Is an exponent of an entire function also an entire function?

I want to prove that if the real part of some complex function $f(z)$ is bounded and $f(z)$ is entire, then the function $f$ is a constant, possibly using the substitution $g(z)=\exp[f(z)]$ and applying Liouville's theorem to $g(z)$.

I'm pretty bad at proofs, but still I've managed to formulate an unnecessarily long version of it. (See below.)

The problem is as follows: I've implicitly assumed that when $f(z)$ is defined everywhere, then also $g(z)=\exp[f(z)]$ is defined everywhere. I can't justify this assumption, and I feel like there's no way around this assumption in one way or another (feel free to judge for yourself from the proof below). So can I simply assume this or do I need an extra step?

Write $g(z)=e^{f(z)}= e^{\operatorname{Re}[f(z)] + i \operatorname{Im}[f(z)]}= e^{\operatorname{Re}[f(z)]}e^{i \operatorname{Im}[f(z)]}$. By assumption, $\lvert\operatorname{Re}[f(z)]\rvert < M$ for all $z$, in other words, $\operatorname{Re}[f(z)]$ is bounded. Since $\operatorname{Re}[f(z)]$ is bounded, $e^{\operatorname{Re}[f(z)]}$ is also bounded. Since this exponent represents the modulus (radius) of $g(z)$, it follows that $g(z)$ is also bounded, say by another constant $M'$, such that $\lvert g(z)\rvert < M'$ for all $z$.

Secondly, we have an analytic function in the exponent. Since the exponent is by definition a power series, we know that $e^z$ is analytic for all $z$. Since $g(z)$ can trivially be expressed in any number $x+iy=z\in\mathbb{C}$, it follows our function $g(z)$ is analytic. Since $g(z)$ is also analytic and bounded, it then follows that \begin{align*} g^{(n)}(z_0) = \frac{n!}{2\pi i} \oint_C \frac{g(z)}{\left(z-z_0\right)^{n+1}} dz. \end{align*} Further, we have the $ML$-inequality, which in our case reads \begin{align*} \left\lvert g^{(n)}(z_0)\right\rvert = \frac{n!}{2\pi} \left\lvert\oint_C \frac{g(z)}{\left(z-z_0\right)^{n+1}} dz\right\rvert \leq \frac{n!}{2\pi} M' \frac1{r^{n+1}}2\pi r \Rightarrow \left\lvert g^{(n)}(z_0)\right\rvert \leq n!\frac{M'}{r^{n}}. \end{align*} Since $g(z)$ is bounded, that is $\lvert g(z)\rvert < M'$, we also have, by the above formula, that $\lvert g'(z_0)\rvert < M'/r$. In the limit $r\rightarrow\infty$ this derivative is zero. We chose $z_0$ arbitrarily, so a derivative in any point $z=z_0$ is zero.

We conclude $g_x= u_x + iv_x =0$ for all $z$ (similar identity holds for derivatives with respect to $y$). Further, we note that $g(z) = e^{\operatorname{Re}[f(z)]+i \operatorname{Im}[f(z)]}$ is a constant, so the argument of the exponent is also a constant. It then follows that $\left(\operatorname{Re}[f(z)]\right)_x+i \left(\operatorname{Im}[f(z)]\right)_x=0$, but we also have $\left(\operatorname{Re}[f(z)]\right)_x=\left(\operatorname{Im}[f(z)]\right)_y$ for our analytic function. It follows that $\left(\operatorname{Im}[f(z)]\right)_y+i \left(\operatorname{Im}[f(z)]\right)_x=0$ and also $\left(1+i\right)\left(\operatorname{Im}[f(z)]\right)_y=0$. Assuming $1+i$ is not zero (safe assumption), we conclude that $\operatorname{Im}[f(z)]$ itself is constant.

Now note that $g(z) = e^{\operatorname{Re}[f(z)]}e^{i \operatorname{Im}[f(z)]}=e^{\operatorname{Re}[f(z)]}\left[\cos\left(\operatorname{Im}[f(z)]\right)+i\sin\left(\operatorname{Im}[f(z)]\right)\right]$. However, since we have previously determined that the imaginary part of $f(z)$ is constant, and the expression for $g(z)$ was also determined to be a constant, it then follows that the real part of $f(z)$ can only be a constant.

In conclusion, both the real and imaginary parts are constant, so $f$ itself is constant.

• You cannot simply assume it, but it is a consequence of results you likely know. The composition of analytic functions is analytic.
– quid
Commented Nov 11, 2015 at 16:46
• @quid Thanks. I guess that was already sufficient (since I used the same argument in the proof elsewhere hehe). Commented Nov 11, 2015 at 17:04