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$$\frac1z+\sum_{n=0}^\infty\left(1+\frac1{2^{n+1}}\right)z^n.$$

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  • $\begingroup$ No. The answer is$$\frac{-1}{z-1}+\sum_{n=0}^\infty2(z-1)^{2n}.$$ $\endgroup$ – José Carlos Santos May 18 at 22:59

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