Proof: $f(z)=Ce^{az}$

Question

Let $f$ be a holomorphic function with: $|f(z)|\leq Me^{aRe(z)},$ with $M,a$ positive real constants. It's obvious that $f$ is the complex exponential function , but to prove it, I assume: $$f(z)=Ce^{az}, \quad C\in \mathbb{C} $$ Then: $$\big|Ce^{az}\big| \leq Me^{aRe(z)} \iff \big|C\big|e^{aRe(z)} \leq Me^{aRe(z)} $$ So: $\big|C\big| \leq M.$

Is this proved if I choose an arbitrary $C$ with its modulus less than M, or is it a bad way to handle this proof? Should I use Liouville's theorem?

Answer 1

Yes , Liouville is a good idea ! Put $g(z):= \frac{f(z)}{e^{az}}$. Then $|g(z)| \le M$ for all $z$. Your turn !