Let $f(z)$ be entire function. Consider the functions $e^{if(z)}$ and $e^{−if(z)}$ and applying the Maximum Modulus Theorem, show that if $f(z)$ is real when $|z| = 1$, then $f(z)$ must be a constant function.
(We take $f(z)=u(z)+iv(z)$)
I am confused as so far I have $|g(z)|=|e^{if(z)}|=|e^{-v(z)}|$ and then since $f(z)$ is real, $f(z)=u(z)$ and $v(z)=0$ so I assumed it would follow that $|g(z)|=|e^{v(z)}|=1$.
Similarly, $|g(z)|=|e^{-if(z)}|=|e^{v(z)}|=1$.
Using Liouville I assumed one could say that both $g(z)$ and $h(z)$ are bounded entire functions, they are constant and so it follows that $v(z)$ is constant, meaning that both its partial derivatives are equal to 0 and, due to Cauchy Riemann, both of the partial derivatives of $u(z)$ are equal to zero. It would then follow that $f(z)$ is constant.
I don't know how to go about the question using the Maximum Modulus Theorem, also I feel I am overlooking the importance of $|z|=1$ perhaps?
Any help would be much appreciated!!