# Proving $f$ is an identically zero entire holomorphic function

Given that $f$ is an entire function such that

$\displaystyle|\,f(z)+e^z| \gt |e^zf(z)|$ for all $z\in \mathbb C$

Show that $f$ is a constant zero function i.e. $f(z) = 0$ for all $z$.

Hint: apply Liouville's Theorem

This is my working:

$|f(z)+e^z| \gt |e^zf(z)| = |e^z||f(z)|$

$|f(z)| \lt \displaystyle\frac{|f(z)+e^z|}{|e^z|} = \displaystyle|\frac{f(z)+e^z}{e^z}|$ $=|\displaystyle\frac{f(z)}{e^z} + 1| \le |\frac{f(z)}{e^z}| + 1$

$|f(z)| - |\frac{f(z)}{e^z}| \lt 1$

Please help me check whether my working is correct because i can prove that it is bounded and by Liouville's Theorem, it is constant but how do i prove that it is all zero, i.e. a zero constant function?

What you have done so far is correct, but from

$$\lvert f(z)\rvert - \left\lvert \frac{f(z)}{e^z}\right\rvert < 1,\tag{1}$$

it is not possible to deduce the boundedness of $f$. On the entire left half-plane, $\lvert e^z\rvert < 1$, and therefore $(1)$ holds there regardless of how $f$ behaves there. If we choose $f$ so that $\lvert f(z)\rvert \leqslant 1$ on the closed right half-plane, e.g. $f(z) = e^{-z}$, we have

$$\lvert f(z)\rvert - \left\lvert \frac{f(z)}{e^z}\right\rvert = \lvert e^{-z}\rvert - \lvert e^{-2z}\rvert = e^{-\operatorname{Re} z} - e^{-2\operatorname{Re} z} < 1$$

on all of $\mathbb{C}$, yet $f$ is not bounded.

The given strict inequality

$$\lvert e^z f(z)\rvert < \lvert f(z)+ e^z\rvert\tag{2}$$

implies that $f(z) + e^z$ has no zeros. Hence

$$g(z) := \frac{e^z f(z)}{f(z)+e^z}\tag{3}$$

is an entire function. By $(2)$ it is bounded, and by Liouville's theorem constant,

$$g(z) \equiv c\tag{4}$$

for some $c\in\mathbb{C}$ with $\lvert c\rvert < 1$. So

$$f(z)e^z = c(f(z) + e^z)\quad \text{equivalently}\quad f(z)(e^z - c) = ce^z$$

for all $z \in \mathbb{C}$, and

$$f(z) = \frac{ce^z}{e^z - c} \tag{5}$$

for those $z$ with $e^z \neq c$. The left hand side of $(5)$ is entire, so the right hand side can have only removable singularities. But the right hand side has a pole at points with $e^z = c$, so $e^z \neq c$ for all $z \in \mathbb{C}$, and that means $c = 0$. Inserting that into $(5)$ immediately yields $f \equiv 0$.

• Can you please explain how to use here the identity theorem? Which two functions should I compare? I have $f(z)=\frac{ce^z}{e^z-c}$, I can not take $z_n=2\pi in$ as the accumulation point is $\infty$ – User3231 Apr 30 '18 at 14:43
• @Kiko I don't remember what I intended then. I've edited to complete the argument. (Sorry for the late reply, been ill and away from my computer.) – Daniel Fischer Jun 6 '18 at 14:51

$$|\,f(z)+\mathrm{e}^z|>|\mathrm{e}^zf(z)|,$$ implies that $\,f(z)+\mathrm{e}^z\ne 0$, for all $z$. Set $$g(z)=\frac{\mathrm{e}^zf(z)}{f(z)+\mathrm{e}^z}.$$ Then $g$ is entire analytic and $|g(z)|<1$, for all $z$, and hence $g$ is constant. (Liouville.) Thus, there exists a $c\in\mathbb C$, with $|c|<1$, such that $$\frac{\mathrm{e}^zf(z)}{f(z)+\mathrm{e}^z}=c,$$ or $$(\mathrm{e}^z-c)\,f(z)=\mathrm{e}^z.$$ If $c\ne 0$, then there exists $z_0\in\mathbb C$, such that $\mathrm{e}^{z_0}=c$, and hence $$0=(\mathrm{e}^{z_0}-c)\,f(z_0)=\mathrm{e}^{z_0}\ne 0.$$ Thus $c=0$, and consequently, $f\equiv 0$.