Computing the logarithmic residue of $\sin(z)/z$

Question

Compute the logarithmic residue of the function $f(z)=\frac{\sin(z)}{z}$ at all poles and zeros.

I know by a proposition that $\text{res}_{z=z_k}\frac{d\ln(f(z))}{dz}$ is equal to the sum of the order of each zero minus the order of each pole. $\frac{\sin(z)}{z}=0\implies z=k\pi\:,\:k\in\mathbb{N}$.

However I have no idea on how to compute the order of zero $\frac{\pi}{2}+k\pi$.

Questions:

Is it right what I have done insofar? How should I solve the problem?

Thanks in advance!

Answer 1

To compute the order of the zero of $\sin(z)/z$ at $\pi$, look at the Taylor series for $\sin(z)$ around $\pi$ (because $1/z$ has no zero or pole at $\pi$); \begin{align} \sin(z)=-\sin(z-\pi)&=-\left[(z-\pi)-\frac{1}{3!}(z-\pi)^3+\dots\right]\\ &=-(z-\pi)\left[1-\frac{1}{3!}(z-\pi)^2+\dots\right], \end{align} and read the order of the zero at $\pi$ to be 1.