Laurent Series of $\frac{1}{(z-1)(z-2)}$ for $|z-1|>1$

Question

I am calculating Laurent series of $$ f(z)=\frac{1}{(z-1)(z-2)} $$ which converges when $|z-1|>1$.

I started as $$ \frac{1}{(z-1)(z-2)}=\frac{1}{z-1}\cdot\frac{1}{(z-1)-1}=\frac{1}{(z-1)^2-(z-1)}. $$ Is this even the right way to transform function to some kind of geometric series, which can be easily expandet as Taylor series? How to calculate this Laurent seires?

Answer 1

You were doing it well:

$\frac{1}{(z-1)-1}=\frac{-1}{1-(z-1)}=\sum_{n=-\infty}^{-1}{(z-1)^n}=\frac{1}{z-1}+\frac{1}{(z-1)^2}+...$

Finally just multiply the series above by $\frac{1}{z-1}$

Answer 2

You are on the right track. Now note that

$$\frac1{(z-1)(z-2)}=\left(\frac1{(z-1)^2}\right)\left(\frac1{1-\frac1{z-1}}\right)$$

Now, expand the second term in a Taylor series of powers of $\frac1{z-1}$ (It's a simple geometric series.).

Can you finish now?

Answer 3

$$f(z) = \frac{1}{(z-1)(z-2)}$$

We write $f(z)$ in its partial fraction expansion, and then we expand $f(z)$ in powers of $\frac{1}{z-1}$:

$$\begin{aligned} f(z) &= \frac{1}{z-2}-\frac{1}{z-1} \\ &= \frac{1}{(z-1)-1} - \frac{1}{z-1} \\ &= \frac{1}{z-1} \frac{1}{1-\frac{1}{z-1}}- \frac{1}{z-1}\\ &= \frac{1}{z-1} \left( \frac{1}{z-1} + \frac{1}{(z-1)^2} + \cdots \right) \\ &= \frac{1}{(z-1)^2} + \frac{1}{(z-1)^3} + \cdots \end{aligned}$$ This is the Laurent series for $f(z)$ in the region $|z-1|>1.$