Image of Complex Exponential Function
Theorem
The image of the complex exponential function is $\C \setminus \set 0$.
Proof
Let $z \in \C \setminus \set 0$.
Let $r = \cmod z$ denote the modulus of $z$.
Let $\theta = \map \arg z$ denote the argument of $z$.
Then $r > 0$.
Let $\ln$ denote the real natural logarithm.
Let $e$ denote the real exponential function.
Then:
| \(\ds \map \exp {\ln r + i \theta}\) | \(=\) | \(\ds e^{\ln r} \paren {\cos \theta + i \sin \theta}\) | Definition of Complex Exponential Function | |||||||||||
| \(\ds \) | \(=\) | \(\ds r \paren {\cos \theta + i \sin \theta}\) | Exponential of Natural Logarithm | |||||||||||
| \(\ds \) | \(=\) | \(\ds z\) | Definition of Polar Form of Complex Number |
Hence, $z \in \Img \exp$.
Suppose instead that $z = 0$.
Let $z_0 = r_0 \paren {\cos \theta_0 + i \sin \theta_0} \in \C$.
From Exponential Tends to Zero and Infinity, it follows that $e^{r_0} \ne 0$.
As $\cmod {\cos \theta_0 + i \sin \theta_0} = 1$, it follows that:
- $\cos \theta_0 + i \sin \theta_0 \ne 0$
Then this equation has no solutions:
- $0 = \exp z_0 = e^{r_0} \paren {\cos \theta_0 + i \sin \theta_0}$
Hence:
- $\Img \exp = \C \setminus \set 0$
$\blacksquare$
Also see
Sources
- 2001: Christian Berg: Kompleks funktionsteori: $\S 1.5$
- 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): complex exponential