In need of theorem recommendations to prove that f(z) maps $\Bbb C \setminus B$ maps to $\Bbb C \setminus \{0\}$

Question

Given a bounded set B of the complex plane, I want to prove that $f(z)=e^z$ maps $\Bbb C \setminus B$ ont $\Bbb C \setminus \{0\}$.

I have no idea where to start with this.

Does anyone have any recommendations on what theorems would be useful here ?

Answer 1

Without loss of generality, assume that $B=D(0,n)$, where $n\in\mathbb{N}$. Indeed there is no loss of generality: since $B$ is bounded, there does exist $n\in\mathbb{N}$ such that $B\subset D(0,n)$, therefore $\mathbb{C}\setminus D(0,n)\subset\mathbb{C}\setminus B$. If we show that $\mathbb{C}\setminus D(0,n)$ is mapped on $\mathbb{C}\setminus 0$, then the hyperset $\mathbb{C}\setminus B$ will be also mapped on $\mathbb{C}\setminus 0$.

But things are simple now: Remember that $z\mapsto e^z$ maps vertical lines to circles, particularly it maps the vertical line $Re(z)=x_0$ to the circle $|z|=e^{x_0}$. Recall also that any interval of length $2\pi$ on the vertical line $Re(z)=x_0$ is mapped onto the circle $|z|=e^{x_0}$ (in other words, the whole circle is the image of any interval on this vertical line of length $2\pi$).

Let $w\in\mathbb{C}\setminus\{0\}$ such that $w=re^{i\theta}$ with $r>0$. Since $D(0,n)$ is a bounded disc, there is at most a finite segment of the horizontal line $Re(z)=\log(r)$ included in this disc. So there exists a segment of this line of length $2\pi$ that is included in $\mathbb{C}\setminus D(0,n)$. As explained above, there exists $z_0$ in this segment such that $e^{z_0}=w$, since $w\in \{z: |z|=r\}$.

Answer 2

Here's a semi-formal argument. Let $B$ be a bounded set in $\mathbb{C}$. Then $\exists M_r,M_i \in \mathbb{R}$ such that forall $b \in B$, $|\text{Im}(b)| < M_i$ and $|\text{Re}(b)| < M_r$. The important part here is that you can bound the imaginary part of every number in $B$. Recall that for some $z = x + iy$, the real part determines the magnitude of $e^z$ and the imaginary part determines the polar angle of $e^z$. Since $e^z = e^x e^{iy}$, we only need a continuous range of $2\pi$ on $y$ to cover all possible polar angles, and $x$ needs to take on all real values in order to cover all possible magnitudes (except $0$ of course).

Can you see now why $e^z$ maps $\{a + bi \mid a \in \mathbb{R} , b \in [M_i, M_i + 2\pi]\} \subseteq \mathbb{C} \setminus B$ onto $\mathbb{C} \setminus \{0\}$?