Why is $\dfrac{z(i-z)}{e^{2\pi z}-1}$ regular at $z=i$?

Question

$f(z):=\dfrac{z(i-z)}{e^{2\pi z}-1} (-1<\text{Im} z<2).$

If denominator is $0$, $e^{2\pi z}=1$.

Let $z=x+iy$ and I get $e^{2\pi x}e^{2\pi y i}=e^0,$ so $z=0,i$.

Therefore, $z=0,i $ are isolated singularities.

But my mathematics book says that $f(z)$ is regular at $z=0,i$.

I understood $f(z)$ is regular at $z=0$ because $f(z)=\dfrac{z(i-z)}{\sum_{k=0}^{\infty} \frac{z^k}{k !}-1} =\dfrac{z(i-z)}{\sum_{k=1}^{\infty} \frac{z^{k}}{k !}}=\dfrac{i-z}{\sum_{k=0}^{\infty} \frac{z^k}{(k+1)!}}$ so $z=0$ is not isolated singularity.

But I cannot understand why $f(z)$ is regular at $z=i$.

I would like you to give me some ideas.

Answer 1

Note that\begin{align}\lim_{z\to i}\frac{e^{2\pi z}-1}{z-i}&=2\pi e^{2\pi i}\\&=-2\pi\end{align}and that therefore\begin{align}\lim_{z\to i}\frac{z(i-z)}{e^{2\pi z}-1}&=-\lim_{z\to i}\frac z{\frac{e^{2\pi z}-1}{z-i}}\\&=\frac i{2\pi}.\end{align}Therefore, yes, $i$ is a regular point (or removable singularity) of $f$.

Answer 2

Taylor series for the numerator at $z=i$ is $$ -i(z-i) - (z-i)^2 $$ Taylor series for the denominator at $z=i$ is $$ 2\pi(z-i)+\dots $$ So the Laurent series for the quotient is $$ \frac{-i}{2\pi} + \dots $$ Therefore the function is regular at the point.

Answer 3

HINT-For a quotient function $f(z)=\frac{p(z)}{(q(z)}$, the zeros $\alpha_i$ of the denominator $q(z)$ are the poles of $f(z)$ iff $p(\alpha_i)\ne 0$ and analytic at $\alpha_i$. Your $p(z)$ doesn't meet the criteria of being non-zero at both $\alpha_i$. Infact you observe that

$\lim_{z\rightarrow i}f(z)$ exists so it is a removable singularity of $f(z)$.