Help with this simple complex integral on the conjugate I am currently trying to calculate
$$\int_{|z| = 2} \overline{z} dz.$$
Take $\overline{z} = f(z).$ I see that for $z = re^{i\theta} = r\cos(\theta) + r\sin(\theta)i,$ We see that
$$\overline{z} = r\cos(\theta) - r\sin(\theta)i
= r\cos(-\theta) + r\sin(-\theta)i
= re^{-i\theta}.$$
Thus,
$$\{z | |z| = 2\} = \{2e^{i\theta}, \theta \in [0,2\pi)\}.$$
Take $q(\theta) = 2e^{i\theta}.$ We can see that
$$\int_{|z| = 2} f(z) dz = \int_{0}^{2\pi} f(q(\theta))q'(\theta) d\theta$$
$$=\int_0^{2\pi} f(2e^{i\theta})(2e^{i\theta} \cdot i) d\theta$$
$$=2i \int_0^{2\pi} 2e^{-i\theta} e^{i\theta} d\theta$$
$$=2i \int_0^{2\pi} 2 d\theta = 2i(4\pi) = 8\pi i.$$
However, I am not entirely sure if this calculation is correct. Any recommendations on how to check this?
 A: A quick way to compute the integral is as follows:
$$
\int_{|z| = 2} \bar z\,dz = 
\int_{|z| = 2} \frac{|z|^2}{z}\,dz = 
\int_{|z| = 2} \frac{4}{z}\,dz
$$
By the residue theorem, this should come out to $4 \cdot 2 \pi i = 8 \pi i$ (assuming the contour is positively oriented).  So, your answer is indeed correct.
A: Another way to evaluate this sort of contour integrals involving $\bar{z}$ is identify $\mathbb{C}$ with $\mathbb{R}^2$ through the parametrization
$$\mathbb{R}^2 \ni (x,y) \quad\mapsto\quad z = x + iy \in \mathbb{C}$$
and then apply Stoke's theorem.
Let $D$ be the disc $\{ z \in \mathbb{C} : |z| \le 2 \}$ in the complex plane, its boundary $\partial D$ is the circle $|z| = 2$.
The contour integral at hand is equivalent to
$$\int_{\partial D} \bar{z}dz
= \int_D d(\bar{z} dz)
= \int_D d\bar{z}\wedge dz
= \int_D (dx - i dy)\wedge(dx + i dy) = 2i\int_D dx\wedge dy
$$
Since $dx \wedge dy$ is the area element of $\mathbb{R}^2$, we find
$$\oint_{|z| = 2} \bar{z} dz = \int_{\partial D} \bar{z}dz = 2i\verb/Area/(D) = 2i(\pi 2^2) = 8\pi i$$
