Use Liouvilles theorem and the fundamental theorem of calculus to prove that for an entire function $f$, if there exists $M \in \mathbb R: Re(f(z)) \leq M$ $ \forall z \in \mathbb C $,
then $f$ is identically constant on all of $\mathbb C$
My solution:
Consider an auxiliary function $ g(z)= e ^{f(z)} $, where $f(z)= u(x,y)+iv(x,y)$ so we have
$g(z)=e^{u(x,y)+iv(x,y)}$ where $e^{iv(x,y)}=1$ as it is a unit circle and $e^{u(x,y)}\geq e^M$
Using Liouville g'(z)=0 and is constant $\implies$ g is bounded, and hence f is bounded since $f=e^g$
$\implies$ f is identically constant on all of $\mathbb C$
Could someone please tell me if my solution is correct or not and if not, where have I gone wrong?
Help much appreciated. Many thanks
This is basically on the right lines though the way you've written it, there are a few mistakes.
Yeah, let $g(z) = e^{f(x)}$, then as $f$ is holomorphic on $\mathbb{C}$ so is $g$.
Then $||g(z)|| = ||e^{f(z)}|| = ||e^{Re(f(z))}|| \leqslant e^M $ which is finite.
So $g$ is holomorphic and bounded on $\mathbb{C}$, so by Liouville's theorem, $g(z)$ is a constant, say $\alpha$. We note $\alpha$ cannot be zero.
So then $f(z) = $log$(g(z)) = $log$(\alpha) = $ constant, as even though log is multivalued on $\mathbb{C}$, we are given that $f$ is holomorphic so $f$ cannot take values $2\pi i$ away from each other only.
As an easy (hopefully) exercise you might like to now prove the same result for a holomorphic function with bounded imaginary part.