[THOUGHT]
Let's tackle each part of the problem step-by-step.
Part I
We need to sketch the region for z z z
z z ˉ + 2 ( z + z ˉ ) + 3 i ( z − z ˉ ) + 2 ≤ 0. z\bar{z} + \sqrt{2}(z + \bar{z}) + 3i(z - \bar{z}) + 2 \leq 0. z z ˉ + 2 ( z + z ˉ ) + 3 i ( z − z ˉ ) + 2 ≤ 0. Let z = x + y i z = x + yi z = x + y i x , y x, y x , y
z ˉ = x − y i , \bar{z} = x - yi, z ˉ = x − y i , z + z ˉ = 2 x , z + \bar{z} = 2x, z + z ˉ = 2 x , z − z ˉ = 2 y i , z - \bar{z} = 2yi, z − z ˉ = 2 y i , z z ˉ = x 2 + y 2 . z\bar{z} = x^2 + y^2. z z ˉ = x 2 + y 2 . Substituting these into the inequality:
x 2 + y 2 + 2 ( 2 x ) + 3 i ( 2 y i ) + 2 ≤ 0 , x^2 + y^2 + \sqrt{2}(2x) + 3i(2yi) + 2 \leq 0, x 2 + y 2 + 2 ( 2 x ) + 3 i ( 2 y i ) + 2 ≤ 0 , x 2 + y 2 + 2 2 x − 6 y + 2 ≤ 0. x^2 + y^2 + 2\sqrt{2}x - 6y + 2 \leq 0. x 2 + y 2 + 2 2 x − 6 y + 2 ≤ 0. This can be rewritten as:
x 2 + 2 2 x + y 2 − 6 y + 2 ≤ 0. x^2 + 2\sqrt{2}x + y^2 - 6y + 2 \leq 0. x 2 + 2 2 x + y 2 − 6 y + 2 ≤ 0. To complete the square:
x 2 + 2 2 x + 2 = ( x + 2 ) 2 − 2 , x^2 + 2\sqrt{2}x + 2 = (x + \sqrt{2})^2 - 2, x 2 + 2 2 x + 2 = ( x + 2 ) 2 − 2 , y 2 − 6 y + 9 = ( y − 3 ) 2 − 9. y^2 - 6y + 9 = (y - 3)^2 - 9. y 2 − 6 y + 9 = ( y − 3 ) 2 − 9. So the inequality becomes:
( x + 2 ) 2 − 2 + ( y − 3 ) 2 − 9 + 2 ≤ 0 , (x + \sqrt{2})^2 - 2 + (y - 3)^2 - 9 + 2 \leq 0, ( x + 2 ) 2 − 2 + ( y − 3 ) 2 − 9 + 2 ≤ 0 , ( x + 2 ) 2 + ( y − 3 ) 2 ≤ 9. (x + \sqrt{2})^2 + (y - 3)^2 \leq 9. ( x + 2 ) 2 + ( y − 3 ) 2 ≤ 9. This represents a disk centered at ( − 2 , 3 ) (-\sqrt{2}, 3) ( − 2 , 3 ) 3 3 3
Part II
(1) Find poles, orders, and residues
The function is:
f ( z ) = z 2 − 2 ( z 2 + 2 i ) z 2 . f(z) = \frac{z^2 - 2}{(z^2 + 2i)z^2}. f ( z ) = ( z 2 + 2 i ) z 2 z 2 − 2 . Poles:
z 2 + 2 i = 0 z^2 + 2i = 0 z 2 + 2 i = 0 z 2 = − 2 i z^2 = -2i z 2 = − 2 i z = ± − 2 i z = \pm \sqrt{-2i} z = ± − 2 i z 2 = 0 z^2 = 0 z 2 = 0 z = 0 z = 0 z = 0
Finding − 2 i \sqrt{-2i} − 2 i
Let z = r e i θ z = re^{i\theta} z = r e i θ
− 2 i = 2 e i ( 3 π / 2 ) . -2i = 2e^{i(3\pi/2)}. − 2 i = 2 e i ( 3 π /2 ) . − 2 i = 2 e i ( 3 π / 4 ) = 2 ( − 2 2 + 2 2 i ) = − 1 + i . \sqrt{-2i} = \sqrt{2}e^{i(3\pi/4)} = \sqrt{2}\left(\frac{-\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i\right) = -1 + i. − 2 i = 2 e i ( 3 π /4 ) = 2 ( 2 − 2 + 2 2 i ) = − 1 + i . Thus, the poles are z = − 1 + i z = -1 + i z = − 1 + i z = 1 − i z = 1 - i z = 1 − i
Orders:
z = 0 z = 0 z = 0 z 2 z^2 z 2 z = − 1 + i z = -1 + i z = − 1 + i z = 1 − i z = 1 - i z = 1 − i
Residues:
Res [ f ( z ) , z = 0 ] = lim z → 0 d d z ( z 2 f ( z ) ) = lim z → 0 d d z ( z 2 − 2 z 2 + 2 i ) . \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). Res [ f ( z ) , z = 0 ] = z → 0 lim d z d ( z 2 f ( z ) ) = z → 0 lim d z d ( z 2 + 2 i z 2 − 2 ) . = lim z → 0 ( 2 z ) ( z 2 + 2 i ) − ( z 2 − 2 ) ( 2 z ) ( z 2 + 2 i ) 2 = lim z → 0 4 i z ( z 2 + 2 i ) 2 = 0. = \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. = z → 0 lim ( z 2 + 2 i ) 2 ( 2 z ) ( z 2 + 2 i ) − ( z 2 − 2 ) ( 2 z ) = z → 0 lim ( z 2 + 2 i ) 2 4 i z = 0.
