Complex analysis - $f$is entire, $f(z)=u+iv$, for all $z\in$ Complex $e^{u(z)}<v(z)$, is $f$ constant? - proof check

Question

I have answer of professor, but I want to check my proof, only to see if it right, if its not, I will just understand my professor answer ( no need to write answer ). let us proof by contradiction. suppose that $f(z)$ is not constant, thus by picard the image is dense.
Thus I Have $|f(z)-f(z_0)|$$<\epsilon $.
$|f(z)|=|f(z)+f(z_0)-f(z_0)|\le |f(z_0)|+\epsilon$ and since $|f(z)|$ is bounded, if $f$ is entire and blocked, so it is constant. in contradiction to my saying that it is not constant. Thus it is constant. The problem with the proof that I did not use any of the info about $e^{u(z)}<v(z)$, which is very weird for me.
where is my logic of my proof goes bad? I guess the proof is maybe wrong, but since something from the basics of contradiction proof is bad.

Answer 1

You just get $|f(z)|<|f(z_0)|+\epsilon$ for some $z$, not for all $z$.

A valid proof: $|e^{(1+i)f(z)}|=e^{\Re (1+i)(u+iv)}=e^{u-v}<\frac 1 e$ because $1+u<e^{u}<v$. Hence, $(1+i)f$ is a bounded entire function.