The question:

Find all solutions $\displaystyle z \in C$ 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 $\displaystyle y = (\frac{1}{2} + k)\pi; k \in Z$

Sub these values into (2):

cosh(0) = 1 (therefore, we need sin(y) = 0)

$\displaystyle sin((\frac{1}{2} + k)\pi) $is never 0.

So it appears that it never equals zero. However, my text states that it does when $\displaystyle z = ik\pi$ Where have I gone wrong? Thanks.