Let $\gamma_1$ be the semicircle from $1$ to $-1$ through $i$ and $\gamma_2$ the semicirlce from $1$ to $-1$ through $-i$. Compute $\int_{\gamma_1}z^2dz$ and $\int_{\gamma_2}z^2dz$. Can you account for the fact that they are equal?
Compute $\int_{\gamma_1}z^2dz$
\begin{align}
\gamma_1(t)=e^{it}\ for\ -\pi \leq t \leq 0\\
\notag \\
\gamma_1'(t)=ie^{it}\\
\notag \\
\int_{\gamma_1}z^2dz = \int^{\pi}_{0}(e^{it})^2 \cdot ie^{it}dt = \dfrac{1}{3}e^{3it} \Big|^{\pi}_{0}=-\dfrac{2}{3}\\
\end{align}
Compute $\int_{\gamma_2}z^2dz$
\begin{align}
\gamma_2(t)=e^{it}\ for\ 0 \leq t \leq \pi\\
\notag \\
\gamma_2'(t)=ie^{it}\\
\notag \\
\int_{\gamma_2}z^2dz = \int^{\pi}_{0}(e^{it})^2 \cdot ie^{it}dt = \dfrac{1}{3}e^{3it} \Big|^{\pi}_{0}=-\dfrac{2}{3}\\
\end{align}
\begin{align}
\gamma_1 + (-\gamma_2) = \gamma\\
\notag \\
\int_{\gamma}z^2dz = \int_{\gamma_1}z^2dz - \int_{\gamma_2}z^2dz =0\\
\notag \\
\therefore\ equal.
\end{align}
