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

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

  • $\begingroup$ Okay, so you have three cube roots, and all the possible logarithms take the form $\log\,z+2\pi i\ell$, $\ell \in \mathbb Z$... $\endgroup$ – J. M. is a poor mathematician Nov 7 '11 at 5:08
  • $\begingroup$ Correct me if I'm wrong but isn't $Log[z]=Log|z|+i*arg[z]+i2πn$? $\endgroup$ – warpstack Nov 7 '11 at 5:22
  • $\begingroup$ You're right, that's the explicit decomposition of the logarithm into real and imaginary parts. $\endgroup$ – J. M. is a poor mathematician Nov 7 '11 at 5:31

Let's denote :

$z=a+b\cdot i$

$e^{a+b\cdot i}-1=1\Rightarrow e^{a+b\cdot i}=2 \Rightarrow e^{a} \cdot e^{bi}=2\Rightarrow$

$\Rightarrow e^{a}(\cos b +i\sin b)=2\Rightarrow e^{a}\cos b+ie^{a}\sin b=2 \Rightarrow$

$\Rightarrow e^{a}\cos b=2 $ and $e^{a}\sin b=0 \Rightarrow b=2k\pi \Rightarrow$

$\Rightarrow e^{a}\cos 2k\pi=2\Rightarrow e^{a}=2 \Rightarrow a=\ln 2\Rightarrow$

$\Rightarrow z=\ln 2 +i\cdot 2k\pi ; k\in \mathbf{Z^*}$

  • $\begingroup$ This solution doesn't involve any roots of unity? When you take the cube root of both sides, there are three posisible values that 1 may take, right? $e^{2\pi ik/3}$, $k=0,1,2$... therefore this is only a part of the solution, right? $\endgroup$ – mathmath8128 Nov 7 '11 at 22:24
  • $\begingroup$ I too am apprehensive about this approach. $\endgroup$ – warpstack Nov 8 '11 at 0:48
  • $\begingroup$ I guess technically there are infinitely many roots of unity but for cube roots there are only 3 unique solutions. $\endgroup$ – warpstack Nov 8 '11 at 1:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.