Tan^(-1)(0) = n(pi)i for n=0,pm 1 pm 2... but I got tan^(-1)(0) = 0 ... where did i go wrong??

Question

The question is.

Find all the values of

Tanh^(-1) (0)

Now in my text the formula for tanh(z) = 1/2 * log[ (1+z)/(1-z)]

so I did tanh(0) = 1/2 * log (1/1)
on my calculator log(1) = 0

so I got zero, but the text book says its vals are n(pi)i (where n = 0, +/- 1 +/- 2.... )

Is there something I'm missing

Answer 1

you are confusing between "tanh" and tan while tanh is the hyperbolic tangent , tan is the trignometric tan

let tanh(z) = y;
y = 1/2 * log[ (1+z)/(1-z)]
log[ (1+z)/(1-z)] = 2y
or (1+z)/(1-z) = exp(2y)

or z = [1 - exp(2y)]/[1 + exp(2y)]

thus the formula for Tanh^(-1) (z) is [1 - exp(2z)]/[1 +
exp(2z)]

put z = 0 Tanh^(-1) (z) = (1-1)/(1+1) = 0

The text book is right when it says tan^-1(0) = n(pi)
BUT that is the trignometric tan, your answer 0 is right for the hyperbolic tan or tanh.

so