Why is $\oint_{|z|=1} \frac{1}{z}dz = \int_0^{2\pi} \frac{ie^{iz}}{e^{iz}}dz$?

Question

I have come across this line $$\oint_{|z|=1} \frac{1}{z}dz = \int_0^{2\pi} \frac{ie^{iz}}{e^{iz}}dz = 2\pi i$$ and I don't understand it. How does it change from being $\frac{1}{z}$ to $ie^{iz}/e^{iz}$? I would have like to write it in polar form as something like $\frac{1}{e^{it}}$, but then the integral would be equal to $0$...

Answer 1

This is simply the defnition of complex line integrals.

The (closed, clock-wise) curve $C = \{|z| = 1\}$ can be parametrized via the function $g(z) = e^{iz}$ for $z \in [0, 2\pi]$. Observe that $g'(z) = i e^{iz}$. Now this means that for a function $f$ that is holomorphic in a neighborhood of $C$ we have $$\oint_C f(z) \, dz = \int_0^{2\pi} i e^{it} f(e^{it}) \, dt$$

Specifically, $f(z) = \frac{1}{z}$ yields $$\oint_C \frac{1}{z} \, dz = \int_0^{2\pi} i e^{it} \frac{1}{e^{it}} \, dt = \int_0^{2\pi} i \, dt = 2\pi i.$$

Answer 2

You are right and wrong:

$\oint_{|z|=1} \frac{1}{z}dz = \int_0^{2\pi} \frac{ie^{it}}{e^{it}}dt =\int_0^{2\pi} i dt= 2\pi i$