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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.

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

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

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

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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? –  ae0709 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
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