Complex Logs and Roots of Unity 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.

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 $e^{a}\sin b=0 \Rightarrow b=2k\pi \Rightarrow$
$\Rightarrow e^{a}\cos 2k\pi=2\Rightarrow e^{a}=2 \Rightarrow a=\ln 2\Rightarrow$
$\Rightarrow z=\ln 2 +i\cdot 2k\pi ; k\in \mathbf{Z^*}$
