$\displaystyle f(z) = \frac{\sin(z)}{1-\cos(\sqrt{z}\,)}$.
The assignment is to find all the singularities of $f$, determine the type of them and the residue.
It is clear that the singularities are $z_k = (2k\pi)^2$ with $k \in \mathbb{Z}$. I'm not allowed to use l'Hôpital so I could try to develop a Laurent series or using the Taylor series of $\cos(\sqrt{z}\,) = \sum\limits_{n = 0}^\infty \frac{(-1)^n }{(2n)!}z^n$ and $\sin(z) = \sum\limits_{n = 0}^\infty \frac{(-1)^n }{(2n+1)!}z^{2n+1}$ but I can't of a way to determine the solution.
You can use the fact that $$ \lim_{z\to0}\frac{1-\cos(\sqrt{z}\,)}{z}=\frac{1}{2} $$ which follows from the Taylor series, so you have $$ \lim_{z\to0}\frac{\sin z}{1-\cos\sqrt{z}}= \lim_{z\to0}\frac{\sin z}{z}\frac{z}{1-\cos\sqrt{z}}=2 $$ and $0$ is a removable singularity.
Let's look at $4k^2\pi^2$, where $k\ne0$. Since $\sin(4k^2\pi^2)\ne0$, because $\pi$ is irrational, we know we have a singularity. However, $$ \lim_{z\to4k^2\pi^2}\frac{(z-4k^2\pi^2)^2}{1-\cos\sqrt{z}}= \lim_{w\to2k\pi}\frac{(w^2-4k^2\pi^2)^2}{1-\cos w}= \lim_{u\to0}\frac{u^2(u+4k\pi)^2}{1-\cos u}=32k^2\pi^2 $$ (by taking a suitable branch for the square root and doing $w-2k\pi=u$).
Thus $4k^2\pi^2$ is a pole of order $2$.