Solution: We will prove that if the function \( f \) satisfies the given two conditions

1. \( f(z + 1) = f(z) \)
2. \( f(z + \iota) = f(z) \)

for all \( z \in \mathbb{C} \), then \( f \) must be bounded and then Liouville's Theorem will imply that it is constant.

First, note that the conditions imply that \( f \) is periodic with periods \( 1 \) and \( \iota \). Consider the square formed by the periods \( 1 \) and \( \iota \) in the complex plane. The function \( f \) takes the same values at all points inside this square that are equivalent modulo the periods. Therefore, \( f \) is completely determined by its values inside this square. If $S$ is the square of side length $1$, then we have $f(\mathbb{C} ) = f(S)$. As $f$ is continuous and $S$ is compact, so the image must be compact and hence bounded.

Since \( f \) is entire, it is analytic everywhere in the complex plane. By Liouville's theorem, a bounded entire function must be constant. To apply Liouville's theorem, we need to show that \( f \) is bounded. Hence, a nonconstant entire function cannot satisfy both \( f(z + 1) = f(z) \) and \( f(z + \iota) = f(z) \) for all \( z \in \mathbb{C} \).