At z = − 1 + i z = -1 + i z = − 1 + i
Res [ f ( z ) , z = − 1 + i ] = lim z → − 1 + i z 2 − 2 ( z − ( − 1 + i ) ) ( z 2 + 2 i ) = ( − 1 + i ) 2 − 2 ( ( − 1 + i ) 2 + 2 i ) ( 2 ( − 1 + i ) ) = − 2 i − 2 ( − 2 i + 2 i ) ( 2 ( − 1 + i ) ) = − 2 i − 2 4 ( − 1 + i ) = − 1 − i − 2 + 2 i = ( − 1 − i ) ( − 2 − 2 i ) ( − 2 + 2 i ) ( − 2 − 2 i ) = 2 + 2 i + 2 i + 2 − 8 = 4 i − 8 = − i 2 . \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}. Res [ f ( z ) , z = − 1 + i ] = z → − 1 + i lim ( z − ( − 1 + i )) ( z 2 + 2 i ) z 2 − 2 = (( − 1 + i ) 2 + 2 i ) ( 2 ( − 1 + i )) ( − 1 + i ) 2 − 2 = ( − 2 i + 2 i ) ( 2 ( − 1 + i )) − 2 i − 2 = 4 ( − 1 + i ) − 2 i − 2 = − 2 + 2 i − 1 − i = ( − 2 + 2 i ) ( − 2 − 2 i ) ( − 1 − i ) ( − 2 − 2 i ) = − 8 2 + 2 i + 2 i + 2 = − 8 4 i = − 2 i . Res [ f ( z ) , z = 1 − i ] = lim z → 1 − i z 2 − 2 ( z − ( 1 − i ) ) ( z 2 + 2 i ) = ( 1 − i ) 2 − 2 ( ( 1 − i ) 2 + 2 i ) ( 2 ( 1 − i ) ) = − 2 i − 2 ( − 2 i + 2 i ) ( 2 ( 1 − i ) ) = − 2 i − 2 4 ( 1 − i ) = − 1 − i 2 − 2 i = ( − 1 − i ) ( 2 + 2 i ) ( 2 − 2 i ) ( 2 + 2 i ) = − 2 − 2 i − 2 i − 2 8 = − 4 i 8 = − i 2 . \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}. Res [ f ( z ) , z = 1 − i ] = z → 1 − i lim ( z − ( 1 − i )) ( z 2 + 2 i ) z 2 − 2 = (( 1 − i ) 2 + 2 i ) ( 2 ( 1 − i )) ( 1 − i ) 2 − 2 = ( − 2 i + 2 i ) ( 2 ( 1 − i )) − 2 i − 2 = 4 ( 1 − i ) − 2 i − 2 = 2 − 2 i − 1 − i = ( 2 − 2 i ) ( 2 + 2 i ) ( − 1 − i ) ( 2 + 2 i ) = 8 − 2 − 2 i − 2 i − 2 = 8 − 4 i = − 2 i . So, the residues are:
Res [ f ( z ) , z = 0 ] = 0 , \text{Res}[f(z), z = 0] = 0, Res [ f ( z ) , z = 0 ] = 0 , Res [ f ( z ) , z = − 1 + i ] = − i 2 , \text{Res}[f(z), z = -1 + i] = -\frac{i}{2}, Res [ f ( z ) , z = − 1 + i ] = − 2 i , Res [ f ( z ) , z = 1 − i ] = − i 2 . \text{Res}[f(z), z = 1 - i] = -\frac{i}{2}. Res [ f ( z ) , z = 1 − i ] = − 2 i . (2) Compute I 1 I_1 I 1
The circle ∣ z + 1 ∣ = 2 |z + 1| = 2 ∣ z + 1∣ = 2 − 1 -1 − 1 2 2 2 z = − 1 + i z = -1 + i z = − 1 + i
Using the residue theorem:
I 1 = 2 π i × Res [ f ( z ) , z = − 1 + i ] = 2 π i × − i 2 = π . I_1 = 2\pi i \times \text{Res}[f(z), z = -1 + i] = 2\pi i \times -\frac{i}{2} = \pi. I 1 = 2 πi × Res [ f ( z ) , z = − 1 + i ] = 2 πi × − 2 i = π . Part III
(1) Show the limit is zero
The integral is:
