How to evaluate residue of $\cot z/z^4$ at $z=0$? How to evaluate residue of $\cot z/z^4$ at $z=0$?
As we know :
$$f(x)=f(0)+f'(0)x+f''(0)x^2/2+...$$
but $\cot(0)\to\infty$ or is undefined? I know that:
$$\tan x=x+x^3/3+2x^5/15+...$$
 A: The function has a pole of order $5$ at zero, so it isn't defined there.
Following @PedroTamaroff's hint and using that $\lim_{z\to 0}\frac{\sin z}{z} = 1$ :
$$
\frac{\cot z}{z^4} = \frac{\cos z}{z^4\sin z}= \frac{z\cos z}{\sin z}\frac{1}{z^5}
$$
Observe that the first fraction is holomorphic in a neighbourhood of $0$. Use Cauchy's theorem to finish.
A: Well,
\begin{align}\cot z = \frac{\cos z}{\sin z} &= \frac{1 - \frac{z^2}{2!} + \frac{z^4}{4!} + O(z^6)}{z - \frac{z^3}{3!} + \frac{z^5}{5!} + O(z^7)}\\
& = \frac{1}{z}\cdot \left(1 - \frac{z^2}{2!} + \frac{z^4}{4!} + O(z^6)\right)\cdot \left(1 + \frac{z^2}{3!} - \frac{z^4}{5!} +\frac{z^4}{3!3!}+ O(z^6)\right)\\
&= \frac{1}{z}\cdot \left(1 + \left(-\frac{1}{2!} + \frac{1}{3!}\right)z^2 + \left(-\frac{1}{2!3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{3!3!}\right)z^4 + O(z^6)\right).\end{align}
Therefore, the coefficient of $\frac{1}{z}$ in the Laurent expansion of $\frac{\cot(z)}{z^4}$ is
$$-\frac{1}{2!3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{3!3!} = -\frac{1}{45}.$$
Hence, $$\text{Res}_{z = 0}\frac{\cot(z)}{z^4} = -\frac{1}{45}.$$
A: 
$$\frac{\cos z}{z^4\sin z}$$

$$\newcommand{\f}[2]{\frac{#1}{#2}}
\newcommand{\r}[1]{\frac{1}{#1}}
\newcommand{\array}[2]{\begin{array}{#1}#2\end{array}}
\newcommand{\b}[1]{\left(#1\right)}
\newcommand{\s}[1]{\left[#1\right]}
\newcommand{\d}[0]{\ldots}
\newcommand{\ul}[1]{\underline{#1}}
\array{r|lll}{
&&\r{z^5}&-\b{\r{3!}-\r{2!}}\r{z^3}&+\s{\b{\r{4!}-\r{5!}}+\r{3!}\b{\r{3!}-\r{2!}}}\r{z}&+\d\\\hline
z^5-\f{z^7}{3!}+\f{z^9}{5!}+\d&&1&-\f{z^2}{2!}&+\f{z^4}{4!}&+\d\\
&&1&-\f{z^2}{3!}&+\f{z^4}{5!}&+\d\\\hline
&&&\b{\r{3!}-\r{2!}}z^2&+\b{\r{4!}-\r{5!}}z^4&+\d\\
&&&\b{\r{3!}-\r{2!}}z^2&+\r{3!}\b{\r{3!}-\r{2!}}z^4&+\d\\\hline
&&&&\s{\b{\r{4!}-\r{5!}}+\r{3!}\b{\r{3!}-\r{2!}}}z^4&+\d\\
&&&&\s{\b{\r{4!}-\r{5!}}+\r{3!}\b{\r{3!}-\r{2!}}}z^4&+\d\\\hline
&&&&&+\d
}\\
\begin{align}
R&=\r{4!}\b{1-\r{5}}+\r{3!}\s{\r{2!}\b{\r{3}-1}}\\
&=\r{24}\b{\f45}+\r{12}\b{\f{-2}3}\\
&=\r{6}\b{\r{5}-\r{3}}\\
&=\f{-2}{6\times15}\\
&=-\r{45}
\end{align}$$
