How would you sketch this subset of $\mathbb{C}$?

Question

How would you sketch the subset $\{z \in \mathbb{C} : |e^{z}| <1 \}$ of $\mathbb{C}$? Is there a general method for this kind of problem?

Also, how is it possible to tell whether a subset of $\mathbb{C}$ is a domain?

Answer 1

First $|e^z|=|e^{Re(z)}|=e^{Re(z)}<1 \iff Re(z)<0$ therefore on the complex plane this set is made by all the points below the real axis (the x-axis on the cartesian plane). This kind of problems are all similar, it's sometimes easier to use either this, the trigonometric representation or the algebraic representation of complex numbers to get the solution. I guess you can use the isomorphism between $\mathbb{C}$ and $\mathbb{R}^2$ to find domains. This one is because, using $\mathbb{R}^2$, it is open and connected.

Answer 2

We need to look at some algebra on $\mathbb C$.

$e^{z}=e^{a+bi}=e^{a}*e^{bi}$.

Therefor $|e^{z}|=|e^{a}|*|e^{bi}|$.

From Euler's formula $|e^{bi}|=1$ so $|e^{z}|=|e^{a}|$.

For $|e^{z}|<1$ you need $|e^{a}|<1$ where a is the real part of z.This happens when $a<0$ so your sketch should shade all complex numbers in the domain of $e^{z}$ where the real component is negative.

I believe that the domain of $e^{z}$ is the entire complex plain.