Which of the following integrals is =0?

Question

$\int_C \bar z dz$ where the curve $C$ is described by $|z|=1$

so the integral of $\bar z=x-iy$, why does it equal zero?

Answer 1

It is not $0$, actually. Parametrize the curve $C$ by $z=e^{it}$ over $[0,2\pi]$. Then $dz=ie^{it}dt$. So $$\int_C \overline{z}\;dz=\int_0^{2\pi}\overline{e^{it}}\;i\; e^{it}dt=i\int_0^{2\pi}1dt=2i\pi.$$

Note: that's, equivalently, $\int_C\frac{1}{z}dz=2i\pi$. This is the only case where $\int_Cz^ndz$ is nonzero, for $n\in\mathbb{Z}$. If you remove one point from the circle, you get a complex logarithm, as an antiderivative of $\frac{1}{z}$. So your integral computes the variation of the argument along the circle. This tends to $2\pi$ when you tend to the removed point.