Does $\int_\Gamma \bar{z} \, dz= \int_\Gamma \frac 1 z \, dz$, where $\Gamma$ is the circle $|z|=1$ traversed counter-clockwise?

Question

I think this is true but I'm not sure. The parametrization for the contour is $z(\theta) = e^{i\theta}$, with $z'(\theta)=ie^{i\theta}$ ($0\leq\theta\leq2\pi$). Calculating each integral separately:

$$\int_\Gamma \bar{z} \, dz = \int_0^{2\pi} \overline{z(\theta)} z'(\theta) \, d\theta = \int_0^{2\pi}(e^{-i\theta})(ie^{i\theta}) \, d\theta$$

Reducing that gives me $2\pi i$

For the next integral, we have

$$ \int_\Gamma \frac{1}{z} \, dz = \int_0^{2\pi} \frac{1}{e^{i\theta}} ie^{i\theta} \, d\theta $$

which also gives $2\pi i.$

Answer 1

On the unit circle, $\bar z = \frac{1}{z}$, so the integrands are equal.