What does "let $f\left(z\right)=\bar{z}$" mean in this context

Question

I'm reading a paper/handout on contour integrals and Cauchy's Theorem which says in an example

Let $f\left(z\right)=\bar{z}$.

$\cdots$

Then \begin{align} \int_{C}\bar{z}\:dz&=\int_{0}^{2\pi}\overline{e^{it}}\cdot ie^{it}\:dt\\ &=\int_0^{2\pi}e^{-it}\cdot ie^{it}\:dt\\ &=\int_0^{2\pi} i\:dt\\ &=2\pi i. \end{align}

What is this operation "$\overline{e^{it}}=e^{-it}$"? Please let me know if it appears I have not given enough context. I've looked over the document and do not see it mentioned though, so I am assuming it is something from complex analysis I have not seen (I haven't taken that class yet).

Answer 1

That notation is complex conjugation. For a complex number $z=a+bi$, the complex conjugate of $z$ is defined to be $$\overline{z} := a-bi$$

Answer 2

It is the complex conjugate. Geometrically, complex conjugation corresponds to a reflection over the real axis in the complex plane. Thus, it is the map that sends a complex number $z = a + ib$ to the complex number $\bar{z} = a - ib$.