Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

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

1 Answer 1

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

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

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

share|improve this answer
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? –  mathmath8128 Nov 7 '11 at 22:24
I too am apprehensive about this approach. –  warpstack Nov 8 '11 at 0:48
I guess technically there are infinitely many roots of unity but for cube roots there are only 3 unique solutions. –  warpstack Nov 8 '11 at 1:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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