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Messages - Yifei Wang

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1
Final Exam / Re: FE-P2
« on: December 18, 2018, 02:31:18 PM »
let $z = {re^{i\theta}}$
$f(Z) = f({re^{i\theta}}) = \frac{1}{2}({re^{i\theta}}+\frac{1}{r}{re^{-i\theta}})$
= $\frac{1}{2}{re^{i\theta}} + \frac{1}{2r}{re^{-i\theta}})$
= $\frac{r}{2}({cos\theta}+i{sin\theta})$ + $\frac{1}{2r}({cos(-\theta)}+i{sin(-\theta)})$

By odd and even function

= $(\frac{1}{2r}+\frac{r}{2}){cos\theta}$ +${isin\theta}(\frac{1}{2r}+\frac{r}{2})$
= $\frac{1+r^2}{2r}{cos\theta}$ + $i\frac{r^2-1}{2r}{sin\theta}$ = $U+iV$
${cos\theta} = \frac{U}{\frac{1+r^2}{2r}}$
${sin\theta} = \frac{U}{\frac{r^2-1}{2r}}$

for a:
$\frac{U}{\frac{1+r^2}{2r}}^2$ + $\frac{U}{\frac{r^2-1}{2r}}^2 = 1$
${\frac{1+r^2}{2r}}^2 - {\frac{r^2-1}{2r}}^2 = 1$
$a^2 = {\frac{1+r^2}{2r}}^2$
$b^2 ={\frac{r^2-1}{2r}}^2$

$---------------------------$
for b:
Similar to part a

= $\frac{1+r^2}{2r}{cos\theta}$ + $i\frac{r^2-1}{2r}{sin\theta}$ = $U+iV$
${\frac{1+r^2}{2r}} = \frac{U}{cos\theta}$
${\frac{r^2-1}{2r}} = \frac{V}{sin\theta}$

$\frac{U}{cos\theta}^2 - \frac{V}{sin\theta}^2 = 1$
$a^2 = {cos\theta}^2$
$b^2 = {sin\theta}^2$
$a^2 + b^2 = 1$

2
Final Exam / Re: FE-P6
« on: December 18, 2018, 01:21:42 PM »
sorry for reposting, the loading for edit seems like taking forever.

$\int_{0}^{\infty} \frac{1}{1+z^n} dz$ = $2 \pi i$ $\sum_{i=1}^{n} Res(f,z_i)$

$z^n = -1 = e^i\pi$

$z = e^{i(\frac{1}{n}+\frac{2k}{n})\pi}$

we can decompose the integral into three parts, where $C_1$ is the straight line on the x-axis, $C_R$ is the curve with radius R and $C_2$ being the line on the second quadrant


    $\int_{C}$  = $\int_{C_1}$  + $\int_{C_R}$  + $\int_{C_2}$

for $C_R$:
  let x = z $dx = dz$
  $\int_{C_R}$ = $\int_{0}^{R} \frac{1}{1+z^n} dz$
  as $\lim_{R\to\infty} $ = $\int_{0}^{\infty} \frac{1}{1+z^n} dz$
  = $2 \pi R *\frac{\alpha}{2 \pi}* \mid \frac{1}{1+z^n} \mid_{Max}$
  = $\alpha R \mid \frac{1}{1+z^n} \mid_{Max}$
  $\leq \alpha R \mid \frac{1}{1+R^n} \mid$
  $\leq \alpha R \mid \frac{1}{R^n} \mid$
  as $\lim_{R\to\infty} $
  $\leq \alpha R \mid \frac{1}{R^n} \mid$ = 0
  $\int_{0}^{R} \frac{1}{1+z^n} dz = 0$
 
