Show that the exists an unbounded open subset of the complex plane where sinz is bounded

Question

$$ \sin z = \frac{e^{iz}-e^{-iz}}{2i} $$

So I think $ \sin z \le 1/2 + 1/2 = 1 $.

But I am not sure how to find an unbounded subset of the complex plane, is it not bounded on the whole

Answer 1

Let $A = \{ z : |\sin z| < 2 \}$. Then $\sin z$ is obviously bounded on $A$, and $A$ is unbounded (since it contains $\mathbb{R}$) and open (since it is the inverse image of the open set $\{ t < 2 \} \subset \mathbb{R}$ under the continuous function $f(z) = |\sin z|$.)