Equivalence of Log and Argument for Complex Functions?

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Discussion Overview

The discussion revolves around the equivalence of the imaginary part of the logarithm of a complex function and the argument of that function, specifically examining the expression Imlog[(1+x)/(1-x)] = arg[(1+x)/(1-x)], where x is a complex number. Participants explore whether this equivalence holds for all forms of complex functions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant questions the general applicability of the equivalence between the imaginary part of the logarithm and the argument of a complex function.
  • Another participant clarifies that Imlog refers to the imaginary part of the logarithm of the complex function.
  • A participant notes that any complex number can be expressed in polar form, suggesting that the argument can be derived from this representation.
  • There is a suggestion that the relevance of the specific function (1+x)/(1-x) may not be critical, as the relationship holds for any complex number in polar form.
  • One participant encourages others to perform calculations to verify the equivalence for specific functions.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the equivalence, with some providing clarifications and others questioning the generality of the statement. The discussion does not reach a consensus on whether the equivalence is universally applicable.

Contextual Notes

Some participants highlight the importance of expressing complex numbers in polar form to derive the argument, but the discussion does not resolve the implications of this representation on the original question.

Who May Find This Useful

This discussion may be of interest to those studying complex analysis, particularly in understanding the properties of logarithmic and argument functions in relation to complex numbers.

M. next
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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.
 
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Yes, indeed. Excuse me for the late reply.
 
Note, a complex number z can always be written in the form z=a+ib: a,b are real.
 
Yes?
 
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).
 
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.
 
M. next said:
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.
 
Last edited:
Okay, I understood it now. Great thanks.
 

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