- #1

fredrick08

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## Homework Statement

express cosh(z) in terms of functions with real variable only, hence find all complex zeroes and mark on argand diagram.

## The Attempt at a Solution

I did cosh(z)=cosh(x+iy) then got down to cosh(z)=cos(x)cos(y)-sin(x)sin(y)... can anyone confirm this plz? also i got told, that cosh(z)=0, when z=pi*k*i+pi*i/2....when k is a interger... I have no idea, how my friend got this, of if its even correct, or what i have done is correct either..... but i would appreciate any help, especially with drawing the zeroes on the argand diagram, becasue with the answer i have, there is only 2 per 2pi