Finding the value of $I=\int_C \overline{z} dz$ along $|z|=2$ from $z=-2i$ to $z=2i$

Question

I want to find the value of the integral $$I=\int_C \overline{z} dz$$ When $C$ is the right hand half of the circle $|z|=2$ from $z=-2i$ to $z=2i$

Refer to beautifully made picture:

Now I am new to this, and I believe this contour integral refers to a line integral, and hence I want to take:

$$\int_C f(z) dz = \int_a^b f[z(t)]z'(t) dt$$ So $f(z)=\overline{z}$ how do I convert this to varying by $t$? Is $t$ just a variable for my parametrisation above? How then do I find $z(t)$ so I can derive it?

Answer 1

Let's reparametrize $z(t) $ as $z(t) = 2 e^ {it} , \ -\pi/2 \leq t \leq \pi/2$. So $dz = 2i e^{it}dt$. Now $$\int _C \bar z dz = \int _ {-\pi/2} ^{\pi/2} 2e^{-it} \ 2 i e^{it} dt= 4i (\pi). $$

Answer 2

\begin{align} z(\theta)=&\quad2e^{i\theta},-\frac\pi 2 \leq \theta \leq \frac\pi 2\\ I=&\quad\int_C \overline{2e^{i\theta}}(2e^{i\theta})' d\theta\\ =&\quad 4\int_{-\frac\pi2}^{\frac\pi2} \overline{e^{i\theta}}(e^{i\theta})'d\theta\\ =&\quad 4i\int_{-\frac\pi2}^{\frac\pi2} d\theta\\ \int_C \overline z\,\, dz=&\quad 4i\pi \end{align}