Laurent Series of $1/\tan z$ How can we find the Laurent series of the function $$f(z)=\frac{1}{\tan z }$$ around 0.
Thank you very much.
 A: I read your comment just now (looking for the general form), but I'll leave the answer below: a way to get the series term by term.

Well there' a formula for the coefficients (see wiki), but I suppose you're looking for another way to find the series expansion of $\cot z$.
Dividing the series for $\sin z$ term by term by $z$ gives:
$$\frac{\sin z}{z} = 1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots$$
Now after recognizing the sum of a geometric series, you get the expansion of $\tfrac{z}{\sin z}$:
$$\begin{array}{rl}
\displaystyle \frac{z}{\sin z} 
& \displaystyle  = \frac{1}{1-\left( \frac{z^2}{3!}-\frac{z^4}{5!}-\cdots \right)} \\[8pt]
& \displaystyle  = 1 +  \left( \frac{z^2}{3!}-\frac{z^4}{5!}-\cdots \right) + \left( \frac{z^2}{3!}-\frac{z^4}{5!}-\cdots \right)^2 + \cdots \\[8pt]
& \displaystyle  = 1 + \frac{1}{6}z^2 + \frac{7}{360}z^4 + \cdots
\end{array}$$
Now you can multiply this series with the one for $\cos z$:
$$\cos z = 1-\frac{z^2}{2!}+\frac{z^4}{4!}-\cdots$$
To get:
$$\begin{array}{rl}
\displaystyle \frac{z \cos z}{\sin z}
& \displaystyle  = \left( 1-\frac{z^2}{2}+\frac{z^4}{24}-\cdots\right)\left(  1 + \frac{1}{6}z^2 + \frac{7}{360}z^4 + \cdots\right)  \\[8pt]
& \displaystyle  = 1-\frac{z^2}{3}-\frac{z^4}{45}-\frac{2z^6}{945}+ \cdots
\end{array}$$
And now you only have to divide by $z$ to get:
$$\cot z = \frac{1}{z}-\frac{z}{3}-\frac{z^3}{45}-\frac{2z^5}{945}+ \cdots$$
A: If you need the general form you may want to use the exponential form of the function:
$$f(z)=\cot z=\frac{\cos z}{\sin z}=i\frac{e^{2iz}+1}{e^{2iz}-1}=i\frac{1+\sum\limits_{n=0}^\infty\frac{2^ni^nz^n}{n!}}{\sum\limits_{n=1}^\infty\frac{2^ni^nz^n}{n!}}$$
Now you can find the coefficients by equating to power series around zero and comparing the coefficients of corresponding powers of $\;z\;$ :
$$i+\sum_{n=0}^\infty\frac{i^{n+1}2^nz^n}{n!}=\sum_{n=1}^\infty\frac{2^ni^nz^n}{n!}\cdot\sum_{j=-k}^\infty a_jz^j=$$
$$=\left(2iz-2z^2-\frac43iz^3+\frac23z^4+\ldots\right)\left(a_{-k}z^{-k}+a_{-k+1}z^{-k+1}+...+a_0+a_1z+a_2z^2+...\right)$$
From the above, and since the first power of $\;z\;$ in the left side is zero, we can see that $\;k=1\;$ , and then comparing powers of $\;z\;$ we get
$$\begin{align*}&z^0:\;2i=\;\;2ia_{-1}\implies a_{-1}=1\\{}\\&z:\;-2=2ia_0-2a_{-1}=2ia_0-2\implies a_0=0\\{}\\
&z^2:\;-2i=2ia_1-2a_0-\frac43i\implies 2ia_1=\left(\frac43-2\right)i\implies a_1=-\frac13\\\text{etc.} \end{align*}$$
