Integrating Complex Functions Over a Path

Question

Let $\gamma : [0,1] \rightarrow \mathbb{C}$ with $\gamma (t) = 2e^{2 \pi t i}$

The function is $f(z) = z + \overline{z}$. I was told that the integral of this function over the path is zero, but I have done it several times and got everything but zero (note I am terrible at calculations). Here is my attempt:

$\int_0^1 (2e^{2 \pi t i} + 2e^{-2 \pi t i}) (4 \pi ie^{2 \pi t i}) dt$

$= 8 \pi i \int_0^1 (e^{4 \pi t i} + 1) dt$

$= 8 \pi i \int_0^1 e^{4 \pi t i} dt + \int_0^1 dt$, where $\int_0^1 e^{4 \pi t i} dt=0$

$= 8 \pi i$

So, what did I do wrong?

Answer 1

Nothing! The integral around the curve $|z|=2$ of $f(z)=\overline{z}$ is $8\pi i$, as you found and the integral around any closed curve of $f(z)=z$ is zero.