Setting z= i x is sinh z = i sin x so that the function vanishes for x= k -> z= i k ...
Kind regards
The question:
Find all solutions of sinh(z) = 0
My attempt:
sinh(z) = sinh(x)cos(y) + icosh(x)sin(y)
Equating real and imaginary parts:
sinh(x)cos(y) = 0 (1)
cosh(x)sin(y) = 0 (2)
For (1):
sinh(x) = 0 when x = 0, and cos(y) = 0 when
Sub these values into (2):
cosh(0) = 1 (therefore, we need sin(y) = 0)
is never 0.
So it appears that it never equals zero. However, my text states that it does when Where have I gone wrong? Thanks.