Laurent expansion with different annuli

Question

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

1. $0 \lt \lvert z \rvert \lt 1 $,
2. $ 1 \lt \lvert z \rvert \lt 2$,
3. $ 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

Answer 1

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!)

Answer 2

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 $$ \sum_{n=-\infty}^0\biggl[\Bigl(\frac{1}{2}\Bigr)^n+(-i)^{n+2}\biggr]z^{n-1} $$