Entire function dominated by another entire function is a constant multiple

Solution 1:

(1) can actually be generalized to show that no entire function $f$ can dominate another entire function $g$ unless $g = \lambda f$ for some constant $\lambda$. Let $Z(f)$ be the set of zeros of $f$. If $|g| \le |f|$, then: $$ \left|\frac{g(z)}{f(z)}\right| \le 1 \quad \forall z \in \Bbb C - Z(f) $$

But $g/f$ is bounded in a deleted neighborhood of each $a \in Z(f)$. Thus each $a \in Z(f)$ is a removable singularity of $g/f$. Therefore $g/f$ is entire and bounded, hence constant by Liouville's theorem. It follows that $g = \lambda f$.

In your particular question, $f(z) = \exp(z)$ and this function doesn't have any zeros, so it isn't necessary to discuss the zeros of $f$.

Solution 2:

Hint for 2) you have written such a function in your post.

Solution 3:

Hint:

1) Liouville + $\,\displaystyle{\frac{f(z)}{e^z}}\,$ is analytic and bounded...

2) Develop the function in power series and extend analytically into the "other" halpf plane.