Proposition 27.6.5.label Let $f \in H(\complex; \complex)$, then the following are equivalent:
- (1)
$f(z) \ne 0$ for all $z \in \complex$.
- (2)
There exists $\ell \in H(\complex; \complex)$ such that $f = e^{\ell}$.
Proof. (1) $\Rightarrow$ (2): For each $z \in \complex$, let
\[g(z) = \int_{0}^{w} \frac{f'(w)}{f(w)}dw\]
then $g \in H(\complex; \complex)$ with $g' = f/f'$, and
\begin{align*}\frac{d}{dz}\braks{f(z)e^{-g(z)}}&= f'(z)e^{-g(z)}- f(z)g'(z)e^{-g(z)}\\&= f'(z)e^{-g(z)}- f'(z)e^{-g(z)}= 0\end{align*}
Therefore $fe^{-g}$ is a non-zero constant. For any $\lambda \in \complex$ with $e^{\lambda} = fe^{-g}$, let $h = g + \lambda$, then $f = e^{h}$.$\square$
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