calcuating residue of a complex function I need to calculate the residue of the function $\frac{(z^6+1)^2}{(z^5)(z^2-2)(z^2-\frac{1}{2})}$ at $z$=0.
z=0 is  a pole of order 5 so I tried using the general formula to calculate the residue but the calculation becomes very tedious since it involves finding the fourth derivative of the complex function. I even tried writing the laurent series but that too got me nowhere. Could anyone please tell how to proceed with this problem?

In order to get the residue of the function at $z=0$ we calculate the coefficient  of $z^{-1}$ of the  Laurent series expansion. We use the coefficient of  operator $[z^n]$  to denote the coefficient of $z^n$.

We obtain
  \begin{align*}
\color{blue}{[z^{-1}]}&\color{blue}{\frac{\left(z^6+1\right)^2}{z^5\left(z^2-2\right)\left(z^2-\frac{1}{2}\right)}}\\
&=[z^4]\frac{\left(z^6+1\right)^2}{(-2)\left(1-\frac{z^2}{2}\right)\left(-\frac{1}{2}\right)\left(1-2z^2\right)}\tag{1}\\
&=[z^4]\left(1+\frac{z^2}{2}+\frac{z^4}{4}\right)\left(1+2z^2+4z^4\right)\tag{2}\\
&=4+1+\frac{1}{4}\tag{3}\\
&\,\,\color{blue}{=\frac{21}{4}}
\end{align*}

Comment:


*

*In (1) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.   We also factor out constants to normalise terms  in the  denominator.

*In (2) we  replace the  term  $(z^6+1)^2$ in  the numerator with  $1$ since other terms do not contribute  to $[z^4]$.   We also  expand $\frac{1}{1-az^2}=1+az^2+a^2z^4+O(z^6)$  up to  terms of $z^4$ since       other     terms do not       contribute.

*In  (3)  we   extract      the        coefficients     of   $z^4$.
