We are supposed to integrate $z^{1/3}$ around the unit circle $|z| = 1$.
If we parameterize $z = e^{i \theta}$ on the unit circle then $z^{1/3} = e^{i\frac{\theta +2 \pi k}{3}}$ for $k = 0, 1, 2$.
The antiderivative is
$$\frac{3}{i} e^{i\frac{\theta +2 \pi k}{3}}= -3ie^{i\frac{\theta +2 \pi k}{3}}$$
So, to evaluate this we are supposed to use branch cuts. After my fiasco with the simple function $z^2$ (in part 1 of today's questions) I am getting a mental block and very flustered so any suggestion how to finish this off neatly and simply would be appreciated.
$\oint z^{\frac13} dz\\ z = e^{it}, dz = i e^{it}dt\\ \int_0^{2\pi} i e^{\frac 43 it} dt\\ \frac 34 e^{\frac 43 it}|_0^{2\pi}\\ \frac 34(\cos \frac {8\pi}{3} - 1 + i\sin\frac {8\pi}{3})$