Integrate $\int_{|z|=r} x \, dz$ I'm looking to integrate
$$\int_{|z|=r} x \, dz$$
My question is:
Given that $z(t)=re^{it}$ is a parametrization for $|z|=r$, then what's $x$? It's not a variable, because $z$ is, but does it refer to the $\cos(t)$ part of $e^{it}$ or is it a constant in the integration?

Contour Integration
On $|z|=r$, $\bar{z}=\frac{r^2}z$ and $x=\frac12(z+\bar{z})$. Therefore,
$$
\begin{align}
\int_{|z|=r}x\,\mathrm{d}z
&=\int_{|z|=r}\frac12\left(z+\frac{r^2}z\right)\mathrm{d}z\\
&=0+r^2\pi i\\[6pt]
&=r^2\pi i
\end{align}
$$

Real Integration
$$
\begin{align}
\int_{|z|=r}x\,\mathrm{d}z
&=\int_{|z|=r}x\,\mathrm{d}(x+iy)\\
&=\int_{|z|=r}x\,\mathrm{d}x+i\int_{|z|=r}x\,\mathrm{d}y\\[4pt]
&=0+i\,\text{Area}(|z|\le r)\\[9pt]
&=\pi r^2 i
\end{align}
$$
