Is it true that $\lvert\sin z\rvert\le 1$ implies that $z\in\mathbb R$?

Question

My book says

Prove that if $z\in\mathbb C$ and $\lvert\sin z\rvert\le 1$, then $z\in\mathbb R$.

But I think this can't be true as $$\lvert\sin z\rvert^2=\sin^2x+\sinh^2y$$ and so if $\lvert\sin z\rvert\le 1$, then, $$\sinh^2y\le1-\sin^2x=\cos^2x.$$

Clearly we can find some $y\neq 0$, such that $\sinh^2y\le\cos^2x$ for some $x$.

Thus I want to know if something went wrong in my explanations?

Answer 1

You're right: just take $y$ such that $-1\le\sinh y\le 1$, which is obviously possible by the intermediate value theorem; then $$ \lvert\sin(iy)\rvert=\lvert\sinh y\rvert\le 1 $$

Answer 2

You are right, indeed we have that

$$\sin z=\sin(x+iy)=\sin x\cos (iy)+\cos x\sin(iy)=\sin x\cosh y+i\cos x\sinh y$$

and thus

$$|\sin z|^2=\sin^2 x\cosh^2 y+\cos^2 x\sinh^2 y=$$$$=\sin^2 x(\cosh^2 y-\sinh^2 y)+\sinh^2 y=\sin^2 x+\sinh^2 y$$

and the given condition is true only for $x=\frac{\pi}2+k\pi \implies \sin^2 x=1$.