Finding Laurent of $ \frac{e^{1/z}}{1-z} $

Question

The question is: Determine the Laurent expansion in the region $0<|z|<1$? $$ \frac{e^{1/z}}{1-z} $$ Here is how i have tried to solve it but my answer is not correct and i don't know why? $$e^{1/z}=\sum_{n=1}^{\infty}\frac{1}{z^kk!}=\frac{1}{z}+\frac{1}{z^22!}+\frac{1}{z^33!}+... $$ $$\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n=1+z+z^2+...$$ $$\left(\frac{1}{z}+\frac{1}{z^22!}+\frac{1}{z^33!}+...\right)\left(1+z+z^2+...\right) $$ Any suggestion would be great, thanks

Answer 1

Hint

The Laurent series centered on zero has the form

$$\sum_{n=-\infty}^\infty a_n z^n.$$

Hence you need to use Cauchy product to write

$$\left(1 +\frac{1}{z}+\frac{1}{z^22!}+\frac{1}{z^33!}+...\right)\left(1+z+z^2+...\right) $$

with the appropriate form.

Alternatively, use the fact that

$$a_n =\frac{1}{2i \pi} \oint_\gamma \frac{f(z) dz}{z^{n+1}}$$ where $\gamma$ is the counterclockwise circle centered on $0$ with radius $1/2$.