If a function is bounded on a set does this mean ut is constant?

Question

Let $H$ be the upper half plane and let $S$ be the union of the first and third quadrants.

(a) Suppose that an entire function $f$ is bounded on $H$. Does it follow that $f$ is constant?

(b) Suppose that an entire function f is bounded on $S$. Does it follow that $f$ is constant?

My attempts:

a) $H$={$x+iy : x \in (-∞,∞),y \in (0,∞)$}

$f$ is bounded on $H$, then there exists a constant $M$ such that for all $z \in H$, $|f(z)| ≤ M$, hence $-M ≤ f(z) ≤ M$.

Let $z_0 \in H$ and so $0≤f(z_0+h)-f(z_0)≤0$

Then lim$_{h → 0} \frac{f(z_0 + h) - f(z_0)}{h}$ = lim$_{h → 0} 0 = 0 = f'(z_0)$ (by squeeze theorem)

So since the derivative is equal to zero ($f'(z) =0$), the function is constant.

b) $S$ = {$x+iy : x\in (-∞,∞), y \in (-∞,∞)$}

And since $f$ is bounded on $S$, then there exists $M$ for all $z \in S$ such that $|f(z)|≤ M$

But isn't this the same as part a)? What's the difference. Please correct me if I'm mistaken on the answerr.

Answer 1

$e^{iz}$ is bounded by $1$ in the upper half-plane and it is nonconstant; $e^{iz^2}$ is bounded by $1$ when $xy \ge 0$ hence in the set $S$

As asked:

$|e^{i(x+iy)}|=e^{-y} \le 1$ when $y \ge 0$

$|e^{i(x+iy)^2}|=|e^{i(x^2-y^2+2ixy)}|=e^{-2xy} \le 1$ when $xy \ge 0$

Answer 2

Hint: $\exp(iz)$ is bounded on the upper half plane.