How to integrate the complex function $f(z) = xy$ over the circle $C = [0, r]$

Question

How do i integrate the complex function $f(z) = xy$ over the circle $C = [0, r]$ ?

What i have tried so far is to pass to polar coordinates, but i keep getting $0$ as a result. But since the function is not holomorphic it doesn't seem to be right.

Answer 1

It is true that the function $$xy = f(x, y) = f(z) = \Re{z}\Im{z}$$ is not holomorphic (since $u(x,y) = xy$ and $v(x,y) = 0$ the CR equations don't work) but that doesn't mean that this particular integral must be non-zero. Indeed, just looking at the graph should help assure you that this integral should be zero. Then to be sure, putting $\gamma(t) = r e^{it}$ we get $$f(z) = \Re{z}\Im{z} = \frac{z + \bar{z}}{2}\frac{z - \bar{z}}{2i} = \frac{r^2}{4i}\left( e^{2it} - e^{-2it} \right)$$ and $\gamma '(t) = i r e^{it}.$ But then $$\int_0^{2\pi} \left( e^{2it} - e^{-2it} \right) ir e^{it} dt = ir \int_0^{2\pi} e^{3it} - e^{-it} dt = 0.$$