how to find the integral of $z$-conjugate along a circular path?

Question

$$\int_\gamma \overline{z}\>dz,\quad \text{where} \quad |z|=2.$$ How to find the integral of $z$-conjugate, where the path is determined as given above?

I saw that we can write $z=x+iy$, but how to continue?

Answer 1

Parametrize $\gamma$ by $z = 2e^{it}$, $0 \le t \le 2\pi$. We have $\bar{z} = 2e^{-it}$ and $dz = 2ie^{it}\, dt$, so $$\int_\gamma \bar{z}\, dz = \int_0^{2\pi} (2e^{-it})(2ie^{it}\, dt) = \int_0^{2\pi} 4i\, dt = 8\pi i.$$

Answer 2

For all $z$ on the circle $\gamma$ of radius $2$ around the origin one has $z\>\bar z=|z|^2=4$. Therefore we may write $$\int_\gamma\bar z\>dz=\int_\gamma {4\over z}\>dz=4\cdot 2\pi i=8\pi i\ .$$