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Messages - Yunfei Xia

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1
MAT334--Lectures & Home Assignments / Re: 2.5 Q19
« on: November 18, 2018, 08:02:05 AM »
Follow the hint: let $r$ be the circle $|z-z_0|=s,0<s<r$. For any $m\in z$.
\begin{align*}
0&=\frac{1}{2\pi i} \int_r(z-z_0)^{-m}\left(\sum\limits_{-\infty}^{+\infty}a_n(z-z_0)^n\right)\text{d}z\\
z-z_0=se^{i\theta}\\
&=\frac{1}{2\pi }\int\nolimits_{0}^{+2 \pi}\left(se^{i\theta}\right)^{-m}\left(\sum\limits_{-\infty}^{+\infty}{a_n}\left(se^{i\theta}\right)^n\right)se^{i\theta}\text{d}\theta\\
&=\sum\limits_{-\infty}^{+\infty}{a_n}s^{-m}s^{n+1}\frac{1}{2\pi }\int e^{-im \theta+i \theta+in}\text{d}\theta\\
&=\sum\limits_{-\infty}^{+\infty}{a_n}s^{-m}s^{n+1}\frac{1}{2\pi} \int e^{i \theta(n+1-m)}\text{d}\theta\\
&\text{Let}~n+1-m=0,n=m-1\\
&=a_{m-1}.
\end{align*}

2
MAT334--Lectures & Home Assignments / 2.5 Q11
« on: November 17, 2018, 10:15:47 PM »
$\frac{az+b}{cz+d}$, $z_0=-\frac{d}{c}$, $(c\neq0)$