I have to calculate the residue of $\exp\left(\frac{z+1}{z-1}\right)$ in every point of $\mathbb{C}$.

So I tried to compute the Laurent Series expansion $\forall z_0 \in \mathbb{C}$.

For $z_0 = 0$ we obtain that $f(z)=\sum_{k \geq 0}\frac{(z+1)^k}{(z-1)^k}$ but I dont understand what the coefficient $a_{-1}$ is.

Thanks in advance.


1 Answer 1


The only singularity of $f$ is at $z=1$, but that's an essential singularity. I get $$f(z)=\sum_{n=0}^\infty\frac{(z+1)^n}{n!(z-1)^n}$$ but that's not a Laurent series as it stands. But also $$f(z)=\exp\left(1+\frac{2}{z-1}\right)=e\exp\left(\frac{2}{z-1}\right) =e\sum_{n=0}^\infty\frac{2^n}{n!(z-1)^n}.$$ Now that is a Laurent series at $z=1$, and the coefficient of $(z-1)^{-1}$ is $2e$.


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