Evaluation of the type of singularity and Residue at that point

Question

$$ \begin{align} f(z)=z^2e^{\frac{1}{z^3}} \end{align} $$

I need to determine the type of singularity and evaluate the Residue at $z_0=0$

We know that $e^{\frac{1}{z}}$ has an essential singularity at $z_0=0$ , but is it enough in order to claim that $f$ will also have ( I think ) essential singularity at the same point. Do I have to try to prove this by taking 2 diferrent sequences $\alpha_n$ and $\beta_n$ for which $\alpha_n\to0 $ and $\beta_n\to0$ , but $f(\alpha_n) \ne f(\beta_n)$ ?

I think I can I evaluate the residue at $0$ by writing $f$ as a series expansion(around $0$) and taking the coefficient of $z^{-1}$ or does the nature of the singularity affect this? Is there also another way to calculate it?

I am a bit confused as to what procedure we follow to determine these. Is there a general approach to address such issues? I 'd welcome any advice or guidance.

Writing f as a series $$f(z)=z^2e^{\frac{1}{z^3}}=\sum_{n=1}^\infty\frac{1}{n!}z^{-3n+2}= z^{-1}+ \frac{z^{-4}}{2}+ ..,0<|z|<1$$

Hence , $Res(f,0)=1$ would be correct or the fact that the singularity is essential tells us that $Res(f,0)$ doesn't exist?

Answer 1

There isn't any general way to solve all problems of this sort. However, if the given function is easy to write as a power series, then this is probably the best approach.

In this case, as I think you noticed, it is pretty easy to write $f$ as a power series, and what you suggested is indeed true - the coefficient of $z^{-1}$ is the residue. The type of the singularity is determined by the least power with nonzero coefficient in the series, if there is such a power. If there isn't it means the singularity is essential.

Answer 2

Multiplying by a power of $z$ can never get rid of an essential singularity. Nor can composition with an integer power (which is what you're doing in going from $\exp(1/z)$ to $\exp(1/z^3)$). So you certainly do have an essential singularity.

As others said, the Laurent series is the way to go to find the residue: fortunately it is easy to find using the Maclaurin series of $\exp$.