Determining a Laurent series with trigonometric functions

Question

I could use some help with the Laurent series around $z_{0}=0$ for all $z\in\mathbb{C}\backslash\{n\cdot\pi;\;n\in\mathbb{N}\} $ of the following function: $$f(z)=\frac{e^{sin(z)}-cos(z)-z}{sin^{2}(z)} $$ In particular, I'm having problems trying to figure out the coefficients of the Laurent-Series and wether or not I can just use the respective Taylor or Fourier series.

Answer 1

Let’s expand near $0$

$$\begin{align} \sin{z}&=z-{z^3\over 6}+\cdots +(-1)^k{z^{2k+1}\over(2k+1)!}+\cdots\\ \cos{z}&=1-{z^2\over 2}+\cdots +(-1)^k{z^{2k}\over (2k)!}+\cdots \end{align}$$

I cannot see a general expression for the general term but assume we want to expand to order $2$, one has

$$\begin{align} \sin^2{z}&=\left(z-{z^3\over 6}\right)^2+o(z^4)=z^2-{z^4\over 3}+o(z^4)\\ e^{\sin{z}}&=1+ \left(z-{z^3\over 6}\right)+ {1\over 2}\left(z-{z^3\over 6}\right)^2+o(z^4)\\ &=1+z+{z^2\over 2}-{z^3\over 6}-{z^4\over 6}+o(z^4)\\ \cos{z}&=1-{z^2\over 2}+{z^4\over 24}+o(z^4) \end{align}$$

Putting all the pieces together

$${e^{\sin{z}}-\cos{z}-z\over \sin^2{z}}={z^2-{z^3\over 6}-{5z^4\over 24}+o(z^4)\over z^2-{z^4\over 3}+o(z^4)}={1-{z\over 6}-{5z^2\over 24}+o(z^2)\over 1-{z^2\over 3}+o(z^2)}$$