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I need to find all the solutions to the following using logarithms:
$(e^z-1)^3=1$ where z is a complex number.

I am told that using roots of unity I can break this equation down but I must be missing something.

So far...
$c=e^z-1$
$c^3=1$
$c=1^{1/3}e^{i(2 π k/3)}$ ; $k={0,1,2}$
$e^z-1=1^{1/3}e^{i(2 π k/3)}$

And from there I'm stuck, assuming I'm actually making progress. A hint would be swell.

  • 0
    Okay, so you have three cube roots, and all the possible logarithms take the form $\log\,z+2\pi i\ell$, $\ell \in \mathbb Z$...2011-11-07
  • 0
    Correct me if I'm wrong but isn't $Log[z]=Log|z|+i*arg[z]+i2πn$?2011-11-07
  • 0
    You're right, that's the explicit decomposition of the logarithm into real and imaginary parts.2011-11-07

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