Equivalence of Log and Argument for Complex Functions?

[M. next]

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.

 

[Simon Bridge]

http://en.wikipedia.org/wiki/Complex_logarithm

By Imlog do you mean the imaginary part of the logarithm of the complex function?

$\Im(\ln(f(z))$

 

[M. next]

Yes, indeed. Excuse me for the late reply.

 

[Simon Bridge]

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

 

[M. next]

Yes?

 

[Office_Shredder]

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]

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]

Yes?

So... you can now answer your own questions...

You asked:

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.

 

[M. next]

Okay, I understood it now. Great thanks.