On a pseudo periodic entire function being an exponential function Let $p_1,p_2\in \mathbb C$ be linearly independent over $\mathbb R$. Let $f: \mathbb C \to \mathbb C$ be an entire function such that $f(0)=f'(0)=1$ and $f(z+p_1)=af(z) $ and $f(z+p_2)=bf(z)$ , $\forall z \in \mathbb C$, where $a,b\in \mathbb C$ are some constants. 
Then is it true that $f(z)=e^z, \forall z \in \mathbb C$ i.e. $f'(z)=f(z),\forall z \in \mathbb C$ i.e. $f(z+w)=f(z)f(w),\forall z,w \in \mathbb C$ ? 
My try: We have $a=f(p_1)$ , $b=f(p_2)$. So $f(n_1p_1+n_2p_2)=a^{n_1}b^{n_2}, \forall n_1,n_2 \in \mathbb Z$. Moreover, we can write any $z\in \mathbb C$ uniquely as $z=x_1p_1+x_2p_2$ for some unique  $x_1,x_2 \in \mathbb R$ . But I am unable to proceed further. Please help. 
 A: Write $a=a_1,b=a_2$ for brevity. Consider
$$ \frac{1}{2\pi i} \int_C \frac{f'(z)}{f(z)} \, dz. $$
This is the famous "zero–pole-counting integral": it gives the number of zeros of $f$ inside $C$, with multiplicity, minus the number of poles of $f$ inside $C$. If we choose $C$ to be a period parallelogram, we see that $f'(z+p_i)/f(z+p_i) = f'(z)/f(z)$, so the integrals along opposite sides cancel, since they have the same values of the function but are taken in opposite directions. Hence the whole integral is equal to zero. Since $f$ is analytic, it has no poles, and therefore this calculation implies that it also has no zeros. We may therefore write $f(z) = e^{g(z)}$ for some analytic function $g$, and we have
$$ g(z+p_i) = g(z)+\log{a_i} =: g(z)+\nu_i $$
for some fixed values of $\log{a_i}$. It remains to determine $g$. Referring to the condition in the question, we know that we can choose $g(0)=0$, and are forced to take $g'(0)=1$, so $h(z)=g(z)/z$ can be made analytic at zero by adding $h(0)=1$. It is hence analytic everywhere. But it is also bounded, since 
$$g(z) = g(z-(k_1p_1+k_2p_2))+k_1 \nu_1 + k_2 \nu_2 $$
 is bounded by an affine function of $z$. Hence by Liouville's theorem, it is constant, so we find $h(z)= 1$, $g(z)= z$, $f(z)=e^z$, and we must also have $a=e^{p_1}$, $b=e^{p_2}$.
The same type of result holds even if we don't have any conditions at $z=0$: the only possible functions are $e^{\lambda z+\mu}$, but we still have $p_i=\lambda\log{a_i}$, so we need it to be possible to choose values for the logarithms so that $p_1\log{a_1}-p_2\log{a_2}=0$ (consider the determinant of the system $(p_i,\log{a_i}) (1,-\lambda)^T = 0$).
