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So I have questions on this equivalence:

*Im*log[(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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- Thread starter M. next
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So I have questions on this equivalence:

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

Thank you.

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Simon Bridge

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##\Im(\ln(f(z))##

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Yes, indeed. Excuse me for the late reply.

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Simon Bridge

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

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Yes?

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Office_Shredder

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

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HallsofIvy

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- #8

Simon Bridge

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

You asked:

... 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.Is it always applicable no matter what form of complex function is under calculation?

So do the math.

It's actually easier in polar form ... put ##x=Re^{i\theta}##

Work it out for your problem first.

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Okay, I understood it now. Great thanks.

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