Complex integral evaluation

Question

I'm trying to solve the following question:

Let $\gamma\subset\mathbb{C}$ be the boundary of the upper (unit) semi-disk (closed path). Calculate: $\int_{\gamma}\left|z\right|\overline{z}dz$.

My approach is to show that $f(z)=\left|z\right|\overline{z}\ $ is holomorphic (maybe using Cauchy-Riemann equation?). Then, by Cauchy's theorem, and conclude that the integral is equal to zero.

I'd love to know if my approach is right.

Thank you.

Answer 1

$f$ is not holomorphic. The integral over the line segment from $-1$ to $1$ is $0$ because $x|x|$ is an odd function. The integral over the semi-cirlular part becomes $\int_0^{\pi} e^{-i\theta} ie^{i\theta}d\theta=\pi i$.

Answer 2

Your approach won't work because we have for the integrand $f(z)=\lvert z\rvert \bar z\implies u(x,y)+v(x,y)i=f(x+iy)=x\sqrt{x^2+y^2}-y\sqrt {x^2+y^2}i\implies u(x,y)=x\sqrt {x^2+y^2},\,v(x,y)=-y\sqrt {x^2+y^2}$.

But then we need $u_x=\sqrt {x^2+y^2}+x\dfrac 12(x^2+y^2)^{-\frac 12}2x=\sqrt {x^2+y^2}+x^2(x^2+y^2)^{-\frac 12}$ to be the same as $v_y=-\sqrt {x^2+y^2}-y^2(x^2+y^2)^{-\frac 12}$, which it is not.

So the Cauchy-Riemann equations aren't satisfied, and the function isn't holomorphic.