What is the class of the singularity $5$ for $\exp(\frac{z}{z-5})$

Question

$$\exp(\frac{z}{z-5})$$ Has a singularity at $5$. The power series representation is $$\sum_{k=0}^\infty z^k(z-5)^{-k}\frac{1}{k!}$$ How can we classify the singularity at $5$? I don't know how to deal with $z^k$. I don't see a way to pull it out or combine it with $(z-5)^{-k}$.

As the exponent has a pole of order $1$ at $5$, I think the result should be that the singularity is essential. But that is only a feeling, I don't know how to check/prove it.

What is the class of the singularity $5$ and how can it be shown?

Answer 1

Since\begin{align}\exp\left(\frac z{z-5}\right)&=\exp\left(1+\frac5{z-5}\right)\\&=e\exp\left(\frac5{z-5}\right)\\&=e\sum_{k=-\infty}^0\frac{5^{-k}}{(-k)!(z-5)^k}\end{align}and therefore $5$ is an essential singularity of $\exp\left(\frac z{z-5}\right)$.

Answer 2

I'm sure that there's a simpler way to do this, but note that you can find different sequences - $z_k \rightarrow 5$ and $z_n \rightarrow 5$, such that $f(z_k)\rightarrow 1$ and $f(z_n)\rightarrow 2$.

You can thus conclude that the singularity is essential.

Answer 3

You can write $$\exp\left(\frac z{z-5}\right)=\exp\left(1+\frac 5{z-5}\right)\\ =e\exp\left(\frac 5{z-5}\right)$$ and expand the last. You will not have the $z^k$ to contend with.

Answer 4

Testing an essential singularity is rather simple without power series. An isolated singularity of $f$ at $z_0$ is essential iff there are two sequences $(x_n),(y_n)$ with $\lim x_n =z_0=\lim y_n$ and $\lim f(x_n) \neq \lim f(y_n)$ - including $\infty$ as limit. In case of $\exp$-function follow the real line to $\pm \infty$.