# Laurent series for $f(z) = \frac{1} {z} + \frac{1} {(1-z)} + \frac{1} {(2-z)}$ around arround $0< |z|<1, 0< |z-1|<1$ and $0< |z|<2$?

Find the Laurent series of the function

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

a)$$\{z \in\Bbb C: 0<|z|<1\}$$ b) $$\{z\in\Bbb C:0<|z-1|<1\}$$ c) $$\{z\in\Bbb C::0<|z-2|<1\}$$

The exercise consists of three lines but my main doubt is to make the series of laurent.

I do: $$\sum \limits_{n=0}^{\infty} z^n +\frac12 \sum \limits_{n=0}^{\infty} (z)^n + \frac12 \sum \limits_{n=0}^{\infty} (z/2)^n$$

is it well done? Can you help me please?

Not, it is not correct. If $$|z|<1$$, then$$\frac1{1-z}=\sum_{n=0}^\infty z^n$$and$$\frac1{2-z}=\sum_{n=0}^\infty\frac{z^n}{2^{n+1}}.$$So, the Laurent series of your function at $$\{z\in\Bbb C\mid0<|z|<1\}$$ is$$\frac1z+\sum_{n=0}^\infty\left(1+\frac1{2^{n+1}}\right)z^n.$$
• No. The answer is$$\frac{-1}{z-1}+\sum_{n=0}^\infty2(z-1)^{2n}.$$ – José Carlos Santos May 18 at 22:59