Laurent expansion with different annuli Find the Laurent expansion about $0$ of $$f(z)= \frac{1}{(z-i)(z-2)}$$ on the annuli:


*

*$0 \lt \lvert z \rvert \lt 1 $,

*$ 1 \lt \lvert z \rvert \lt 2$,

*$ 2 \lt \lvert z \rvert \lt \infty $.


So far I have put the function into partial fractions so $ \frac{i+2}{5} \left(\frac{1}{z-i} - \frac{1}{z-2} \right)$ but I'm not sure how to continue
 A: First, we need to address your partial fraction decomposition since it is missing a multiplication by $-1$. You should have obtained 
$$
\frac{1}{(z-i)(z-2)}=\frac{2+i}{5}\Bigl[\frac{1}{z-2}-\frac{1}{z-i}\Bigr]
$$
Recall that for $\lvert z\rvert < 1$, the geometric series $\sum_{n=0}^{\infty}z^n$ converges to $\frac{1}{1-z}$.
For $0<\lvert z\rvert < 1$, we have
\begin{align}
\frac{1}{z-2}&=\frac{-1}{2(1-z/2)}\\
&= -\frac{1}{2}\sum_{n=0}^{\infty}\Bigl(\frac{z}{2}\Bigr)^n\\
\frac{-1}{z-i}&= \frac{1}{i(1-z/i)}\\
&=-i\sum_{n=0}^{\infty}\Bigl(\frac{z}{i}\Bigr)^n\\
&=\sum_{n=0}^{\infty}(-i)^{n+1}z^n
\end{align}
Our series is then
$$
\frac{2+i}{5}\sum_{n=0}^{\infty}z^n\biggl[(-i)^{n+1}-\Bigl(\frac{1}{2}\Bigr)^{n+1}\biggr]
$$
For $1<\lvert z\rvert < 2$, we have $1/\lvert z\rvert < 1$ and $\lvert z\rvert/2<1$ which means we can use our first geometric series again.
\begin{align}
\frac{1}{z-2}&=\frac{-1}{2(1-z/2)}\\
&= -\frac{1}{2}\sum_{n=0}^{\infty}\Bigl(\frac{z}{2}\Bigr)^n\\
\frac{-1}{z-i}&= \frac{-1}{z(1-i/z)}\\
&=\frac{-1}{z}\sum_{n=0}^{\infty}\Bigl(\frac{i}{z}\Bigr)^n\\
&=\sum_{n=-\infty}^{0}(-i)^{n+2}z^{n-1}
\end{align}
Our series is then
$$
\sum_{n=-\infty}^{0}(-i)^{n+2}z^{n-1}-\frac{1}{2}\sum_{n=0}^{\infty}\Bigl(\frac{z}{2}\Bigr)^n
$$
where if we peel of the zero term of the first series, we have what is referred to as the principal part of the Laurent series. That is,
$$
\frac{1}{z}+\underbrace{\sum_{n=-\infty}^{-1}(-i)^{n+2}z^{n-1}}_{\text{principal part}}
$$
For the final region $1<2<\lvert z\rvert$, we have $1/\lvert z\rvert < 1$ and $2/\lvert z\rvert < 1$.
\begin{align}
\frac{1}{z-2}&=\frac{1}{z(1-2/z)}\\
&= \frac{1}{z}\sum_{n=0}^{\infty}\Bigl(\frac{2}{z}\Bigr)^n\\
&=\sum_{n=-\infty}^0\Bigl(\frac{1}{2}\Bigr)^nz^{n-1}\\
\frac{-1}{z-i}&= \frac{-1}{z(1-i/z)}\\
&=\frac{-1}{z}\sum_{n=0}^{\infty}\Bigl(\frac{i}{z}\Bigr)^n\\
&=\sum_{n=-\infty}^0(-i)^{n+2}z^{n-1}
\end{align}
Our series is then
$$
\sum_{n=-\infty}^0\biggl[\Bigl(\frac{1}{2}\Bigr)^n+(-i)^{n+2}\biggr]z^{n-1}
$$
A: Let's do it at $\;0<|z|<1\;$ , you do the other ones:
$$\frac1{z-i}=-\frac1i\frac1{1-\frac zi}=i\left(1+\frac zi-\frac{z^2}1+\ldots\right)$$
$$\frac1{z-2}=-\frac12\frac1{1-\frac z2}=\frac12\left(1+\frac z2+\frac{z^2}4+\ldots\right)$$
Now use your decomposition in partial fractions (but be careful with its sign!)
