# Show that singularities of a specific complex function are poles.

Consider the function $$g:\mathbb{C}\longrightarrow\mathbb{C}$$ defined by $$g(z):=\dfrac{1}{2}\Bigg(\dfrac{1}{1-e^{z(i-1)}}+\dfrac{1}{1-e^{z(-1-i)}}\Bigg)=\dfrac{1}{2}\Bigg(\dfrac{1-e^{z(-1-i)}+1-e^{z(i-1)}}{(1-e^{z(i-1)})(1-e^{z(-1-i)})}\Bigg).$$

It has singularities at $$z_{k}=(1-i)k\pi\ \ \text{and}\ \ \omega_{k}=(1+i)k\pi\ \ \text{for}\ \ k\in\mathbb{Z}.$$

I want to classify these singularities. What I can tell so far is the following:

1. Note that for $$k=0$$, we have a double zeros in the denominator, but we have a simple zero in the numerator, so even though we have double zeros, the singularity for $$k=0$$, is still of order 1. And clearly there is no way to remove it.

2. For all other $$k$$, we can only have one component of the product in the denominator to be $$0$$ at one time, and $$z_{k}$$ and $$\omega_{k}$$ for each $$k$$, cannot individually make the numerator $$0$$, and thus for all other $$z_{k}$$ and $$\omega_{k}$$, they are simple singularities that cannot be removed.

I want to say more than these. If they are not removable, are they poles or essential singularities? I can prove neither.

So I expect that they are poles, but I am having a hard to to write $$g(z)$$ in the form of $$g(z)=(z-z_{k})^{-1}h(z)$$ for some $$h(z)$$ holomorphic non-vanishing in a neighborhood of $$z_{k}$$. This is difficult since we don't have things like $$z^{4}+1$$ so that we can factorize.

For $$z=0$$, for sure I can use power series expansion of $$e^{z(i-1)}$$ and $$e^{z(-1-i)}$$ around $$0$$, and conclude what I want. For all other $$z_{k}$$, it seems really time-consuming to do the power series around them.

I don't have any idea about how to prove them are essential either..(I don't believe they are essential actually)

Any idea? Thank you so much!

• Try evaluating $\lim_{z \to z_k} (z - z_k) g(z)$ using L'Hospital's rule (and the same thing for $w_k$). Since this limit exists, the $z_k$ and $w_k$ are all simple poles. (If they were poles of a higher order, you would need to consider $\lim_{z \to z_k} (z - z_k)^n g(z)$.) – sasquires Dec 9 '20 at 2:45
• @sasquires I am having a little trouble to get a solid reference of identifying simple poles. Someone says I should prove that $$g(z)=(z-z_{k})^{-k}h(z)$$ for some non-vanishing holomorphic function $h$ around $z_{k}$, and I also saw your version that I can just show that $(z-z_{k})^{n}$ can remove the singularity in $g(z)$. They are equivalent? – JacobsonRadical Dec 9 '20 at 2:53
• @sasquires yeah I guess they are the same, removable singularity is basically a pole of order $0$. Okay, thank you for the help. I will answer my own post. – JacobsonRadical Dec 9 '20 at 3:01

## 1 Answer

For $$z_{k}\neq 0$$, we use L'Hopital's Rule to get \begin{align*} \lim_{z\rightarrow z_{k}}(z-z_{k})g(z)&=\dfrac{1}{2}\lim_{z\rightarrow z_{k}}\dfrac{(z-z_{k})(2-e^{z(-1-i)}-e^{z(i-1)})}{(1-e^{z(i-1)})(1-e^{z(-1-i)})}\\ &=\dfrac{1}{2}\lim_{z\rightarrow z_{k}}\dfrac{(2-e^{z(-1-i)}-e^{z(i-1)})+(z-z_{k})((1+i)e^{z(-1-i)}-(i-1)e^{z(i-1)})}{-(i-1)e^{z(i-1)}(1-e^{z(-1-i)})+(1-e^{z(i-1)})(1+i)}\\ &=\dfrac{1}{2}\dfrac{2-e^{z_{k}(-1-i)}}{(1+i)}<\infty. \end{align*}

A similary computation shows that $$\lim_{z\rightarrow\omega_{k}}(z-z_{k})g(z)<\infty,$$ so all $$z_{k},\omega_{k}\neq 0$$ are simple poles.

A similar computation shows that $$\lim_{z\rightarrow z_{k}}(z-z_{k})g(z)=\dfrac{1}{(1+i)},$$ similar for $$\omega_{k}$$.

So all the $$z_{k}$$ and $$\omega_{k}$$ are simple poles.