 $----------------------------------------------------$
for $C_1$:
  let $x = t e^{i\theta}, dx = e^{i\theta}dt $ $ t \in [0,R] $ and $\theta$ be the angle between the line and x-axis
  $\int_{C_1}$ = $\int_{0}^{R} \frac{1}{1+x^n} dz$
  as $\lim_{R\to\infty} $ = $\int_{0}^{\infty} \frac{1}{1+x^n} dx$
  = $\int_{0}^{\infty} \frac{e^{i\theta}}{1+{t e^{i\theta}}^n} dt$
  = $\int_{0}^{\infty} \frac{e^{i\theta}}{1+t^n e^{ni\theta}}dt$
  Since $\theta$ = 0
  = $\int_{0}^{\infty} \frac{1dt}{1+t^n}$ = $I$
  = $I$
 
 
  $-------------------------------------------------------$
  for $C_2$:
  let $x = t e^{i\alpha}, dx = e^{i\alpha}dt $ $ t \in [R,0] $
  $\int_{C_1}$ = $\int_{R}^{0} \frac{1}{1+x^n} dz$
  as $\lim_{R\to\infty} $ = $\int_{\infty}^{0} \frac{1}{1+x^n} dz$
  =$\int_{\infty}^{0} \frac{e^{i\alpha}}{1+{te^{i\alpha}}^n} dt$
  =$\int_{\infty}^{0} \frac{e^{i\alpha}}{1+{t^ne^{i\alpha n}}} dt$
  = -$e^{i\alpha}\int_{0}^{\infty} \frac{1}{1+{t^ne^{i\alpha n}}} dt$
  = -$e^{i\alpha}I$
 
 
  $\int_{C}$ = $I$ -$e^{i\alpha}I$
  = $1-e^{i\alpha}$I
  =$e^{i(\frac{1}{n}+\frac{2k}{n})\pi}$
  $I = \frac{e^{i(\frac{1}{n}+\frac{2k}{n})\pi}}{1-e^{i\alpha}}$


since we are in the principal branch, we let K = 0 Then you do not need $k$ at all anywhere

$I = \frac{e^{i(\frac{1}{n})\pi}}{1-e^{i\alpha}}$


3
Final Exam / Re: FE-P3
« on: December 18, 2018, 12:55:26 PM »
let $w = \frac{1}{z}$ then $ Z = \frac{1}{w}$
${Z\to\infty} $ ${w\to0}$

$f(z) = f(\frac{1}{w}) = \frac{sin{\frac{1}{w}}}{cos\frac{1}{w}}+\frac{1}{w}\frac{cos^2{\frac{1}{w}}}{sin^2{\frac{1}{w}}}$

at ${w\to0}$, we have a non-isolated singularity

4
Final Exam / Re: FE-P6
« on: December 18, 2018, 12:16:13 PM »
$\int_{0}^{\infty} \frac{1}{1+z^n} dz= 2 \pi i \sum_{i=1}^{K} Res(f,z_i)$

$z^n = -1 = e^{i\pi}$

$z = e^{i(\frac{1}{n}+\frac{2k}{n})\pi}$

we can decompose the integral into three parts, where $C_1$ is the straight line on x-axis, $C_R$ is the curve with radius R and $C_2$ being the line on the second quadrant

    $\int_{C}$  = $\int_{C_1}$  + $\int_{C_R}$  + $\int_{C_2}$

for $C_R$:
  let x = z $dx = dz$
  $\int_{C_R}$ = $\int_{0}^{R} \frac{1}{1+z^n} dz$
  as $\lim_{R\to\infty} $ = $\int_{0}^{\infty} \frac{1}{1+z^n} dz$
  = $2 \pi R *\frac{\alpha}{2 \pi}* \mid \frac{1}{1+z^n} \mid_{Max}$
  = $\alpha R \mid \frac{1}{1+z^n} \mid_{Max}$
  $\leq \alpha R \mid \frac{1}{1+R^n} \mid$
  $\leq \alpha R \mid \frac{1}{R^n} \mid$
  as $\lim_{R\to\infty} $
  $\leq \alpha R \mid \frac{1}{R^n} \mid$ = 0
  $\int_{0}^{R} \frac{1}{1+z^n} dz = 0$
 
