Equivalence of Definitions of Complex Natural Logarithm

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:

Thus:
 * $w \in G$

and so:
 * $f \subseteq G$

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:

Thus $w$ is of the form:
 * $\ln r + i \theta + k \pi i$

where $k = 0$.

Therefore:
 * $w \in F$

and so:
 * $G \subseteq F$

So by definition of set equality:
 * $F = G$