I have a question, if $f(z)$ is a entire function such that there exists a $C > 0$ such that $|\Im f(z)| \le C$. Show that $f(z)$ is a constant function. Have i made a mistake
I said let $$ g(z)= e^z $$ then $$ e^{\Re f(z)}e^{\Im f(z)}|=|e^{\Im f(z)}|=e^a=c $$
(now I think you apply Liouville’s theorem to show $e^{f(z)}$ is constant)
To show f must also be a constant function, differentiate $e^{f(z)}$ to get $0 = (e^{f(z)})′ = f′(z)e^{f(z)}$. Since $e^{f(z)}$ is never zero, we get $f′(z) = 0$ and so $f$ must be a constant function
Apply $g(z):=e^{iz}$ to $f$ and consider its absolute value: $$ |g(f(z))|=|e^{if(z)}|=|e^{i\Re f(z)-\Im f(z)}|=e^{-\Im f(z)}\le e^C $$ Since both $f$ and $g$ are entire, then $g\circ f$ is entire too. But by computations above it's bounded and thus (by Liouville) is constant.
Hence, since $g$ is clearly nonconstant, we deduce $f$ has to be constant.