TITLE: Find Complex Roots
QUESTION [1 upvotes]: 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?

REPLY [0 votes]: Only $$z^{12}-1$$ must equal 0.
Also, find the roots of $$z^{4}-1$$ and $$z^{3}-1$$ and remove them from your final solution, since the denominator ≠ 0
{"set_name": "stack_exchange", "score": 1, "question_id": 1599114}