I know what the third roots of unity are but I want to solve this exercise:
Complex numbers can be written as $a + bi$. Simplify $(a + bi)^3$ and
equate the real part to $1$ and the imaginary part to $0$ to find the
three roots of unity.
So I expanded $(a + bi)^3$ to get $(a^3 – 3ab^2) + (3a^2b – b^3)i$.
I know that $a = 1$ and $b = 0$ is a solution just by looking at it. Now is there some sort of clever way to get another root of unity? I tried to do it with just algebra but it ended up being very messy. I was wondering whether there is some neat solution to this.
Best Answer
You have the equations $$a^3-3ab^2=1\\3a^2b-b^3=0$$ Having identified $a=1,b=0$ as one solution, you can now note that $b\neq 0$ in any other solutions. You can then divide the second one by $b$ and get $$3a^2=b^2$$ and plug that into the first to get $$a^3-9a^3=1\\a=-\frac 12\\b^2=\frac 34\\b=\pm \frac{\sqrt 3}2$$