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