Equivalence of Definitions of Complex Natural Logarithm

Theorem

The following definitions of the concept of Complex Natural Logarithm are equivalent:

Definition 1

Let $z = r e^{i \theta}$ be a complex number expressed in exponential form such that $z \ne 0$.

The complex natural logarithm of $z \in \C_{\ne 0}$ is the multifunction defined as:

$\map \ln z := \set {\map \ln r + i \paren {\theta + 2 k \pi}: k \in \Z}$

where $\map \ln r$ is the natural logarithm of the (strictly) positive real number $r$.

Definition 2

Let $z \in \C_{\ne 0}$ be a non-zero complex number.

The complex natural logarithm of $z$ is the multifunction defined as:

$\map \ln z := \set {w \in \C: e^w = z}$

Proof

Let $z = r e^{i \theta}$ such that $z \ne 0$.

Let $F = \set {\ln r + i \theta + 2 k \pi i: k \in \Z}$.

Let $G = \set {w \in \C: e^w = z}$.

We will demonstrate that $F = G$.

Definition 1 implies Definition 2

Let $w = x + i y$ such that $w \in F$.

Then:

$x + i y = \ln r + i \theta + 2 k \pi i$

for some $k \in \Z$.

Equating real and imaginary parts:

$x = \ln r$
$y = \theta + 2 k \pi$

Then:

 $\displaystyle e^w$ $=$ $\displaystyle e^{x + i y}$ $\displaystyle$ $=$ $\displaystyle e^x e^{i y}$ Exponential of Sum $\displaystyle$ $=$ $\displaystyle e^{\ln r} e^{i \paren {\theta + 2 k \pi} }$ Definition of $w$ $\displaystyle$ $=$ $\displaystyle e^{\ln r} e^{i \theta}$ Periodicity of Complex Exponential Function $\displaystyle$ $=$ $\displaystyle r e^{i \theta}$ Exponential of Natural Logarithm $\displaystyle$ $=$ $\displaystyle z$ Definition of $z$

Thus $w \in G$ and so $f \subseteq G$.

$\Box$

Definition 2 implies Definition 1

Let $w \in G$.

By definition:

$\exists z \in \C_{\ne 0}: z = e^w = r e^{i \theta}$

Thus:

 $\displaystyle r e^{i \theta}$ $=$ $\displaystyle e^{\ln r} e^{i \theta}$ Exponential of Natural Logarithm $\displaystyle$ $=$ $\displaystyle e^{\ln r + i \theta}$ Exponential of Sum $\displaystyle$ $=$ $\displaystyle e^w$ by definition

Thus $w$ is of the form $\ln r + i \theta + k \pi i$, where $k = 0$.

Therefore $w \in F$, and so $G \subseteq F$.

$\Box$

So by definition of set equality:

$F = G$

$\blacksquare$