Problem 3. Consider $f(z)=\frac{3z}{(z+1)(z-2)}$.
(i)Decompose it into Laurent's series $f(z) = \sum_{n=?}^{n=?}a_nz^n$ (find r and R)
(a) In the disc {$z:|z|<r$};
(b) In the annulus {$z:r<|z|<R$};
(c) In the disc exterior {$z : |z| > R$}.
(ii) Calculate $Res(f(z),0)$ and $Res(f(z),\infty)$.
Answer:
(i) Let $f(z)=\frac{A}{(z+1)} +\frac{B}{(z-2)}$, and let $r=1$, $R=2$.
Then, $A(z-2)+B(z+1)=3z$
$\left\{ \begin{align*}
A+B=3\\
-2A+B=0\\
\end{align*}
\right.$ $\Rightarrow$ $\left\{ \begin{align*}
A=1\\
B=2\\
\end{align*}
\right.$
For $|z|<1: \frac{1}{z+1} = \sum_{n=0}^{\infty}(-z)^n$.
For $|z|>1: \frac{1}{z+1} = \frac{1}{z}\frac{1}{1+\frac{1}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}(-\frac{1}{z})^n=\sum_{n=0}^{\infty}(-1)^nz^{-n-1}=\sum_{n=-\infty}^{-1}(-1)^{-n-1}z^n$
For $|z|<2: \frac{1}{z-2}=\frac{1}{2}\frac{1}{\frac{z}{2}-1}=-\frac{1}{2}\sum_{n=0}^{\infty}(\frac{z}{2})^n = - \sum_{n=0}^{\infty}2^{-n-1}z^n$
For $|z|>2: \frac{1}{z-2}=\frac{1}{z}\frac{1}{1-\frac{2}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}(\frac{2}{z})^n = \sum_{n=0}^{\infty}2^nz^{-n-1}=\sum_{n=-\infty}^{-1}2^{-n-1}z^n$
Thus,
\begin{align*}
f(z)&=\frac{1}{(z+1)} +\frac{2}{(z-2)}\\
\\
&=
\begin{cases}
\sum_{n=0}^{\infty}((-1)^n-2\cdot 2^{-n-1})z^n, & |z|< 1\\
\\
\sum_{n=-\infty}^{-1}(-1)^{-n-1}z^n-2\cdot \sum_{n=0}^{\infty}2^{-n-1}z^n, &1<|z|<2\\
\\
\sum_{n=-\infty}^{-1}((-1)^{-n-1}+2\cdot 2^{-n-1})z^n, &|z|>2\\
\end{cases}
\end{align*}
(ii)From part(i), we know the Laurent's series expression of $f(z)$.
For $z_0=0, f(z)=\sum_{n=0}^{\infty}((-1)^n-2\cdot 2^{-n-1})z^n$,
then $Res(f(z),0)=a_{-1}=(-1)^{-1}-2\cdot2^{1-1}=-1-2=-3$.
For $z_0=\infty, f(z)=\sum_{n=-\infty}^{-1}((-1)^{-n-1}+2\cdot 2^{-n-1})z^n$,
then $Res(f(z),\infty)=a_{-1}=(-1)^{1-1}+2\cdot2^{1-1}=1+2=3$
