Laurent series expansion of given function

Question

How to find the Laurent series for |z|>1 for $$\dfrac{e^{1/z}}{z^2-1} ?$$ Firstly, how to decide if given function has laurent series expansion in specified domain or not? I just did the long division .but I am not sure if that is the expected answer or not. Please help

Answer 1

$$|z|>1\implies\frac1{|z|}<1\implies\frac{e^{1/z}}{z^2-1}=\frac1{z^2}\cdot e^{1/z}\cdot\frac1{1-\frac1{z^2}}=$$

$$=\frac1{z^2}\left(1+\frac1z+\frac1{2!z^2}+\frac1{3!z^3}+\ldots\right)\left(1+\frac1{z^2}+\frac1{z^4}+\ldots\right)=$$

$$\frac1{z^2}+\frac1{z^3}+\frac3{2z^4}+\frac4{3z^5}+\ldots$$

Answer 2

Let $z=1/u$ to get

$$\frac{e^uu^2}{1-u^2}$$

since we want $|z|>1$, then we want $|u|<1$, that is, the Laurent series at $u=0$. This is very easy, just notice that:

$$e^u=1+u+\frac12u^2+\frac16u^3+\dots$$

$$\frac1{1-u^2}=1+u^2+u^4+u^6+\dots$$

and multiply it all together, then let $u=1/z$ to turn it back in terms of $z$.