Zeros of sinh()

1. May 27, 2008

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

2. May 27, 2008

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)

3. May 27, 2008

HallsofIvy

Staff Emeritus
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.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook