Image of domain under map $f(z) = z^6$? I am given the domain $\Omega := \{z \in \mathbb{C}: \operatorname{Im}(z) > \sqrt{3}|\operatorname{Re}(z)|\}$. I want to know what the image $f(\Omega)$ of this domain is under the mapping $f(z)=z^6$. I'm not sure if my method is correct, here is my attempt:
Having sketched the domain, I think I can rewrite $\Omega$ as:
$$\Omega = \{z=re^{i\theta} \in \mathbb{C}: r>0, \frac{\pi}{3} < \theta \leq \frac{\pi}{2}\} \cup \{z=re^{i\theta} \in \mathbb{C}: r>0, \frac{\pi}{2} < \theta < \frac{2\pi}{3}\}$$
Then, using the reasoning that $f$ sends $z=re^{i\theta}$ to $z^6=r^6e^{6\theta i}$, I have said that the domain is mapped to:
$$f(\Omega) = \{z=re^{i\theta} \in \mathbb{C}: r>0, 2\pi < \theta \leq 3\pi\} \cup \{z=re^{i\theta} \in \mathbb{C}: r>0, 3\pi < \theta < 4\pi\}$$
which I am interpreting as: if $f(\Omega)$ was a sketch on the complex plane, it would be the whole plane minus the positive real axis.
Is this correct? If it is, is there a quicker or more intuitive/interesting way to arrive here?
 A: You have the right idea, but I'm a bit confused by your description of the domain. Surely the union of those two sets is simply
$$\Omega=\{re^{i\theta}:\ r>0,\ \tfrac13\pi<\theta<\tfrac23\pi\}.$$
Either way, your reasoning from here is correct, and then the image $f(\Omega)$ can similarly be described simply as
$$f(\Omega)=\{re^{i\theta}:\ r>0,\ 2\pi<\theta<4\pi\},$$
which is indeed the entire complex plane minus the positive real axis. Also, don't forget that you're missing $0\in\Bbb{C}$ is well; perhaps it would be less ambiguous to say that it's the entire complex plane minus the nonnegative real axis.

Another way to see this, though less rigorously, is to note that $\Omega$ is an open region bounded by two half-lines. These two half-lines are both mapped to the nonnegative real axis, and so $\Omega$ is mapped to an open region bounded by the nonnegative real axis. This must then be precisely $\Bbb{C}$ minus the nonnegative real axis. This can of course be made rigorous as well.
Here's an interactive piece of geometry that shows what $f(\Omega)$ looks like for $f(z)=z^a$, as $a$ varies. The red region is the part of $f(\Omega)$ that is inside the unit disk. It breaks at $a=6$, unfortunately.
