An explanation of how the range of a complex function works? Specifically $f(z)=z^2$ for$Re(z)>0$, $Im(z)>0$ and in the first quadrant.. I'm going through this complex analysis textbook, and it tells me that the range of the aforementioned function is $Im(w) \geq 0$.  To me, that makes no sense. Could someone explain that, by chance? The complex plane isn't clicking, and I have a feeling these answers are going to make me feel dumb.
To add to this question, I have another problematic question:
Find the range of $h(z)=1/z$ for $0<|z|\leq1$.
The answer, apparently, is $|w|\geq1$. Again, I have no idea how that answer was found.
Thanks!
 A: In the first quadrant, the argument (angle) obeys
$$0<\theta<\frac{\pi}2$$
When you square a complex number, you square its modulus (distance from the origin in the complex plane) and double its angle, due to De Moivre's formula
$$[r\cdot \mathrm{cis}(\theta)]^2=r^2\cdot \mathrm{cis}(2\theta)$$
Therefore, while the range of the modulus stays the same (greater than zero), the argument now obeys
$$0<2\theta<\pi$$
The argument is between zero and $\pi$ for the imaginary part positive (not just non-negative, as you stated). So $Im(z^2)>0$ and your textbook is almost correct.
For your added, second question, if $0<|z|\le 1$, then $z$'s argument $\theta$ has no restriction but its modulus is restricted by $0<r\le 1$. Then by De Moivre's theorem,
$$\frac 1z=z^{-1}=r^{-1}\mathrm{cis}(-\theta)$$
so
$$|z^{-1}|=r^{-1}\ge 1$$
We easily see the argument $-\theta$ can be anything, so the range of $h(z)=1/z$ for $0<|z|\le 1$ is indeed $|w|\ge 1$.
A: For your second question, this is just a matter of solving. Suppose $x \in \mathbb{C}$ is not zero, then $\frac{1}{y} = x \Leftrightarrow y=\frac{1}{x}$. So $x$ will be in the image if and only if $0 < \left | \frac{1}{x} \right | \leq 1$. The first inequality will always hold (why?), while the other can be solved by taking reciprocals, giving $|x| \geq 1$.
