### Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

### Messages - Huanglei Ln

Pages: [1]
1
##### Term Test 2 / Re: TT2B Problem 3
« on: November 25, 2018, 02:45:19 AM »
\begin{align*}
f(z)&=\frac{sin(\pi z)}{\pi z^3}\\
sin(\pi z^3)&=0\\
\Rightarrow \pi z^3&=k\pi,k\in Z\\
\Rightarrow z^3=k&z_0=\sqrt[3]{k}
\left\{
\begin{array}{lr}
k\ is\ a\ perfect\ cubic,\sqrt[3]{k}\in {z-\{0\}}&  \\
k\ is\ not\ a\ perfect\ cube& \\
k=0, &
\end{array}
\right.
\end{align*}

case1:

\begin{align*}
f(z)&=\underbrace{sin(\pi z)}_{g(z)}\underbrace{\frac{1}{\pi z^3}}_{h(z)}\\
\sqrt[3]{k}&=Z\Rightarrow z\in Z\Rightarrow g(\sqrt[3]{k})=0\\
g'(z)&=\pi cos(\pi z)\Rightarrow g'(\sqrt[3]{k})=\pm \pi\neq0
\end{align*}

$\Rightarrow$ g(z) has a zero of order 1 at $\sqrt[3]{k}$

\begin{align*}
\frac{1}{h(z)}&=sin(\pi z^3)\\
\frac{d}{dz}sin(\pi z^3)&=cos(\pi z^3)3\pi z^2
\end{align*}

$cos(\pi k)3\pi k^{\frac{2}{3}}\neq 0$

$\Rightarrow$ h(z) has a pole of order 1 at $\sqrt[3]{k}$

Thus, f(z)has a removable singularity at $z=\sqrt[3]{k}\in Z-\{0\}$.

case2:

\begin{align*}
\sqrt[3]{k}\notin Z \Rightarrow g(z_0)\neq0
\end{align*}

h(z) has a pole of order 1 at $\sqrt[3]{k}$ as shown in case1

Thus, f(z) has a pole of order 1 at$z=\sqrt[3]{k} \notin Z$.

case3:

$g(0)=0,g'(0)\neq 0,\Rightarrow g(z)$ has zero od order 1 at z=0

$\frac{d}{dz}sin(\pi z^3)=xos(\pi z^3)3\pi z^2$

$\frac{d^2}{dz^2}sin(\pi z^3)=cos(\pi z^3)6\pi z+3\pi z^2(-3z^2\pi sin(\pi z^3))$

$\frac{d^3}{dz^3}sin(\pi z^3)=cos(\pi z^3)6\pi +6\pi z(-3\pi z^2sin(\pi z^3)+\cdots$

Note $\frac{d^3}{sz^3}|_{z=0}\neq0\Rightarrow h(z)$ has pole of order 3 at z=0

Thus f(z) has pole of order 2 at zero.

2
##### Term Test 2 / Re: TT2 Problem 2
« on: November 25, 2018, 01:39:58 AM »
\begin{aligned}
a)f(z)&=\frac{1}{1-z}=\sum^{\infty}_{n=o}z^n \\
\frac{1}{R}&=lim_{n\rightarrow {\infty} }|\frac{1}{1}|=1\Rightarrow R=1
\end{aligned}
\end{displaymath}
\begin{displaymath}
\begin{aligned}
b)f(-z^2)&=\frac{1}{1+z^2}=\sum^{\infty}_{n=o}{-z^2}^n=\Sigma^{\infty}(-1)^nz^{2n} \\
\Rightarrow \int f(-z^2)dz&=\sum^{\infty}_{n=o}(-1)^n\int^{2n}dz\\
\Rightarrow \int \frac{1}{1+z^2}dz&=\sum^{\infty}_(n=0)(-1)^n\int z^{2n}dz\\
\Rightarrow artan(z)+c&=\sum^{\infty}_{n=0}(-1)^n \frac{z^{2n+1}}{2n+1}\\
\Rightarrow artan(z)&=\sum^{\infty}_{n=0}\frac{(-1)^n}{2n+1}z^{2n+1} +c
\end{aligned}

3
##### Term Test 2 / Re: TT2 Problem 2
« on: November 25, 2018, 01:34:47 AM »
\begin{aligned} a)f(z)&=\frac{1}{1-z}=\sum^{\infty}_{n=0}z^n \\ \frac{1}{R}&=\lim_{n\rightarrow {\infty} }|\frac{1}{1}|=1\Rightarrow R=1 \end{aligned}
\begin{aligned} b)f(-z^2)&=\frac{1}{1+z^2}=\sum^{\infty}_{n=o}{-z^2}^n=\Sigma^{\infty}(-1)^nz^{2n} \\ \Rightarrow \int f(-z^2)dz&=\sum^{\infty}_{n=o}(-1)^n\int^{2n}dz\\ \Rightarrow \int \frac{1}{1+z^2}dz&=\sum^{\infty}_(n=0)(-1)^n\int z^{2n}dz\\ \Rightarrow \arctan(z)+c&=\sum^{\infty}_{n=0}(-1)^n \frac{z^{2n+1}}{2n+1}\\ \Rightarrow \arctan(z)&=\sum^{\infty}_{n=0}\frac{(-1)^n}{2n+1}z^{2n+1} +c \end{aligned}

4
##### Quiz-6 / Re: Q6 TUT 0301
« on: November 17, 2018, 07:18:50 PM »
\begin{align*}
\frac{\sin z}{(z-\pi)^2}=\frac{-\sin(z-\pi)}{(z-\pi)^2}&=-(z-\pi)^{-2}\sin(z-\pi)\\
&=-(z-\pi)^{-2}\sum\limits_{n=0}^\infty{\frac{(-1)^n(z-\pi)^{2n+1}}{(2n+1)!}}\\
&=\sum\limits_{n=0}^\infty{\frac{(-1)^{n+1}(z-\pi)^{2n-1}}{(2n+1)!}}
\end{align*}

residue at $z_0=\pi$ is $-1$.

Pages: [1]