a contour integration involving $5$th roots of unity

Question

Let $\gamma$ be the circle of radius $2$ with centre at the origin in the complex plane oriented in the anti-clockwise direction, then the integral $$\oint_\gamma \frac{dz}{(z-3)(z^5-1)}$$ equals

$(a)$ $\displaystyle{\frac{2\pi i}{3^5-1}}$.

$(b)$ $\displaystyle{\frac{2\pi i}{3^4-1}}$.

$(c)$ $\displaystyle{-\frac{2\pi i}{3^5-1}}$.

$(d)$ $0$.

Clearly by Rouche's theorem, all the zeroes of $z^5-1$ lies on the circle $|z|=1$. So if we count those zeroes by $z_k$ where $k=0,1,2,3,4$ we have, by Residue theorem $$\oint_\gamma \frac{dz}{(z-3)(z^5-1)}=-\frac{2\pi i}{5}\sum_{k=0}^4 \frac{1}{3{z_k}^4-1}$$ which is a nightmare to simplify algebraically. Can someone help in this regard if I have done something wrong? Thanks in advance.

on a side note, rouche's theorem was overkill. if $z^5-1-0$, then $z^5=1$, so obviously they lie on $|z|=1$ — Saketh Malyala, Dec 06 '19 at 07:00
My side note, as this is complex analysis, I usually think everything complex..:) — am_11235..., Dec 06 '19 at 07:02

Robert Z · Accepted Answer · 2019-12-06T07:12:09.650

5

Hint. By using the residue at infinity we have that $$\oint_{|z|=2} \frac{dz}{(z-3)(z^5-1)}=-2\pi i(\text{Res}(f,3)+\text{Res}(f,\infty)). $$ and those two residues are quite easy to evaluate for $f(z)=\frac{1}{(z-3)(z^5-1)}$.

Recall that on the Riemann sphere, the algebraic sum of the complex residues of a meromorphic function is zero. See also the Inside-Outside Theorem.

edited Dec 06 '19 at 07:12

answered Dec 06 '19 at 07:06

Robert Z

145,942

Why would I calculate the residue at $z=3$, whereas this point lies completely outside $\gamma$? Also why do I need that residue at infinity? – am_11235... Dec 06 '19 at 07:08
Please see my edit. – Robert Z Dec 06 '19 at 07:11
what the heck! how that would've helped ME during my analysis class. +1! – Saketh Malyala Dec 06 '19 at 07:11

a contour integration involving $5$th roots of unity

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