Taylor series of $\frac{1}{\sin(z)}$ and $\frac{1}{\tan(z)}$

Question

What is Taylor Series for $\frac{1}{\sin(z)}$ and $\frac{1}{\tan(z)}$

Note that $z$ is complex, $z = x+iy$

Answer 1

If one is interested in an expansion at any order near $0$ then one may observe that the even function $(-\pi,\pi) \ni z \mapsto \dfrac {z}{\tan z}$ rewrites $$ \dfrac {z}{\tan z}=z\cdot \frac{\cos z}{\sin z}=iz\cdot\frac{e^{iz}+e^{-iz}}{e^{iz}-e^{-iz}}=iz\cdot\frac{e^{2iz}+1}{e^{2iz}-1}=iz+\frac{2iz}{e^{2iz}-1} \tag1$$ then using the classic generating function of the Bernoulli numbers $$ \sum_{k=0}^{\infty} B_k \frac{z^k}{k!} = \frac{z}{e^z-1}, \quad |z|<2\pi, $$ $B_0=1,\,B_1=-1/2, \,\ldots$ one gets the following Laurent series around $0$,

$$ \dfrac {1}{\tan z}= \frac1z+\sum_{n=1}^{\infty} B_{2n} \frac{(-1)^n 4^n z^{2n-1}}{(2n)!},\quad 0<|z|<\pi. \tag2 $$

Similarly, by writing $$ \dfrac {z}{\sin z}=\frac{2iz}{e^{iz}-e^{-iz}}=2iz\cdot\frac{e^{iz}}{e^{2iz}-1}=\frac{2iz}{e^{iz}+1}+\frac{2iz}{e^{2iz}-1} $$one gets the following Laurent series around $0$,

$$ \dfrac {1}{\sin z}= \frac1z+\sum_{n=1}^{\infty} B_{2n} \frac{(-1)^{n-1}2( 2^{2n-1}-1) z^{2n-1}}{(2n)!},\quad 0<|z|<\frac{\pi}2. \tag3 $$

Answer 2

I assume you mean centered at $z=0$. Functions that have a singularity at their center of expansion do not have a Taylor Series, but a Laurent Series.

For example, $$ \frac1{\sin(z)}=\frac1z+\frac{z}6+\frac{7x^3}{360}+\frac{31x^5}{15120}+\frac{127x^7}{604800}+\dots $$ and $$ \frac1{\tan(z)}=\frac1z-\frac{z}3-\frac{z^3}{45}-\frac{2z^5}{945}-\frac{z^7}{4725}-\dots $$