Residue of $f(z) = \frac{z}{1-\cos(z)}$ at $z=0$

Solution 1:

The first question you should ask is what type of singularity $f(z)$ has at $z = 0$. Since $z$ has a zero of order $1$ and $1 - cos(z)$ has a zero of order $2$, $f(z)$ has a simple pole. This means $zf(z)$ has a removable singularity, and relating the power series expansion at $0$ of $f(z)$ and $zf(z)$, you know that the residue of $f(z)$ at $z = 0$ is the value of $zf(z)$ at $z = 0$. Now you need to find this value.

Solution 2:

Write the (first few terms of the) Taylor series for $\cos z$, and subtract this from 1. You should find that the result can be written in the form $1-\cos z=az^2+bz^4+\cdots$ for some constants $a$ and $b$, and where the dots represent higher powers of $z$. Then think about using the binomial theorem to pin down the coefficient of $z^{-1}$ in $$ f(z) = \frac{z}{1-\cos z}=\frac{z}{az^2(1+\frac{b}{a}z^2+\cdots)}.$$

Solution 3:

Using the power series $\cos(z)=1-\frac12z^2+O(z^4)$ yields $$ \begin{align} f(z) &=\frac{z}{1-(1-\frac12z^2+O(z^4))}\\ &=\frac{1}{\frac12z+O(z^3)}\\ &=\frac{1}{z}\frac{1}{\frac12+O(z^2)}\\ &=\frac{1}{z}(2+O(z^2)) \end{align} $$ Thus the residue is $2$

Another thought is to write $$ \begin{align} \frac{z}{1-\cos(z)} &=\frac{z^2}{\sin^2(z)}\frac{1+\cos(z)}{z}\\ &=\frac{1}{1+O(z^2)}\cdot\frac{2+O(z^2)}{z} \end{align} $$ again giving a residue of $2$.