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**arbolis** Oh, that would be a counter example?

I realize that $\displaystyle e^{1/z}=\sum _{n=0}^{+\infty} \frac{1}{z^n n!}=1+ \sum _{n=1}^{+\infty} \frac{1}{z^n n!}$.

But I'm not sure how to reach $\displaystyle e^{1/z}-\frac1z = 1+\sum_{n=2}^\infty \frac{1}{n!z^n}$.

Also, $\displaystyle Res(e^{1/z},z=0)=1$ while $\displaystyle Res(-1/z,z=0)=-1$ so indeed $\displaystyle Res(f,z=0)=0$. So the singularity of $\displaystyle e^{1/z}$ is essential (there's an infinity of negative powers of z in the series expansion of the function) and so should be the one of f... So that after all, z isn't bounded nor analytic in a neighborhood of $\displaystyle z=0$?

Should I use Cauchy-Riemann equations to show that f isn't analytic? How would you do it?