Can I get help with this question? z^6 + (i+1)z^3 + i =0
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let w = z^{3}, and use the quadratic formula to solve for w. then find the cube roots of w (there will be 3 for each solution you find for w, use demoivre).
How do I use the quadratic formula on w^2 + (i +1)w+ i? Would it be a=1,b=(i +1) and c=i?
Another approach: I would factor as follows: Can you proceed from here?
yes thank you
Originally Posted by frigid How do I use the quadratic formula on w^2 + (i +1)w+ i? Would it be a=1,b=(i +1) and c=i? yes, the quadratic formula works in ANY field, as long as 1+1 ≠ 0. so we have: the two square roots of -2i are: -1+i and 1-i (check this by squaring them). this gives: . since w = z^{3}, we are solving which is exactly where MarkFL2's solution leads you to.
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