Proving $f$ is an identically zero entire holomorphic function Please help me with this question:
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?
 A: 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 $f \equiv 0$.
A: $$
|\,f(z)+\mathrm{e}^z|>|\mathrm{e}^zf(z)|,
$$
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, $f\equiv 0$.
