Find the singulartiy of $\frac{1}{(\pi + z)\sin(z)}-\frac{1}{\pi z}$

Question

Find the singulartiy of $\frac{1}{(\pi + z)\sin(z)}-\frac{1}{\pi z}$.

I'm trying to use the strategy shown in the following question (2nd answer):

Locate and classify the singularity of $\frac{e^{z-1}-1}{z^4-1}$

So my singularities are: $-\pi, 0, \pi k$ for $k \in \mathbb{Z}$. If I rewrite the function as follows:

$\frac{\pi z - (\pi+z)\sin(z)}{(\pi + z)\sin(z)\pi z}$. I do see that $0$ is a zero of the function in the numerator. But how can I derive the multiplicity of 0?

Answer 1

Near $0$, you have$$\pi z-(\pi+z)\sin(z)=-z^2+\frac\pi6z^3+\cdots$$and$$(\pi+z)\sin(z)\pi z=\pi^2z^2+\pi z^3+\cdots$$and therefore$$\frac{\pi z-(\pi+z)\sin(z)}{(\pi+z)\sin(z)\pi z}=\frac{-z^2+\frac\pi6z^3+\cdots}{\pi^2z^2+\pi z^3+\cdots}=\frac{-1+\frac\pi6z+\cdots}{\pi^2+\pi z+\cdots}.$$So, $0$ is a removable singularity, since, near $0$, you can express your function as a quotient of two analytic functions such that $0$ is not a zero of the denominator.

Near $-\pi$ you have$$\pi z-(\pi+z)\sin(z)=-\pi^2+\pi(z+\pi)+\cdots$$and$$(\pi+z)\sin(z)\pi z=\pi^2(z+\pi)^2-\pi(z+\pi)^3+\cdots$$Since $0$ is not a zero of the numerator and it is a double zero of the denominator, it is a double pole of the quotient.

Answer 2

Note that$$\lim_{z\to 0}(z-0)\cdot\left(\frac{1}{(\pi + z)\sin(z)}-\frac{1}{\pi z}\right)=0,$$ so there is no pole, but a removable singularity. That's the Riemann's principle about removable singularities.