∫ C R e i a z g ( z ) d z , \int_{C_R} e^{iaz} g(z)\,dz, ∫ C R e ia z g ( z ) d z , where C R C_R C R R R R
On C R C_R C R z = R e i θ z = Re^{i\theta} z = R e i θ θ \theta θ 0 0 0 π \pi π
e i a z = e i a R e i θ = e i a R ( cos θ + i sin θ ) = e i a R cos θ ⋅ e − a R sin θ . e^{iaz} = e^{iaRe^{i\theta}} = e^{iaR(\cos\theta + i\sin\theta)} = e^{iaR\cos\theta} \cdot e^{-aR\sin\theta}. e ia z = e ia R e i θ = e ia R ( c o s θ + i s i n θ ) = e ia R c o s θ ⋅ e − a R s i n θ . Since g ( z ) → 0 g(z) \to 0 g ( z ) → 0 ∣ z ∣ → ∞ |z| \to \infty ∣ z ∣ → ∞ M M M ∣ g ( z ) ∣ < M / R |g(z)| < M/R ∣ g ( z ) ∣ < M / R R R R
∣ e i a z g ( z ) ∣ ≤ ∣ e i a R cos θ ∣ ⋅ e − a R sin θ ⋅ M R = e − a R sin θ ⋅ M R . \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}. e ia z g ( z ) ≤ e ia R c o s θ ⋅ e − a R s i n θ ⋅ R M = e − a R s i n θ ⋅ R M . The integral:
∣ ∫ C R e i a z g ( z ) d z ∣ ≤ ∫ 0 π e − a R sin θ ⋅ M R ⋅ i R d θ = M ∫ 0 π e − a R sin θ d θ . \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. ∫ C R e ia z g ( z ) d z ≤ ∫ 0 π e − a R s i n θ ⋅ R M ⋅ i R d θ = M ∫ 0 π e − a R s i n θ d θ . Since sin θ ≥ 0 \sin\theta \geq 0 sin θ ≥ 0 [ 0 , π ] [0, \pi] [ 0 , π ] sin θ ≥ 2 θ π \sin\theta \geq \frac{2\theta}{\pi} sin θ ≥ π 2 θ [ 0 , π / 2 ] [0, \pi/2] [ 0 , π /2 ]
∫ 0 π e − a R sin θ d θ = 2 ∫ 0 π / 2 e − a R sin θ d θ ≤ 2 ∫ 0 π / 2 e − 2 a R θ / π d θ = 2 ⋅ π 2 a R ( 1 − e − a R ) ≤ π a R . \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}. ∫ 0 π e − a R s i n θ d θ = 2 ∫ 0 π /2 e − a R s i n θ d θ ≤ 2 ∫ 0 π /2 e − 2 a Rθ / π d θ = 2 ⋅ 2 a R π ( 1 − e − a R ) ≤ a R π . Thus:
∣ ∫ C R e i a z g ( z ) d z ∣ ≤ M π a R → 0 as R → ∞ . \left| \int_{C_R} e^{iaz} g(z)\,dz \right| \leq M \frac{\pi}{aR} \to 0 \quad \text{as} \quad R \to \infty. ∫ C R e ia z g ( z ) d z ≤ M a R π → 0 as R → ∞. (2) Compute I 2 I_2 I 2
We use the residue theorem with a semicircular contour in the upper half-plane. Consider:
f ( z ) = e i z z . f(z) = \frac{e^{iz}}{z}. f ( z ) = z e i z . The contour integral is:
∮ C e i z z d z = ∫ − R R e i x x d x + ∫ C R e i z z d z . \oint_{C} \frac{e^{iz}}{z} \,dz = \int_{-R}^R \frac{e^{ix}}{x} \,dx + \int_{C_R} \frac{e^{iz}}{z} \,dz. ∮ C z e i z d z = ∫ − R R x e i x d x + ∫ C R z e i z d z . The residue at z = 0 z = 0 z = 0 Res [ f ( z ) , z = 0 ] = 1 \text{Res}[f(z), z = 0] = 1 Res [ f ( z ) , z = 0 ] = 1