for $C_1$:
  let $x = t e^{i\theta}, dx = e^{i\theta}dt $ $ t \in [0,R] $ and $\theta$ be the angle between the line and x-axis
  $\int_{C_1}$ = $\int_{0}^{R} \frac{1}{1+x^n} dz$
  as $\lim_{R\to\infty} $ = $\int_{0}^{\infty} \frac{1}{1+x^n} dx$
  = $\int_{0}^{\infty} \frac{e^{i\theta}}{1+{t e^{i\theta}}^n} dt$
  = $\int_{0}^{\infty} \frac{e^{i\theta}}{1+t^n e^{ni\theta}}dt$
  Since $\theta$ = 0
  = $\int_{0}^{\infty} \frac{1dt}{1+t^n}$ = $I$
  = $I$
 
 
  for $C_2$:
  let $x = t e^{i\alpha}, dx = e^{i\alpha}dt $ $ t \in [R,0] $
  $\int_{C_1}$ = $\int_{R}^{0} \frac{1}{1+x^n} dz$
  as $\lim_{R\to\infty} $ = $\int_{\infty}^{0} \frac{1}{1+x^n} dz$
  =$\int_{\infty}^{0} \frac{e^{i\alpha}}{1+{te^{i\alpha}}^n} dt$
  =$\int_{\infty}^{0} \frac{e^{i\alpha}}{1+{t^ne^{i\alpha n}}} dt$
  = -$e^{i\alpha}\int_{0}^{\infty} \frac{1}{1+{t^ne^{i\alpha n}}} dt$
  = -$e^{i\alpha}I$
 
 
  $\int_{C}$ = $I$ -$e^{i\alpha}I$
  = $1-e^{i\alpha}$I
  =$e^{i(\frac{1}{n}+\frac{2k}{n})\pi}$
  $I = \frac{e^{i(\frac{1}{n}+\frac{2k}{n})\pi}}{1-e^{i\alpha}}$

5
Term Test 2 / Re: TT2A Problem 1
« on: November 25, 2018, 02:38:21 PM »
Thank you for the correction!

6
Term Test 2 / Re: TT2 Problem 2
« on: November 24, 2018, 05:38:47 AM »
I think the power is -1/2 inside of the -1

Correct V.I.  It was actually Test2B

7
Term Test 2 / Re: TT2A Problem 1
« on: November 24, 2018, 05:36:15 AM »

ZhenDi Pan

I think you are missing the $z$ on the numerator.

8
Term Test 2 / Re: TT2A Problem 1
« on: November 24, 2018, 05:34:25 AM »
We can rewrite the fraction as:

$let  f(z) =\frac{z}{z^2-4z+5}$

as

$\frac{z}{(z-(2+i))(z-(2-i))}$

a. When $2+i$ is inside
$f(z) = \int \frac{\frac{z}{z-2+i}}{z-2-i}~dz$

By Cauchy's thm

$f(z) = \int \frac{\frac{z}{z-2+i}}{z-2-i}~dz= 2i\pi *f(2+i) = \pi * (2+i)$

b. When $2-i$ is inside

$f(z) = \int \frac{\frac{z}{z-2-i}}{z-2+i}~dz$

By Cauchy's thm

$f(z) = \int \frac{\frac{z}{z-2-i}}{z-2+i}~dz= 2i\pi *f(2-i) = -\pi * (2-i)$

c. When both points are inside

$f(z) = \int \frac{z}{(z-(2+i))(z-(2-i))}~dz = 2i\pi (Res(f, 2+i) + Res(f, 2-i)) = 2\pi i$


9
Quiz-3 / Re: Q3 TUT 0301
« on: October 15, 2018, 10:39:25 PM »
I have a different approach to this question (without using the hint).


Let W as cos(x)cosh(y)-isin(x)sinh(y)
When X = 0:
sin(x) = 0, cos(x) = 1
W = cosh(y)

When X = pi
sin(x) = 0, cos(x) = -1
W =- cosh(y)

As we know cosh(y) in [1, limits) , and sin(x) in [1, -1].
Therefore, we can conclude y in (-limits, limits) and x in [-1, 1]

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