Image of Complex Exponential Function
Theorem
The image of the complex exponential function is $\C \setminus \left\{ {0}\right\}$.
Proof
Let $z \in \C \setminus \left\{ {0}\right\}$.
Let $r = \cmod z$ be the modulus of $z$, and let $\theta = \arg \left({z}\right)$ be the argument of $z$.
Then $r>0$.
Let $\ln$ denote the real natural logarithm, and let $e$ denote the real exponential function.
Then:
\(\ds \exp \left({\ln r + i \theta}\right)\) | \(=\) | \(\ds e^{ \ln r } \left({\cos \theta + i \sin \theta }\right)\) | Definition of Complex Exponential Function | |||||||||||
\(\ds \) | \(=\) | \(\ds r \left({\cos \theta + i \sin \theta }\right)\) | Exponential of Natural Logarithm | |||||||||||
\(\ds \) | \(=\) | \(\ds z\) | Definition of Polar Form of Complex Number |
Hence, $z \in \operatorname{Im} \left({\exp}\right)$.
Suppose instead that $z=0$.
Let $z_0 = r_0 \left({\cos \theta_0 + i \sin \theta_0 }\right) \in \C$.
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:
- $0 = \exp z_0 = e^{ r_0 } \left({\cos \theta_0 + i \sin \theta_0 }\right)$
Hence, $\operatorname{Im} \left({\exp}\right) = \C \setminus \left\{ {0}\right\}$-
$\blacksquare$
Also see
Sources
- 2001: Christian Berg: Kompleks funktionsteori: $\S 1.5$