$$\int_{-1}^i |z|^2dz$$ The hint says to use the parametrization $z=e^{i\theta}$ where $\pi \ge \theta \ge \pi/2$. Below is my attempt:
$$|z|=1$$ $$|z|^2=1$$ $$dz=ie^{i\theta}d\theta$$
Thus,
$$\int_{\pi/2}^\pi (1)ie^{i\theta}d\theta$$ $$=e^{i\theta}|_{\pi/2}^{\pi}$$ $$=e^{i\pi}-e^{i\pi /2}$$ $$=-1-i$$
But the answer given is $1+i$. I'm not sure how they didn't get minus signs. Any idea where I might have gone wrong?
The reason is, as Jane points out, because you should swap the order of integration. Now the question is why should you swap the order of integration?
In an informal way you can think of it this way: you are integrating from $-1$ to $i$ in the original integral. Visualize the complex plane. The complex numbers represent the $y$ axis and the real numbers represent the $x$ axis.
Consider what you are doing when substituting $z = e^{i\theta}$. Graphing $e^{i\theta}$ in the complex plane you get the unit circle. Informally speaking, you are traversing from $-1$ to $i$ so you should start at $e^{i\pi} = -1$ then go $e^{i\frac{\pi}{2}} = i$.