Prove that $f(z) = e^z$ maps $\mathbb{C} \setminus B $ onto $\mathbb{C} \setminus \{0\}$.

Question

Prove that $f(z) = e^z$ maps $\mathbb{C} \setminus B $ onto $\mathbb{C} \setminus \{0\}$ where $B$ is a bounded subset of $\mathbb{C}$.

I have no idea how to even attempt this question so I would greatly appreciate some hints.

Answer 1

Hints:

What is the image of the exponential function as a function $\mathbb R\to\mathbb R$?

What is the image under $f$ of a closed vertical line segment in $\mathbb C$ of height $2\pi$?

Answer 2

Let $z=x+yi$ and $w=u+iv=\rho(\cos\theta+i\sin\theta)$ ($\rho>0$). Assume that $e^z=w$ and then $$ e^x(\cos y+i\sin y)=\rho(\cos\theta+i\sin\theta) $$ from which one has $$ e^x\cos y=\rho\cos\theta, e^x\sin y=\rho\sin \theta. $$ Thus $$ x=\frac12\ln\rho,y=\theta+2n\pi, n\in Z.$$ So for any $w=\rho(\cos\theta+i\sin\theta)\in\mathbb{C}\setminus\{0\}$, one always can choose $n\in Z$ such that $z=\ln\rho+i(\theta+2n\pi)\in\mathbb{C}\setminus B$; namely $f(z)=e^z$ is from $\mathbb{C}\setminus B$ onto $\mathbb{C}\setminus\{0\}$.