sinhz = 0 iff z=n(pi)i (n=0, +/- 1, +/- 2...) ~ a question about this?

Question

I have this weird quesiton in my text book I'm not even sure which part is the question and what i need to do...

It says

"Give details showing that the zeros of sinhz and coshz are as in statements (14) and (15) "


(14) sinhz = 0 iff z=n(pi)i (n=0, +/- 1, +/- 2...)

(15) coshz = 0 iff z=(pi/2 + n(pi)) (n=0, +/- 1, +/- 2...)

Other relevant equations I know that i'll prob need are


-sin(iz) = sinh(z)
cos(iz) = cosh(z)
sinh(z) = [e^z - e^(-z) ] /2
cosh(z) = [e^z + e^(-z)]/2


I haven't made a start on the question because I don't understand the words and what it wants me to calc?

Answer 1

2sinh(z) = e^(a+bi)-e^-(a+bi) = e^a*e^bi -e^-a*e^-bi=

e^a(cos b +i sin b) -e^-a( cos b -i sin b)=
=cos (b) *(e^a-1/e^a) + sin(b) (e^a+1/e^a)

Both parts must be =0 As e^a+1/e^a is never 0 sin b =0
so b= n*pi
sin and cos never are 0 for the same angle so for the real part being 0

e^2a =1 so a= 0
So the roots are for z= n*pi*(i)