# Complex: Argument and log

1. Oct 6, 2013

### 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.

2. Oct 6, 2013

3. Oct 9, 2013

### M. next

Yes, indeed. Excuse me for the late reply.

4. Oct 9, 2013

### Simon Bridge

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

5. Oct 9, 2013

Yes?

6. Oct 9, 2013

### Office_Shredder

Staff Emeritus
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).

7. Oct 9, 2013

### HallsofIvy

Staff Emeritus
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.

8. Oct 9, 2013

### Simon Bridge

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

... 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: Oct 9, 2013
9. Oct 10, 2013

### M. next

Okay, I understood it now. Great thanks.