Solution: Let us write $z = x + \iota y$. Then we can find integers $m \text{ and } n$ such that \[ \vert x - n a \vert \leq a \text{ and } \vert y - m a \vert \leq a. \] Since $f$ is periodic we have \begin{align*} f(z) & = f(z + \iota a) = f(x - na + \iota (y - ma)). \end{align*} Therefore, the image of $f$ is lying in the square of side length $a$ (look at the figure below). Thus, the image of $f$ is compact and hence $f$ is bounded. As $f$ is an entire function and it is bounded also, so by the Liouville's theorem, $f$ must be constant. Hence, for any $z \in \mathbb{C} $, \[ f(z) = f(\iota ) = \iota \implies f(a) = \iota . \]