# Zeros of sinh()

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

## Answers and Replies

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
Science Advisor
Homework Helper
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 ex= e-x or e2x[/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.

• Erebus_Oneiros