Find Complex Roots 

Question:
Find the complex roots of $$ {(z^{12} -1)\over (z^4-1)(z^3-1)}  = 0  $$


What I have attempted:
$$ {(z^{12} -1)\over (z^4-1)(z^3-1)} = 0  $$
$$ {(z^{6} -1)(z^{6} +1)\over (z^2-1)(z^2+1)(z^3-1)} = 0  $$
$$ {(z^{3} -1)(z^3 +1)(z^6+1)\over (z+1)(z-1)(z^2+1)(z^3-1)} = 0  $$
$$ {(z^3 +1)(z^6+1)\over (z+1)(z-1)(z^2+1)} = 0  $$
$$ {(z+1)(z^2-z+1)(z^6+1)\over (z+1)(z-1)(z^2+1)} = 0 $$
$$ {(z^2-z+1)(z^6+1)\over (z-1)(z^2+1)} = 0 $$
How should I continue?

A fraction is zero if and only if its numerator is zero, and well-defined if and only if its denumerator is non-zero.
Therefore
$$
  \frac{z^{12} - 1}{(z^4 - 1)(z^3 - 1)} = 0
  \iff
  z^{12} = 1, z^4 \neq 1, z^3 \neq 1.
$$
The solutions of the first condition are given by the $12$-th root of unity, namely
$$
  z = e^{k 2 \pi i / 12}
  \qquad
  k = 0, 1, \dotsc, 11.
$$
We have to exlude the fourth roots of unity and third roots of unity for the other two conditions.
These are
$$
  z
  = e^{k 2 \pi i / 4}
  = e^{(3k) 2 \pi i /12}
  \quad\text{for $k = 0, 1, 2, 3$}
$$
and
$$
  z
  = e^{k 2 \pi i / 3}
  = e^{(4k) 2 \pi i / 12}
  \quad\text{for $k = 0, 1, 2$}.
$$
Thus all we are left with is
$$
  z
  = e^{k 2 \pi i / 12}
  \qquad
  k \in \{1,2,5,7,10,11\}.
$$
