Image of Complex Exponential Function

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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:

\(\displaystyle \exp \left({\ln r + i \theta}\right)\) \(=\) \(\displaystyle e^{ \ln r } \left({\cos \theta + i \sin \theta }\right)\) Definition of Complex Exponential Function
\(\displaystyle \) \(=\) \(\displaystyle r \left({\cos \theta + i \sin \theta }\right)\) Exponential of Natural Logarithm
\(\displaystyle \) \(=\) \(\displaystyle z\) 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