# Isolated Singularity Behavior

Tasked with locating each of the isolated singularities of this function , telling whether it is removable, pole or an essential and if it's removable giving the value at the singularity and if it's a pole the order.

The function: $$\frac{z^2}{sin(z)}$$

So to check what kind of singularity it is, my understanding is we evaluate the limit.

Clearly there are singularities at $$0$$ and $$k\pi$$ for $$k$$ integer.

How do I evaluate the limit? Do I use L'Hospital's Rule?

Taking the derivative of the numerator and denominator gives

$$\frac{2z}{cos(z)}$$

which evaluating at 0 gives us 0. Does that mean it's a removable singularity at $$z_0 = 0$$ with a value of $$0$$? What do I do for all of the singularities at integer multiples of $$\pi$$?

• correct for $0$, simple poles with residues that are easy to compute at $k\pi, k \ne 0$ – Conrad Apr 24 '19 at 17:31

You do not need L'Hospital's rule. All you have to know that the zeroes $$z_k = k\pi$$ of $$\sin z$$ have order $$1$$ because $$\sin' z_k = \cos z_k \ne 0$$. This means that $$\sin z = (z - z_k) \cdot f_k(z)$$ with a holomorphic function $$f_k : \mathbb C \to \mathbb C$$ such that $$f_k(z_k) \ne 0$$.
This show that $$\lim_{z \to 0} \dfrac{z^2}{\sin z} = \lim_{z \to 0} \dfrac{z}{f_0(z)} = 0$$. Thus you have a removable singularity at $$0$$.
For $$k \ne 0$$ you have $$\dfrac{z^2}{\sin z} = \dfrac{z^2}{f_k(z)} \cdot \dfrac{1}{z - z_k}$$ which shows that there is pole of order $$1$$ at $$z_k$$.