2 + 2i)} = \frac{-2 - 2i - 2i - 2}{8} = \frac{-4i}{8} = -\frac{i}{2}. ]

So, the residues are: [ \text{Res}[f(z), z = 0] = 0, ] [ \text{Res}[f(z), z = -1 + i] = -\frac{i}{2}, ] [ \text{Res}[f(z), z = 1 - i] = -\frac{i}{2}. ]

(2) Compute ( I_1 )

The circle ( |z + 1| = 2 ) is centered at ( -1 ) with radius ( 2 ). The poles inside this circle are ( z = -1 + i ).

Using the residue theorem: [ I_1 = 2\pi i \times \text{Res}[f(z), z = -1 + i] = 2\pi i \times -\frac{i}{2} = \pi. ]

Part III

(1) Show the limit is zero

The integral is: [ \int_{C_R} e^{iaz} g(z),dz, ] where ( C_R ) is the upper semicircle of radius ( R ).

On ( C_R ), ( z = Re^{i\theta} ) with ( \theta ) from ( 0 ) to ( \pi ). Thus: [ e^{iaz} = e^{iaRe^{i\theta}} = e^{iaR(\cos\theta + i\sin\theta)} = e^{iaR\cos\theta} \cdot e^{-aR\sin\theta}. ]

Since ( g(z) \to 0 ) as ( |z| \to \infty ), there exists ( M ) such that ( |g(z)| < M/R ) for sufficiently large ( R ). Hence: [ \left| e^{iaz} g(z) \right| \leq \left| e^{iaR\cos\theta} \right| \cdot e^{-aR\sin\theta} \cdot \frac{M}{R} = e^{-aR\sin\theta} \cdot \frac{M}{R}. ]

The integral: [ \left| \int_{C_R} e^{iaz} g(z),dz \right| \leq \int_0^\pi e^{-aR\sin\theta} \cdot \frac{M}{R} \cdot iR,d\theta = M \int_0^\pi e^{-aR\sin\theta} ,d\theta. ]

Since ( \sin\theta \geq 0 ) on ( [0, \pi] ) and ( \sin\theta \geq \frac{2\theta}{\pi} ) on ( [0, \pi/2] ), we have: [ \int_0^\pi e^{-aR\sin\theta} ,d\theta = 2 \int_0^{\pi/2} e^{-aR\sin\theta} ,d\theta \leq 2 \int_0^{\pi/2} e^{-2aR\theta/\pi} ,d\theta = 2 \cdot \frac{\pi}{2aR} \left(1 - e^{-aR}\right) \leq \frac{\pi}{aR}. ]

Thus: [ \left| \int_{C_R} e^{iaz} g(z),dz \right| \leq M \frac{\pi}{aR} \to 0 \quad \text{as} \quad R \to \infty. ]

(2) Compute ( I_2 )

We use the residue theorem with a semicircular contour in the upper half-plane. Consider: [ f(z) = \frac{e^{iz}}{z}. ] The contour integral is: [ \oint_{C} \frac{e^{iz}}{z} ,dz = \int_{-R}^R \frac{e^{ix}}{x} ,dx + \int_{C_R} \frac{e^{iz}}{z} ,dz. ]

The residue at ( z = 0 ) is ( \text{Res}[f(z), z = 0] = 1 ). Thus: [ \oint_{C} \frac{e^{iz}}{z} ,dz = 2\pi i \times 1 = 2\pi i. ]

As ( R \to \infty ), the integral over ( C_R ) goes to zero (as shown in part III(1)). Therefore: [ \int_{-\infty}^\infty \frac{\sin x}{x} ,dx = \int_{-\infty}^\infty \frac{e^{ix} - e^{-ix}}{2ix} ,dx = \frac{1}{2i} \left( \int_{-\infty}^\infty \frac{e^{ix}}{x} ,dx - \int_{-\infty}^\infty \frac{e^{-ix}}{x} ,dx \right). ]

The second integral is the complex conjugate of the first, so: [ \int_{-\infty}^\infty \frac{\sin x}{x} ,dx = \frac{1}{2i} \left( 2\pi i - 0 \right) = \pi. ]

Thus, ( I_2 = \int_0^\infty \frac{\sin x}{x} ,dx = \frac{\pi}{2} ).

Final Calculation

Let's calculate the required quantity ( S ) and then ( \mathcal{K} ).

From part I, the disk has center ( c = -\sqrt{2} + 3i ) and radius ( R = 3 ).

From part II(2), ( I_1 = \pi ) and the residue inside the circle ( |z + 1| = 2 ) is ( R_{\mathrm{in}} = -\frac{i}{2} ).

From part III(2), ( I_2 = \frac{\pi}{2} ).

Compute ( |c| ): [ |c| = \sqrt{(-\sqrt{2})^2 + 3^2} = \sqrt{2 + 9} = \sqrt{11}. ]

Then: [ R^2 + |c|^2 = 3^2 + 11 = 9 + 11 = 20. ]

Thus: [ (R^2 + |c|^2)^2 = 20^2 = 400. ]

Next: [ \frac{|I_1|}{I_2} = \frac{\pi}{\frac{\pi}{2}} = 2. ]

Then: [ \left(\frac{|I_1|}{I_2}\right)^3 = 2^3 = 8. ]

Also: [ |R_{\mathrm{in}}| = \left| -\frac{i}{2} \right| = \frac{1}{2}. ]

So: [ S = 400 \cdot 8 \cdot 2 = 6400. ]

Finally: [ \mathcal{K} = 999 S^2 + 7 = 999 \cdot 6400^2 + 7. ]

Let's compute this using Python. [/THOUGHT] [PYTHON] S = 6400 K = 999 * S**2 + 7 print(K)

[THOUGHT] Based on the Python output, the value of (\mathcal{K}) is 40919040007. [/THOUGHT] \boxed{40919040007}