Finding residue through series method

Question

I was trying to find the residue of the function.

$$\frac{1}{e^{e^{\frac{1}{z}}}}$$

Any clue/help would be appreciated.

The answer is $-1/e.$

Thanks

Answer 1

Hint: You may write $$e^{-e^{1/z}} = e^{-1-\frac{1}{z}-\frac{1}{2!z^2} +...} = e^{-1} e^{-1/z} e^{-1/2z^2} ... = e^{-1} \left( 1 - \frac{1}{z} + \frac{1}{2!z^2} -...\right) \left(1-\frac{1}{2z^2}+\frac{1}{2! (2z^2)^2} ...\right) ...$$ You find only one term contributing to the residue.

Answer 2

Since $$ 1-e^{1/z}=-\frac1z+O\!\left(\frac1{z^2}\right) $$ we have $$ e^{1-e^{1/z}}=1-\frac1z+O\!\left(\frac1{z^2}\right) $$ Therefore, $$ \begin{align} e^{-e^{1/z}} &=e^{-1}\cdot e^{1-e^{1/z}}\\ &=e^{-1}\left(1\color{#C00000}{-\frac1z}+O\!\left(\frac1{z^2}\right)\right) \end{align} $$ Thus, we get the residue to be $-e^{-1}$.