Find the residue of $\frac {\cot(z)\coth(z)}{z^3}$ at $z=0$.
In my notebook, i expanded the functions and took the coefficient of $1/z$ as residue, which is $-7/45$
I couldn't recall what type of singularity is this, why we have to take coefficient of $1/z$.
Can you please tell me what type of singularity is this and why we take coefficient of $1/z$.
You have\begin{align*}\cot(z)\coth(z)&=\frac{\cos(z)\cosh(z)}{\sin(z)\sinh(z)}\\&=\frac{\left(1-\frac{z^2}{2!}+\frac{z^4}{4!}+\cdots\right)\left(1+\frac{z^2}{2!}+\frac{z^4}{4!}+\cdots\right)}{\left(z-\frac{z^3}{3!}+\cdots\right)\left(z+\frac{z^3}{3!}+\cdots\right)}\\&=\frac{1-\frac{z^4}6+\cdots}{z^2-\frac{z^6}{90}+\cdots}\end{align*}In the numerator, all the exponents are multiples of $4$ and in the denomitor they are all of the type $4n-2$. So, this quotient can be written as $\frac{a_0+a_1z^4+a_2z^8+\cdots}{z^2}$. It is easy to see that $a_0=1$ and that $a_1=-\frac7{45}$. So$$\frac{\cot(z)\coth(z)}{z^5}=\frac1{z^5}-\frac7{45z}+\cdots$$and the residue is equal to $-\frac7{45}$. Furtermore, this singularity is a pole.