The Laurent Series of $\dfrac{e^z}{z^2-1}$
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At $z=1$

As we seek for powers of $z-1$, note that:

$$e^z=e\cdot e^{z-1}=e(1+(z-1)+\dfrac{(z-1)^2}{2!}+\dfrac{(z-1)^3}{3!}+...)$$

So: $f(z)=\dfrac{ee^{z-1}}{(1+(z-1))(z-1)}=e(\sum\_{n=0}^{\infty}{(-1)^n(z-1)^n})(\dfrac{1}{z-1}+1+\dfrac{(z-1)}{2!}+\dfrac{(z-1)^2}{3!}+...)$

But I don't know what to do, becasue of the series multiplication, or I should have done it other way?