Calculate residue of function $f(z) = \frac{(z^2+1)^5(z^{10}+1)}{64z^{11}i}$

Question

I need to calculate the residue of $$f(z) = \frac{(z^2+1)^5(z^{10}+1)}{64z^{11}i}$$ where $z_0 = 0$.

Basically I know the formula $$Res_{z_0}(f)=\lim_{z\to z_0}\frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}(z-z_0)^mf(z)$$

and based on this formula, I need to calculate the tenth derivative, right? But whenever I calculate it, I get $0$ in the denominator after I set z to $0$. Wolfram Alpha tells me the right solution, so somehow there must be a mistake on my side.

Could somebody please give me a hint? I am stuck.

Thanks

Answer 1

You want $1/(64 i)$ times the coefficient of $z^{10}$ in $(z^2+1)^5 (z^{10}+1)$. If you use the $z^{10}$ in the factor on the right, you need the coefficient of $z^{0}$ in $(z^2+1)^5$, i.e. the constant term $1$. If you use the $1$ in that factor, you need the coefficient of $z^{10}$ in $(z^2+1)^5$, i.e. of the leading term in that expansion, another $1$. So the residue is $2/(64 i) = - i/32$.

Answer 2

The residue is equal to $\frac{1}{10!}\cdot\frac{d^{10}}{dz^{10}} \frac{{(z^2+1)}^5(z^{10}+1)}{64i}$

I would use binomial theorem to expand the numerator, but note that you're evaluating the 10th derivative at $z=0$ so you only want to look at the $z^{10}$ coefficient in $f(z)$ because all other components are 0. The residue is then $\frac{1}{10!} \cdot \frac{2\cdot 10!}{64i} =\boxed{\frac{1}{32i}}$