Determine all isolated singularities and their type: $$g(z)=\frac{\sin(z)}{1-\tan(z)}$$
I want to do a Laurent series expansion for this function. So I got this: $$g(z)=\frac{z-\frac{1}{3!}z^3+\frac{1}{5!}z^5±...}{1-\tan(z)}$$
I'm not sure, how I can go further with this $\tan(z)$. The denominator is 0 for $z=\frac{\pi}{4}+\pi n, n\in\mathbb{N}$. So there are multiple singularities. Or is this the wrong idea?
The isolated singularities of $g(z)=\frac{\sin(z)}{1-\tan(z)}$ occur at the zeroes of $1-\tan(z)$, which lead to simple poles of $g$.
Note that there are also singularities at the zeroes of $\cos(z)$ (i.e., $z=\frac{\pi}{2}+n\pi$) since $\tan(z)$ is undefined there, but these are removable since we may define $g$ as $0$. And once removed, $g$ is analytic in a neighborhood of $z=\frac{\pi}{2}+n\pi$.
As you correctly identified, the poles of $g$ are located at values of $z$ such that $\tan(z)=1$. These values are roots of $e^{i2z}=i$ which implies that
$$z=\frac\pi4+n\pi$$
for all integers $n$.
We can construct Laurent series in each annulus for which $\frac\pi4+n\pi<|z|<\frac\pi4+(n+1)\pi$. Note that when $n=0$, the Laurent series is a Taylor series.