As the title says, I have to compute $$\int_{|z|=r} x^2dz$$ for the circle traversed anti-clockwise in 2 different ways.
If I use a parametrisation I quickly get to 0 - which might be wrong though -, but I'm not really sure how to go about the other way. I got to $x=\frac{1}{2}(z-\frac{r^2}{z})$, but does this help? The only thing I could do is use $re^{it}$ but that would be the very same way as the first one. Any help please?
Setting $z=re^{it}$, you have $$\oint_{|z|=r}\mathrm{Re}(z)^2\,\mathrm{d}z=ir^3\int_0^{2\pi}\cos^2t\,e^{it}\,\mathrm{d}t=\frac{ir^3}{2}\int_0^{2\pi}\bigg(1+\cos2t\bigg)e^{it}\,\mathrm{d}t$$ which agrees with your other approach, giving an answer of $0$.