Exercises: Exponential and Logarithm functions – Serlo

Range of the exponential function

Exercise

Show that $\exp ^{-1}\{0\}=\emptyset$

You are supposed to show that $\exp(z)\neq 0$ holds for all $z\in \mathbb {C}$ .

How to get to the proof?

We have already shown that $\exp(a)\neq 0$ holds for all $a\in \mathbb {R}$. Let $z$ be an arbitrary complex number with $z=a+ib$, $a,b\in \mathbb {C}$.

Question: How can we express $\exp(z)$ using $a$ and $b$?

Using the exponential functional equation, we get

$\exp(z)=\exp(a+ib)=\exp(a)\exp(ib)$

As $\exp(a)\neq 0$, $\exp(z)\neq 0$ holds if and only if $\exp(ib)\neq 0$. A good trick to show that a number is not equal to $0$ is to show that its absolute value (or the square of its absolute value) is unequal to $0$.

Question: What is $|\exp(ib)|^{2}$?

Using the computation rules for complex numbers, we get

$\left|\exp(ib)\right|^{2}=\exp(ib)\cdot {\overline {\exp(ib)}}=\exp(ib)\cdot \exp(-ib)=\exp(ib-ib)=\exp(0)=1$

Hence, we have $\exp(ib)\neq 0$ and the claim follows.

Proof

We have already shown that $\exp(a)\neq 0$ holds for all $a\in \mathbb {R}$. Furthermore, for all $b\in \mathbb {R}$, holds that $\left|\exp(ib)\right|^{2}=\exp(ib)\cdot {\overline {\exp(ib)}}=\exp(ib)\cdot \exp(-ib)=\exp(ib-ib)=\exp(0)=1$. Hence, for all $b\in \mathbb {R}$ follows that $\exp(ib)\neq 0$.

Let $z$ be an arbitrary complex number. Then there exist $a,b\in \mathbb {R}$ such that $z=a+ib$. It follows that $\exp(z)=\exp(a+ib)=\underbrace {\exp(a)} _{\neq 0}\cdot \underbrace {\exp(ib)} _{\neq 0}\neq 0$.