Messages

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

MAT334--Lectures & Home Assignments / 2.5 Q19

« on: November 18, 2018, 04:01:10 AM »

Can anyone help me this question? Thanks!
Suppose that the Laurent series $\sum\nolimits_{-\infty}^{+\infty}a _n (z-z_0)^n $ converges for  $ f<| z-z_0|<R $ and
$$
\sum \limits_{-\infty}^{+\infty}  a_n (z-z_0)^n=0, 0<|z-z_0|<r.
$$

Show that $a_n=0,n=0, \pm 1, \pm 2,\cdots$, (Hint:Multiply the series by $(z-z_0)^{-m}$ and integrate around the circle $|z-z_0|=s$, $r<s<R$ with respect to $z$. The result must be zero, but it is also $a_{m-1}$.)

5

MAT334--Lectures & Home Assignments / Re: 2.5 Q11

« on: November 17, 2018, 10:19:43 PM »

\begin{align*}
f(z)&=\frac{\frac{a}{c}(cz+d)+b-\frac{ad}{c}}{cz+d}=\frac{a}{c}+\frac{b-\frac{ad}{c}}{cz+d}\\
&=\frac{a}{c}+(b-\frac{ad}{c})\cdot\frac{1}{cz+d}=\frac{a}{c}+(\frac{bc-ad}{c})\frac{1}{c}\cdot\frac{1}{z+\frac{d}{c}}\\
&=\frac{a}{c}+\frac{bc-ad}{c^2}(z+\frac{d}{c})^{-1}
\end{align*}

residue at $z_0=-\frac{d}{c}$ is $\frac{bc-ad}{c^2}$ which is coefficient of $(z-z_0)^{-1}$.

6

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 $-1$.

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MAT334--Lectures & Home Assignments / Re: Question 2 from 2.5

« on: November 14, 2018, 03:18:38 PM »

the answer may like this

8

MAT334--Lectures & Home Assignments / Re: Question 2 from 2.5

« on: November 14, 2018, 03:12:53 PM »

Sin(z)=0
∴z=0  or  z=kπ
When z=0,
Numerator: f(z)=z^2  ,f(0)=0
            f'(z)=2z,f'(0)=0
          f^'' (z)=2z ,f^'' (0)≠0
∴order=2
Denominator: g(z)=sinz ,g(0)=0
            g^' (z)=cosz ,g^' (0)≠0
∴order=1
2-1=1
Order of zero=1


When  z=kπ   (k≠0)
Numerator: f(z)=z^2,f(kπ)=(kπ)^2≠0
∴order=0
Denominator: g(z)=sinz ,g(kπ)=0
            g^' (z)=cosz ,g^' (kπ)≠0
∴order=1
1-0=1
It is simple pole.