Find the Laurent series of the function $$f(z) =\frac1{(z−1)(z−2)}$$ in each of the following domains:
(a) $\{z ∈C : 1 < |z| < 2\} $
(b) $\{z ∈C : 0 < |z−1| < 1\}$
I am very new to this but this is what I did for (a):
You can rewrite the function as $$f(z)=-\frac1{2-z}+\frac1{1-z}=-\frac12\cdot\frac1{1-z/2}+\frac1{1-z}=\frac12 \sum \limits_{n=0}^{\infty} (-z/2)^n+\sum \limits_{n=0}^{\infty} z^n $$
$$=\sum \limits_{n=0}^{\infty} (-z)^n/2^{n+1}+\sum \limits_{n=0}^{\infty} z^n=\sum \limits_{n=0}^{\infty}\bigg(z^n-\frac{(-z)^n}{2^{n+1}} \bigg)$$
Is this correct? I will attempt (b) if it is. Please advise me.
Your series for $-1/(2-z)$ is incorrect. We are given that $\lvert z\rvert < 2\iff \lvert z\rvert/2<1$ so $$ \frac{-1/2}{1 - z/2} = -\frac{1}{2}\sum_{n=0}^{\infty}\Bigl(\frac{z}{2}\Bigr)^n $$ Therefore, you should have obtained for part $(a)$ $$ \sum_{n=0}^{\infty}\biggl[z^n-\frac{1}{2}\Bigl(\frac{z}{2}\Bigr)^n\biggr] $$ You picked up an additional minus sign. For part $(b)$, write \begin{gather} \frac{-1}{2-z} = \frac{-1}{1+1-z}=\frac{-1}{1 - (z-1)}=\mbox{what geometric series?}\tag{1}\\ \frac{1}{1-z}+\text{geometric series obtained from }(1) \end{gather}