Solution: In order to compute the integration, we will take $z = 2e^{\iota \theta }$. Since the integration is taken from $(2,0)$ to $(0,2)$, we have $0\leq \theta \lt \pi /2$ (look at the figure below ). So we have \[ z = 2e^{\iota \theta } \implies \mathrm{d} z = 2\iota e^{\iota \theta }\mathrm{d} \theta \] We now will evaluate the integral by substituting these values. \begin{align*} \int _C (z^2 + 3z)\mathrm{d} z & = \int _0^{\pi /2} \left( 4e^{2\iota \theta } + 6 e^{\iota \theta } \right) \left( 2\iota e^{\iota \theta } \right) \mathrm{d} \theta \\ & = 8\iota \int _0^{\pi /2} e^{3\iota \theta }\mathrm{d} \theta + 12\iota \int _0^{\pi /2} e^{2\iota \theta }\mathrm{d} \theta \\[1ex] & = 8\iota \left[ \frac{e^{3\iota \theta }}{3\iota } \right]_0 ^{\pi /2} + 12\iota \left[ \frac{e^{2\iota \theta }}{2\iota } \right]_0 ^{\pi /2} \\[1ex] & = \frac{8}{3} \left[ e^{\frac{3\pi}{2}} - 1 \right] + 6 \left[ e^{\pi \iota } - 1 \right] \\[1ex] & = \frac{8}{3}(-i - 1) + 6 (-1 - 1) \\ & = \frac{-8\iota}{3} - \frac{44}{3}. \end{align*} Thus, the value of the integration will be \[ \int _C (z^2 + 3z) \mathrm{d}z = \frac{-8\iota}{3} - \frac{44}{3}. \]