# Complex cosh

## 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

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HallsofIvy
Homework Helper
Well, what you have can't be right because cosh(z) is NOT a real number for all complex z, but cos(x)cos(y)-sin(x)sin(y) is real for all real x and y.

$$cosh(x+iy)= \frac{e^{x+iy}+ e^{-x-iy}}{2}= \frac{e^xe^{iy}+ e^{-x}e^{-iy}}{2}$$
$$= \frac{e^xe^{iy}+ e^xe^{-iy}- e^xe^{-iy}+ e^{-x}e^{-iy}}{2}$$
$$= e^x\frac{e^{iy}+ e^{-iy}}{2}- e^{iy}\frac{e^x+ e^{-x}}{2}$$
$$= e^x cos(y)- cosh(x)e^{-iy}$$

Of course, $e^{ix}= cos(x)+ i sin(x)$ so $e^{-iy}= cos(y)- i sin(y)$.

That is, cosh(z)= e^xcos(y)- cosh(x)(cos(y)- i sin(y))= [e^x- cosh(x)]cos(y)- cosh(x)cos(y)]- i cosh(x)sin(y).

since $cosh(x)= (e^x+ e^{-x})/2$,
$$e^x- cosh(x)= e^x- (e^x+e^{-x})/2= (e^x- e^{-x})/2= sinh(x)$$

$$cosh(z)= sinh(x)cos(y)- i cosh(x)sin(y)$$

That will be 0 when sinh(x)cos(y)= 0 and cosh(x)sin(y)= 0.

ahh i see what i did wrong... i forgot about the ix, and converted the wrong terms... ok.. so buy solving those 2 equations for zero, give x=0 and y=(2k+1)pi/2 and second eq gives, y=k*pi, and x=(2k+1)i*pi/2 where k is a integer. is that right... but how do i put that on an argand diagram?

so cosh(z)=0, when z=(2k+1)i*pi/2.... coz that would mean cos((2k+1)pi/2)... which give zero for all half pi's? so is my diagram just a point at iy=I*pi.2 and -I*pi/2.. for the principal values? or is there special way of drawing it?