### Nonvanishing holomorphic function

If $f(x)$ is a holomorphic function on some disk $\Delta$ and $f$ has a zero, it is clear that $f$ cannot be expressed in the form $e^g$ for any function $g$ on this region, for the complex exponential never vanishes. The inverse, however, is also (nontrivially) true. That is, if $f$ is holomorphic and nonvanishing in a disk $\Delta$, then $f=e^g$ for some holomorphic function $g$ defined on $\Delta$.

To see why this is true, we may employ Cauchy's Theorem on disks . In particular, if $f$ is nonvanishing, then its logarithmic derivative is also holomorphic because the denominator never vanishes. This means that for all closed curves $\gamma\subset\Delta$. This implies that if we fix $z_0\in \Delta$, then represents a function of $z$ that does not depend on the choice of path from $z_0$ to $z$. Further, if we consider the function we may easily calculate that $h'(z)=0$ for all $z\in\Delta$, meaning that $h$ is a constant function, and furthermore a nonzero constant since both $f$ and $e^{-g}$ are nonvanishing on $\Delta$. Hence, we have that for all $z\in\Delta$, which may be rewritten in the form for some constant $c\in\mathbb C$. This yields the desired representation of $f$.

# complex-analysis

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