∮ C e i z z d z = 2 π i × 1 = 2 π i . \oint_{C} \frac{e^{iz}}{z} \,dz = 2\pi i \times 1 = 2\pi i. ∮ C z e i z d z = 2 πi × 1 = 2 πi . As R → ∞ R \to \infty R → ∞ C R C_R C R
∫ − ∞ ∞ sin x x d x = ∫ − ∞ ∞ e i x − e − i x 2 i x d x = 1 2 i ( ∫ − ∞ ∞ e i x x d x − ∫ − ∞ ∞ e − i x x d x ) . \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). ∫ − ∞ ∞ x sin x d x = ∫ − ∞ ∞ 2 i x e i x − e − i x d x = 2 i 1 ( ∫ − ∞ ∞ x e i x d x − ∫ − ∞ ∞ x e − i x d x ) . The second integral is the complex conjugate of the first, so:
∫ − ∞ ∞ sin x x d x = 1 2 i ( 2 π i − 0 ) = π . \int_{-\infty}^\infty \frac{\sin x}{x} \,dx = \frac{1}{2i} \left( 2\pi i - 0 \right) = \pi. ∫ − ∞ ∞ x sin x d x = 2 i 1 ( 2 πi − 0 ) = π . Thus, I 2 = ∫ 0 ∞ sin x x d x = π 2 I_2 = \int_0^\infty \frac{\sin x}{x} \,dx = \frac{\pi}{2} I 2 = ∫ 0 ∞ x s i n x d x = 2 π
Final Calculation
Let's calculate the required quantity S S S K \mathcal{K} K
From part I, the disk has center c = − 2 + 3 i c = -\sqrt{2} + 3i c = − 2 + 3 i R = 3 R = 3 R = 3
From part II(2), I 1 = π I_1 = \pi I 1 = π ∣ z + 1 ∣ = 2 |z + 1| = 2 ∣ z + 1∣ = 2 R i n = − i 2 R_{\mathrm{in}} = -\frac{i}{2} R in = − 2 i
From part III(2), I 2 = π 2 I_2 = \frac{\pi}{2} I 2 = 2 π
Compute ∣ c ∣ |c| ∣ c ∣
∣ c ∣ = ( − 2 ) 2 + 3 2 = 2 + 9 = 11 . |c| = \sqrt{(-\sqrt{2})^2 + 3^2} = \sqrt{2 + 9} = \sqrt{11}. ∣ c ∣ = ( − 2 ) 2 + 3 2 = 2 + 9 = 11 . Then:
R 2 + ∣ c ∣ 2 = 3 2 + 11 = 9 + 11 = 20. R^2 + |c|^2 = 3^2 + 11 = 9 + 11 = 20. R 2 + ∣ c ∣ 2 = 3 2 + 11 = 9 + 11 = 20. Thus:
( R 2 + ∣ c ∣ 2 ) 2 = 20 2 = 400. (R^2 + |c|^2)^2 = 20^2 = 400. ( R 2 + ∣ c ∣ 2 ) 2 = 2 0 2 = 400. Next:
∣ I 1 ∣ I 2 = π π 2 = 2. \frac{|I_1|}{I_2} = \frac{\pi}{\frac{\pi}{2}} = 2. I 2 ∣ I 1 ∣ = 2 π π = 2. Then:
( ∣ I 1 ∣ I 2 ) 3 = 2 3 = 8. \left(\frac{|I_1|}{I_2}\right)^3 = 2^3 = 8. ( I 2 ∣ I 1 ∣ ) 3 = 2 3 = 8. Also:
∣ R i n ∣ = ∣ − i 2 ∣ = 1 2 . |R_{\mathrm{in}}| = \left| -\frac{i}{2} \right| = \frac{1}{2}. ∣ R in ∣ = − 2 i = 2 1 . So:
S = 400 ⋅ 8 ⋅ 2 = 6400. S = 400 \cdot 8 \cdot 2 = 6400. S = 400 ⋅ 8 ⋅ 2 = 6400. Finally:
K = 999 S 2 + 7 = 999 ⋅ 6400 2 + 7. \mathcal{K} = 999 S^2 + 7 = 999 \cdot 6400^2 + 7. K = 999 S 2 + 7 = 999 ⋅ 640 0 2 + 7. Let's compute this using Python.
[/THOUGHT]
[PYTHON]
S = 6400
K = 999 * S**2 + 7
print(K)