Can an entire, non-constant function map the complex plane to a an open bounded set? Here is a question from Conway.

Let $f$ be entire and non-constant. For any positive real number $c$ show that the closure of $\{z: |f(z)| < c\}$ is the set $\{z: |f(z)| ≤ c\}$.

If an entire, non-constant function can map the complex plane to a an open bounded set, then this is not true.  The only counter example I can think of is $f(z) = \arctan(x) + i\arctan(y)$.  This is a counter example, right?  Specifically this is continuously differentiable,right?  Since the derivative of $\arctan(x)$ is $\frac{1}{1+x^2}$.

NO, A bounded entire function is constant.
