# Residue, singularity and range where function is holomorphic?

Find residue, singularity (type) and (maximal) range where the given function is holomorphic?

$f(z)=\frac{z}{sin z}$

Function has singularity in $z_0=0$ (removable). But how do I find residue and range?

I tried to find residue with this formula but can't finish.

$Res(f;0)=\frac{1}{(m-1)!}\lim_{z \to 0} \frac{d^{m-1}}{dz^{m-1}}(z-z_0)f(z) =\lim_{z \to 0} \frac{z^2}{sin z}$

• $z$ is analytic everywhere and $\sin(z)$ is analytic and non-zero on $\mathbb{C}\setminus \pi \mathbb{Z}$ so $\frac{z}{\sin(z)}$ is analytic on $\mathbb{C}\setminus \pi \mathbb{Z}$. Also $z$ has only one zero at $z=0$ cancelling the zero of $\sin(z)$ so that $\frac{z}{\sin(z)}$ is analytic at $0$. Altogether means that $\frac{z}{\sin(z)}$ is analytic on $\mathbb{C}\setminus \pi \mathbb{Z}^*$ – reuns Apr 12 '17 at 16:27

That is not the only singularity, there are others at the real multiples of $\pi$. To find the residue is the same for all the poles, LHopitals rule holds for complex limits as well, so use that.
For example, to compute the limit you found, $$\lim_{z\to 0} \frac{z^2}{\sin(z)}=\lim_{z\to 0} \frac{2z}{\cos(z)}=0$$
• I have one more question, how do I calculate limit if singularity is $z_0=n$? – Ana Matijanovic Apr 12 '17 at 16:25
• @AnaMatijanovic something is holomorphic when it's derivative exists, and you can take the derivative of the quotient when it is not divided by zero, so whenever $\sin(z)\neq 0$ it is holomorphic. – user416426 Apr 12 '17 at 16:40