I have to find the type of singularities and their residues of $f(z)=\frac{z}{\sin(z)}$.
I know $z=0$ is a removable singularity, so residue is zero. The other singularities are at $z=k\pi$, and I wanted to calculate the residues by limit:
$$\lim_{z\to k\pi} (z-k\pi)f(z) = \lim_{z\to k\pi} z \lim_{z\to k\pi} \frac{z-k\pi}{\sin(z)} = k\pi \lim_{z\to k\pi} \frac{1}{\cos(z)} = k\pi (-1)^k$$
Which is wrong since the answer is just $(-1)^k$ according to the teacher. So my question is where do I go wrong? 🙂
Thanks in advance.
Best Answer
You result is true. In $z=k\pi$ the residues by l'hospital rule are $$\lim_{z\to k\pi} (z-k\pi)f(z) = \lim_{z\to k\pi} \frac{z^2-k\pi z}{\sin z} = \lim_{z\to k\pi} \frac{2z-k\pi}{\cos z} = (-1)^k k\pi$$ for $k\in\mathbb{Z}$.