# Calculate the residue of $\exp\left(\frac{z+1}{z-1}\right)$ in every point of $\mathbb{C}$

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.

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 $$2e$$.