Exercises: Exponential and Logarithm functions – "Math for Non-Geeks"

Range of the exponential function

Exercise

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

You are supposed to show that $\exp (z)\ne 0$ holds for all $z\in ℂ$ .

How to get to the proof?

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

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 $\exp (a)\ne 0$, $\exp (z)\ne 0$ holds if and only if $\exp (ib)\ne 0$. 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 ${|\exp (ib)|}^2=\exp (ib)\cdot \overline{\exp (ib)}=\exp (ib)\cdot \exp (-ib)=\exp (ib-ib)=\exp (0)=1$

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

Proof

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

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