$\int |z| \bar z dz$ over contour

Question

I was wondering whether I can receive some form of confirmation to my method as I feel it's very informal in the way it is worked out.

$$\int_{\Gamma+} |z| \bar z dz $$

Where $\Gamma +$ is a closed contour consisting of the upper semi-circle $|z|=1$, $Im(z)\geq 0$ and the segment $-1\leq Re(z) \leq 1$, $Im(z)=0$ traversed in a positive direction.

The way I worked this out is essentially by splitting the closed contour into 2 contours and evaluating them seperately. My final result is that of $\pi i$. Is my method correct? Or am I able to use some theorems such as Cauchy's Theorem or Green's Theorem? (Both of which I did not use as I am unaware of an easy way to show $f(z)=|z|\bar z$ is either holomorphic, or continuously differentiable)

Answer 1

The integrand is not analytic, by brute force:

\begin{align*} \oint_{C} \bar{z}|z| \, dz &= \int_{-1}^{1} x|x| dx+\int_{0}^{\pi} e^{-it} \times 1 \, d(e^{it}) \\ &= 0+\int_{0}^{\pi} i \, dt \\ &= i\pi \end{align*}

Your result is right.

Answer 2

Since $|z|=1$ on the integration path, you also have $|z|^2=z\overline z=1$ and thus $|z|\overline z=(1)(1/z)=1/z$. So your differential is just $dz/z$ and the definite integral is $\ln [(-1)/(+1)]$.