Holomorphic function such that $\theta\left(z+\omega_j\right)=a_j\theta\left(z\right)$ satisfies $\theta\left(z\right)=ae^{bz}$. I am trying to solve the following exercise:

Let $\mathbb{Z}\omega_1+\mathbb{Z}\omega_2$ be a lattice in $\mathbb{C}$ and let $\theta$ be an entire function such that there exist $a_1,a_2\in\mathbb{C}$ such that $\theta\left(z+\omega_j\right)=a_j\theta\left(z\right)$ for all $z\in\mathbb{C}$ and $j\in\left\{1,2\right\}$. Prove that $\theta\left(z\right)=ae^{bz}$ for some $a,b\in\mathbb{C}$.

So, if $\theta=0$, there is nothing to prove. Assume $\theta\neq 0$. Observe that $a_1\neq0\neq a_2$ because $\theta\left(z\right)=a_j\theta\left(z-\omega_j\right)$ and $\theta\neq 0$.
If $a_1=a_2=1$ then $\theta$ is an entire elliptic function and therefore it is constant, and this proves what we want (take $b=0$). Assume that at least one $a_j$ is not equal to $1$. It follows that $\theta\neq 0$ cannot be constant, since otherwise $a_1=a_2=1$.
Since $\theta'$ satisfies the same conditions as $\theta$ does, assume we could prove what we want for $\theta'$. If $\theta'=ae^{bz}$, we cannot have $b\neq 0$ since otherwise $\theta=az+c$ for some $c\in\mathbb{C}$, which shows that $a\neq 0$ (because $\theta$ is not constant) and then, using the hypothesis on $\theta$ and comparing coefficients, we would obtain $a=a_ja$, which implies $a_1=a_2=1$, a contradiction. So $\theta'=ae^{bz}$ where $b\neq 0$ (and $a\neq 0$ because $\theta$ is not constant), and therefore $\theta=\frac{1}{b}\theta'+d$ for some $d\in\mathbb{C}$. Using that both $\theta$ and $\theta'$ satisfy the conditions of the statement, one easily obtains $d=a_jd$, so $d=0$ because at least one $a_j$ is not equal to $1$. So $\theta=\frac{1}{b}ae^{bz}$ also has the desired form.
The above paragraph shows that if we prove the problem for $\theta'$ then it is also solved for $\theta$ (assuming that $\theta\neq 0$ and at least one $a_j$ is not equal to $1$). I do not know whether it helps.
Another idea I had was the following: keep assuming $\theta\neq 0$ and at least one $a_j$ is not equal to $1$ (the other cases were already handled) and consider the meromorphic function $\psi=\frac{\theta'}{\theta}$. Since both $\theta$ and $\theta'$ satisfy the conditions of the statement and $a_1\neq 0\neq a_2$ (because $\theta\neq 0$), one easily shows that $\psi$ is an elliptic function. So if we could show that it is entire, we would be done: it would be constant and that would give what we want. But of course, $\psi$ may not be holomorphic: $\theta$ may have some zeros. I thought of using the above paragraph and take enough derivatives of $\theta$ to get rid of the possible zeros, but of course it does not work because taking derivatives may introduce new zeros.
How would you solve this problem?
 A: First choose $b \in \Bbb C$ such that $e^{\omega_1 b} = a_1$ and define $g(z) = e^{-bz} \theta(z)$. Then
$$
 g(z+\omega_1) = e^{-b(z+\omega_1)} a_1 \theta(z) = g(z) \, ,
$$
i.e. $g$ is $\omega_1$-periodic. It follows that
$$
 g(z) = h\left( e^{2\pi i z/\omega_1}\right)
$$
where $h$ is holomorphic in $\Bbb C\setminus \{ 0 \}$. $h$ can be developed into a Laurent series
$$
 h(w) = \sum_{n=-\infty}^\infty c_n w^n \, .
$$
Now we use that $\theta(z+\omega_2) = a_2 \theta(z)$, so that
$$
 g(z+\omega_2) = e^{-b(z+\omega_2)} a_2 \theta(z) = B g(z) 
$$
with $B= e^{-b\omega_2}a_2$. It follows that
$$
 h(e^{2\pi i \omega_2/\omega_1} w) = B h(w)
$$
for all $w \ne 0$. Substituting this into the Laurent series of $h$ implies that
$$
 c_n \left( e^{2\pi i n\omega_2/\omega_1} - B\right) = 0
$$
for all $n$. The absolute value of $e^{2\pi i n\omega_2/\omega_1}$ is not one because $\omega_2/\omega_1$ is not a real number. It follows that $ e^{2\pi i n\omega_2/\omega_1} - B=0$ can hold only for at most one index $n$. So we have
$$
 h(w) = c_m w^m
$$
for some $m \in \Bbb Z$, so that
$$
 g(z) = c_m e^{2\pi i m z/\omega_1}
$$
and
$$
 \theta(z) = c_m e^{bz }e^{2\pi i m z/\omega_1}
$$
has the desired form.
