Image of a space under the map $f(z)=z^2$ This is a very basic question. If $G=\{z\mid \operatorname{Im}(z)>0, z\notin [0,i]\}$ and $f(z)=z^2$, is $f(G)=\{z\mid z\notin [-i,i], \operatorname{Im}(z) \neq0\}$? Here $[0,i]$ is the vertical line segment between $0,i$ and $[-i,i]$ is the vertical line segment between $-i,i.$ Is this correct?
 A: No, it's not correct. For example, your condition $\operatorname{Im}(z) \ne 0$ excludes -4, however $-4 \in f(G)$:
$$z = 2i \Rightarrow z^2 = -4$$
and $2i \in G$.

To find the answer, you can write $z$ in polar form: $z = r e^{i \theta}$.


*

*The condition $\operatorname{Im}(z) > 0$ becomes $0 < \theta < \pi$ and $r \ne 0$.

*The condition $z \notin [0, i]$ becomes $r > 1$ whenever $\theta = \frac{\pi}2$.


If we look at $z^2 = r^2 e^{2 i \theta} = s e^{i \gamma}$ the two conditions change as follows:


*

*$2 \cdot 0 < 2 \theta < 2 \pi$ and $r \ne 0$, or equivalently $0 < \gamma < 2 \pi$ and $s \ne 0$ (this condition excludes the positive real line).

*$r^2 > 1^2$ whenever $2 \theta = 2 \cdot \frac{\pi}2$, or equivalently $s > 1$ whenever $\gamma = \pi$ (this condition excludes the segment $[-1, 0]$)


Putting everything together, we can say that:
$$f(G) = \mathbb{C} \backslash [-1, +\infty]$$

You can also visualize the transformation by considering that the map $z \rightarrow z^2$ changes the modulus from $|z|$ to $|z|^2$ and doubles the argument of $z$.
If you take a piece of paper, draw some points from $G$ and rotate them so that their argument is doubled (ignore the change of modulus as it is not very important): you will see that you'll be covering all the complex plane.
If you try to rotate the points on $\mathbb{R}$ and $[0, i]$ (which are some of the points that are excluded from $G$), you will see that they will land on the line $[-1, +\infty]$ (which is in fact what we have excluded from $f(G)$).
