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I want to show that $|\sinh(az)|\leq|\sinh(z)|$ for all $z\in\mathbf{C}$ (or at least for all $z\in\mathbf{H}$, the upper half plane), provided that $0<a<1$ However, I am not even certain whether this holds or not! I am asking, because I want to show that a certain integral vanishes, and it will do if I can prove an upper bound for $|\sinh(az)|/|\sinh(z)|$. Now I ask: Does there exists some (constant) upper bound for all $z$ in question? (I suppose it be 1) If yes, how does one prove it?

My approach is to write $|\sinh(az)|=|\sin(aiz)| = \sin(-ay)\cosh(ax)$ (similarly for $\sinh z$, taking $a=1$) using the identity relating $\sinh$ to $\sin$. But I don't see how to proceed. Any help appreciated.

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1 Answer 1

The inequality doesn't hold in the upper (or the lower) half planes. The ratio $\sinh(az)/\sinh(z)$ has simple poles at $z=n\pi i$ so it blows up near there.

Here's a plot showing some of the regions where $|\sinh(9z/10)/\sinh(z)| > 1$.

enter image description here

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Well, I think this was a silly question of mine. –  user55315 Jan 21 '13 at 11:17
    
Nah :) ${}{}{}$ –  Antonio Vargas Jan 21 '13 at 15:56

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