An entire function satisfies $f(az+b)=f(z)$ Here is a problem that I got stuck on while preparing for an upcoming exam: 
If $a,b\in \mathbb{C}$ and $f:\mathbb{C}\to\mathbb{C}$ is non-constant and entire with $f(az+b)=f(z)$ for all $z\in \mathbb{C}$, prove that there exists a positive integer $n$ such that $a^n=1$. 
I proved the first part of the problem which is the same thing but with $b=0$. I proved this by breaking into the three cases of $|a|<1$, $|a|=1$ and $|a|>1$. The first and last case, I got a contradiction that $f$ is constant (by analytic continuation and Liouville theorem respectively). I am not sure however, how to do it with $b\neq 0$. I would really appreciate a hint. 

Hint: if $a\neq 1$ then $g(z):=f\big(z+b/(1-a)\big)$ satisfies $g(az)=g(z)$.
