Complex numbers with trig functions

[thanksie037]

Homework Statement

prove: arctanh(z) = 1/2 ln( (1+z)/(1-z) )

Homework Equations

cosh z = (e^z + e^-z)/2
sinh z = (e^z - e^-z)/2
e^z = e^{x + iy} = e^x(cosy + siniy)

The Attempt at a Solution

cosh z / sin hz = e^z+e^-z/e^z-e^-z

 

[Dick]

tanh(z)=sinh(z)/cosh(z). Not cosh(z)/sinh(z). This is just like the problem of finding the inverse of y=f(x). First you try and solve for x in terms of y then interchange x and y. Try and solve y=(e^x-e^(-x))/(e^x+e^(-x)) for x in terms of y and then interchange x and y. Hint: multiply numerator and denominator by e^x.

 

[thanksie037]

sorry, the problem specifies artanh . . .
solving for e^x won't give me the same properties that using z implies (since z = xi + y)

 

[Dick]

The inverse function of y=tanh(z) is arctanh. You don't need to break z down to it's real and imaginary components. You need to solve that equation for z in terms of y and then replace y with z.

 

[thanksie037]

thanks! I finally got it. i appreiate your help. cool problem.