Complex Equation Solution: Finding Values for Z using Trig Identities

[Phillipe]

Homework Statement

Find all the values of Z which satisfy the equation, Z being a complex number in the form x+i*y.

Homework Equations

Every trig identity out there.

The Attempt at a Solution

Here's what I got so far:

Cosh(z) = i*Cos(3)
Cosh(x)*Cos(y)+i*(Sinh(x)*Sin(y)) = i*Cos(3)

Therefore,

Cosh(x)*Cos(y) = 0
Sinh(x)*Sin(y) = i*Cos(3)

y = Pi/2 (since there's no value of x which can give Cosh(x) = 0). y may also be Pi/2 + 2nPi, where n is an integer.

Therefore,
Sinh(x) = i*Cos(3)

But I don't know how to proceed from there.

 

[Samuelb88]

Use the identity sinh(x) = .5[exp(x)-exp(-x)]. Identities for sinh(x), and cosh(x) can be found on wikipedia.

 

[Phillipe]

Use the identity sinh(x) = .5[exp(x)-exp(-x)]. Identities for sinh(x), and cosh(x) can be found on wikipedia.

I have thought of using that, but I don't know how to get the values for x from there. Would you be so kind to tell me how?

 

[Samuelb88]

I have thought of using that, but I don't know how to get the values for x from there. Would you be so kind to tell me how?

Sure. After we replace sinh(x) with .5[exp(x)-exp(-x)], let's first multiply both sides of the equation by 2, then by e^x. Set everything equal to zero (isolate all terms on one side) which should yield:
e^2^x-something*e^x-1=0
Can you see what to do from here?

 

[Phillipe]

Sure. After we replace sinh(x) with .5[exp(x)-exp(-x)], let's first multiply both sides of the equation by 2, then by e^x. Set everything equal to zero (isolate all terms on one side) which should yield:
e^2^x-something*e^x-1=0
Can you see what to do from here?

I think I just saw it. I get:

e^2^x-e^x*2cos(3)-1 = 0

which stinks of quadratic equation.Things could get messy with that cos(3), though, but I'll take a shot. Thanks a lot, dude!