I was doing my homework:
If $w_1, w_2\in\mathbb{C}$ satisfies $\mathrm{Im}\frac{w_1}{w_2}\neq0$ , $f$ is an entire function and there are $a,b$ such that
\begin{array}{ll} f(z+w_1)=af(z) \\ f(z+w_2)=bf(z) \end{array}
Show that $f=Ae^{Bz}$.
I gave the following solution and my teacher said that is wrong and I don't know why.
If $f$ has no zeros, then $\frac{f'}{f}$ is double period entire function and hence $\frac{f'}{f}=B$, so $f=Ae^{Bz}$.
For the general case, according the previous discussion, $g=e^{Bz}$ sastifies
\begin{array}{ll} g(z+w_1)=ag(z) \\ g(z+w_2)=bg(z) \end{array}
so $\frac{f}{g}$ is double period entire funciton and hence $f=Ag$, i.e, $f=Ae^{Bz}$.
I don't understand why my solution is wrong. I am wondering if anyone could point the error. Any help will be appreciated.
From \begin{align}f(z+w_1)&=af(z)\\f(z+w_2)&=bf(z)\end{align} it follows that \begin{align}f(z+w_1)e^{-\frac{\log(a)}{w_1}(z+w_1)}&=e^{-\frac{\log(a)}{w_1}(z+w_1)}e^{\log(a)}f(z)\\f(z+w_2)e^{-\frac{\log(b)}{w_2}(z+w_2)}&=e^{-\frac{\log(b)}{w_2}(z+w_2)}e^{\log(b)}f(z)\end{align} and \begin{align}f(z+w_1)e^{-\frac{\log(a)}{w_1}(z+w_1)}&=e^{-\frac{\log(a)}{w_1}z}f(z)\\f(z+w_2)e^{-\frac{\log(b)}{w_2}(z+w_2)}&=e^{-\frac{\log(b)}{w_2}z}f(z).\end{align}
Rewriting gives \begin{align}f(z+w_1)e^{-\frac{\log(a)}{w_1}(z+w_1)}&=e^{-\frac{\log(a)}{w_1}z}f(z)\\f(z+w_2)e^{-\frac{\log(b)}{w_2}(z+w_2)}&=e^{-\frac{\log(b)}{w_2}z}f(z)\end{align}
Excuse me, I don't know the answer yet. Hopefully an edit follows.