Compute $\int_{|z|=1}|z^5-1|^2|dz|$

Question

Question: Compute $\int_{|z|=1}|z^5-1|^2|dz|$.

My attempt: Let $z=e^{i\theta}$, then $|dz|=|ie^{i\theta}|d\theta=1$. So,

\begin{align} \int_{|z|=1}|z^5-1|^2|dz| &= \int_0^{2\pi}|e^{i5\theta}-1|^2d\theta\\ &= \int_0^{2\pi}[\cos^25\theta-2\cos5\theta+1+\sin^25\theta]d\theta\\ &= \int_0^{2\pi}(2-2\cos5\theta)d\theta\\ &= [2\theta-\frac{2}{5}\sin5\theta]|_0^{2\pi}\\ &= 4\pi-\frac{2}{5}\sin10\pi\\ &= 4 \pi \end{align}

I suppose I was thinking I made some algebra/calculus error somewhere because I got the same answer when I computed $\int_{|z|=1}|z-1|^2|dz|$, but I couldn't find an error.

Answer 1

The exact same computation holds for any non-zero integer $n$ (negative as well).

Answer 2

For an arbitrary integrable function $f$ and non-zero integer $n$ is the value of the integral $$ \int_0^{2\pi} f(e^{in\theta}) \, d\theta = \frac 1n \int_0^{2n\pi}f(e^{i\tau}) \, d\tau = \int_0^{2\pi}f(e^{i\tau}) d\tau $$ independent of $n$. (First substitute $n\theta = \tau$, then use that $\tau \mapsto e^{i\tau}$ is $2\pi$-periodic.)

Applied to your case, $$ \int_0^{2\pi}|e^{in\theta}-1|^2 \, d\theta = \int_0^{2\pi}|e^{i\tau}-1|^2d \, \tau = \int_0^{2\pi} (2-2\cos(\tau)) \, d\tau = 4 \pi + 0 $$ for $n \in \Bbb Z \setminus \{ 0 \}$.