Entire functions such that $f(z+\lambda_1)=af(z) \; , \; f(z+\lambda_2)=bf(z) $

Question

Let $\lambda_1$ and $\lambda_2$ be complex numbers with nonreal ratio. Let $f(z)$ be an entire function and assume there are constants $a$ and $b$ such that $$f(z+\lambda_1)=af(z) \;\;\;\;,\;\;\;\; f(z+\lambda_2)=bf(z) $$ for all $z$. Prove that $f(z)=Ae^{Bz}$ , where $A$ and $B$ are constants.

I think the idea is to build a doubly periodic entire function, but I'm not sure if it is $\displaystyle \frac{f(z)}{e^{Bz}}$.

This is exercise 10 from chapter 1 of the book "Modular forms and Dirichlet series" by Tom Apostol.

Any help would be appreciated.

Answer 1

Hint: show there exists a $B$ such that $e^{B\lambda_1} = a$ and $e^{B\lambda_2} = b$. Then consider the function $f(z)/e^{Bz}$, and use what you know about doubly periodic entire functions.

Answer 2

First we solve the equations $$ Ae^{B\lambda_1}=a,\quad Ae^{B\lambda_2}=b, $$ and get $$ B=\frac{\ln (a/b)}{\lambda_1-\lambda_2}. $$ Then consider the doubly periodic entire function $$ f(z)e^{-Bz}. $$ It is constant according to Liouville's theorem.