# Find all residues of all isolated singular points of the function $f(z) = \frac{1+z^{2n}}{z^n(z-a)}$, where $a \neq 0$ and $n \in \mathbb{Z}^+$

Singular points are $$0, a,\infty$$ and all of them are isolated. If $$n=1$$, then both $$0,a$$ are simple poles, so calculating residues are easy and by the Residue Theorem residue at $$\infty$$ is also can be found. My problem is to calculate the residue at $$0$$ when $$n>1$$. What I have tried so far is the following:

$$\small\frac{1+z^{2n}}{z^n(z-a)}=\left(\frac{1}{z^n}+z^n\right)\left(-\frac{1}{a}\right)\frac{1}{1-\frac{z}{a}}=\left(\frac{-1}{z^n}-z^n\right)\sum_{n=0}^{\infty} \frac{z^n}{a^{n+1}}=\sum_{n=0}^{\infty} \frac{-1}{a^{n+1}}-\sum_{n=0}^{\infty} \frac{z^{2n}}{a^{n+1}}$$

But this expansion around $$0$$ implies that residue is zero. Is this correct? Am I missing something?

Also, I want to know how do we find the Laurent Expansion of this function around $$\infty$$?

Thanks for any help.

In your computations, you used the letter $$n$$ for two distinct things.

The residue at $$0$$ of $$f$$ is the residue at $$0$$ of $$\frac1{z^n(z-a)}$$, since $$\frac{z^{2n}}{z^n(z-a)}=\frac{z^n}{z-a}$$, which has a removable singularity at $$0$$.

Besides,\begin{align}\frac1{z^n(z-a)}&=-\frac1a\times\frac1{z^n}\times\frac1{1-\frac za}\\&=-\frac1a\times\frac1{z^n}\times\left(1+\frac za+\frac{z^2}{a^2}+\frac{z^3}{a^3}+\cdots\right)\\&=-\frac1{az^n}-\frac1{a^2z^{n-1}}-\frac1{a^3z^{n-2}}-\cdots\end{align}and the coefficient of $$\frac1z$$ in this Laurent series is $$-\frac1{a^n}$$ . In other words,$$\operatorname{res}_{z=0}f(z)=-\frac1{a^n}.$$

• Thanks for the correction, I will edit now. Commented Dec 26, 2019 at 11:31

This answers your last question. To determine the Laurent series at $$\infty$$, let $$t=z^{-1}$$. Then $$f(z):=\frac{1+z^{2n}}{z^n(z-a)}\equiv\frac{1+t^{-2n}}{t^{-n}(t^{-1}-a)}=\frac{1+t^{2n}}{t^{n-1}(1-at)}=\frac{1+t^{2n}}{t^{n-1}}\cdot\left(1+at+a^2t^2+\cdots\right)$$ and collecting terms gives \begin{align}f(t)=\sum_{k_1=1}^{2n}\frac{a^{k_1-1}}{t^{n-k_1}}+\sum_{k_2=1}^\infty(a^{2n}+1)a^{k_2-1}t^{n+k_2}.\end{align} Since $$dt=-dz/z^2=-t^2\,dz$$, the residue at $$\infty$$ occurs when $$t^2\cdot t^{-(n-k_1)}=t^{-1}$$. Thus $$k_1=n-3$$ so that \begin{align}\operatorname{Res}(f(z),\infty)=\operatorname{Res}(-t^2f(t),0)=a^{n-4}.\end{align}

• Should not we calculate the residue of the differential $f(1/z)(-1/z^2)dz$ when we change the variables? Commented Dec 26, 2019 at 14:20
• Corrected, thanks. Commented Dec 26, 2019 at 14:30
• Is this correct? The coefficient of $t^{-1}$ should come exclusively from $1/t^{n-1}$ times the series, which would be $a^{n-2}$ if I'm not mistaken. Commented Dec 26, 2019 at 16:44
• @Fimpellizieri As OP pointed out we have to multiply by $-1/z^2=-t^2$ which causes the change from $a^{n-2}$ to $a^{n-4}$. Commented Dec 26, 2019 at 19:35
• Huh. I am not very familiar with this. Can I read more about it elsewhere? Commented Dec 26, 2019 at 19:42