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?

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    $\begingroup$ I think what this is trying to say is that the sin function is bounded as a function from R to R, but is not bounded as a function from C to C. I am not sure how to reconcile that with the statement of the problem though $\endgroup$ – Ant Jun 19 '18 at 9:46
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    $\begingroup$ You are correct. In fact, since $\sin(z)$ is continuous and $\sin0=0$, we have $|\sin(z)|\le1$ for any $z$ sufficiently close to $0$. The statement in your textbook is obviously wrong. $\endgroup$ – Cave Johnson Jun 19 '18 at 9:48
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    $\begingroup$ $z = i/2$ is an example that you are right and your book is wrong. $\endgroup$ – Alex Silva Jun 19 '18 at 9:50
  • $\begingroup$ No idea whether this is the intended meaning, but if the statement is changed to "if $z \in \mathbb C$ and $-1 \leq \sin z \leq 1$, then $z \in \mathbb R$" (that is, the condition requires $\sin z \in \mathbb R$), then the statement is true. $\endgroup$ – Maxim Jun 19 '18 at 14:01

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


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

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

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


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