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Taking the usual convention that $z=x+iy$, show that $|\sinh(y)|\leq|\sin(z)|\leq|\cosh(y)|$

What I tried so far was to expand for $\sin(z)$ in terms of complex exponentials and substituting $z=x+iy$ and got the following identity:

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

This means that the absolute value of $\sin(z)$ is given by

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

Given that this identity relates all the terms covered by the inequality I suspect that I may be on the right track however I do not really know how to proceed from here (assuming that I can).

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    $\begingroup$ Pretty good so far. Now recall that $\cosh^2 y = 1 + \sinh^2 y$. $\endgroup$ – Daniel Fischer Jan 25 '18 at 21:35
  • $\begingroup$ Ok I think I got to the inequality but I took some steps that I'm not really sure are legitimate mind if I run them by you? I used the fact that (for real number arguments) $f(x)^2=|f(x)|^2$ where $f$ is any one of the many functions I have in the proof. My reasoning was that since the square maps negatives as though they were positives any way this should not alter the value of $f(x)^2$. Is this reasonable or can I not assume this when taking components of complex numbers? $\endgroup$ – Jepsilon Jan 25 '18 at 21:52
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    $\begingroup$ That's no problem. We have $\lvert a\rvert^2 = a^2$ for all real $a$, regardless of the context in which they arise. And $\lvert w\rvert^2 = (\operatorname{Re} w)^2 + (\operatorname{Im} w)^2$ is correct, no need to take the absolute value of the real and imaginary part before squaring. $\endgroup$ – Daniel Fischer Jan 25 '18 at 21:56
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$\sin^2+\cos^2=1$, so let $θ = \cos^2 x \in [0,1]$ to see that $|\sin z|^2 = θ \sinh^2 y + (1-θ) \cosh^2 y $ is expressed as a convex combination of $\sinh^2 y$ and $\cosh^2 y$. Finally, on $\mathbb R$, $|\sinh| ≤ |\cosh|$, so the inequality follows from the fact that convex combinations are in between the two endpoints.

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