# Complex: Argument and log

Hello!

So I have questions on this equivalence:

Imlog[(1+x)/(1-x)] = arg [(1+x)/(1-x)] where x: complex number

How is this true? Is it always applicable no matter what form of complex function is under calculation?

Thank you.

Yes, indeed. Excuse me for the late reply.

Simon Bridge
Homework Helper
Note, a complex number z can always be written in the form z=a+ib: a,b are real.

Yes?

Office_Shredder
Staff Emeritus
Gold Member
2021 Award
If I ask you for the imaginary part of log(z) can you tell me what it is?

From there it should be fairly obvious what the imaginary part of log(f(z)) is in terms of f(z).

HallsofIvy
Homework Helper
I don't see that the "(1+ x)/(1- x)" is really relevant. If z is any complex number, $z= re^{i\theta}$ where "$\theta$" is the "argument" of z. Then $log(z)= log(re^{ix\theta})= log(r)+ i \theta$. That is, $Im(log(z))= \theta$, the argument of z.

Simon Bridge
Homework Helper
Yes?

Is it always applicable no matter what form of complex function is under calculation?
... you want to know if ##\Im\big[ \log[f(a+ib)]\big]=f(a+ib)## for any function f of complex number x=a+ib: a,b, real.

So do the math.
It's actually easier in polar form ... put ##x=Re^{i\theta}##
Work it out for your problem first.

Last edited:
Okay, I understood it now. Great thanks.