Evaluate the complex integration $(z^2 + 3z)$ wrt $z$ along the circle $|z| = 2$, from $(2,0)$ to $(0,2)$ in a counterclockwise direction.
As far as I understand, this can be solved by taking $x = 2 \cos \theta$, $y = 2 \sin \theta$, and then integrating wrt $\theta$ from $0$ to $π/2$.
But on making these substitutions, the integration becomes quite lengthy and clumsy. Is there any other way to solve this, which I might be missing right now?
On your circle, $z=2e^{it}$, $0\leq t\leq\pi/2$. Then your integral is (never forget to include the change of variable, $dz=2ie^{it}\,dt$) \begin{align} \int_0^{\pi/2} ((2e^{it})^2+6e^{it})\,2ie^{it}\,dt &=2i\int_0^{\pi/2}(4e^{3it}+6e^{2it})\,dt =2i\left[ \left.\frac{4e^{3it}}{3i}+\frac{3e^{2it}}{i}\right|_0^{\pi/2}\right]\\ \ \\ &=-\frac {8i}3-6-\frac83-6=-\frac {44}3-\frac {8i}3. \end{align}