What Defines the Zeros in Sinh() Function?

[phioder]

Hello

How are the sinh() zeros defined?

Unfortunately I don't understand the definitions found, google and wikipedia was searched,
any help would be appreciated

Thank you and best Regards
phioder

 

[mhill]

you could use the fact that sin(-ix)(i)=sinh(x) so the zeros of hyperbolic sine are in\pi for every integer 'n' (complex zeros)

 

[HallsofIvy]

It is not clear to me what you mean by "defining" zeros.

sinh(x), for real numbers, is defined as
\frac{e^x- e^{-x}}{2}
The zeros are 'defined', of course, by
sinh(x)= \frac{e^x- e^{-x}}{2}= 0
which is the same as saying e^x= e^-x or e<sup>2x[/itex]= 1. That leads immediately to x= 0 as the only real zero of sinh(x).

If you expand to complex numbers, it is not to easy to see, from the fact that sinh(x) is defined as
sinh(z)= \frac{e^z- e^{-z}}{2}
and the Cauchy formula
sin(x)= \frac{e^{iz}- e^{-iz}}{2i}
that sinh(z)= (1/i) sin(iz). Since sin(z) is 0 if and only if z is an integer multiple of \pi, sinh(z) is 0 if and only if z is an integer multiple of 2\pi, as mhill said.</sup>