What are the complex roots of $x^3-1$?

Question

What are the complex roots of $x^3-1$?

Work I've done so far:

I've set $x = a + bi$. Since $x^3-1=0$, I set $x^3 = (a+bi)^3=1$.

This gives me the following:

(1) $(-ab^2 + a^3) + (2ab^2 + 2a^2b + a^2b - b^3)i$

Which means that I set $(-ab^2 + a^3) = a(a^2-b^2)= 1$ which is also equivalent to

(2) $a(a-b)(a+b)=(a-b)(a^2+ab)=1$.

I also set

(3) $(2ab^2 + 2a^2b + a^2b - b^3) = 0$.

I simplify (3) to

(4) $2b(a^2 + ab) + (a^2 -b^2)b = 0 $

which gives me

(5) $\frac{2b}{a-b} + \frac{b}{a} = 0$ using (2).

Then I get

(6) $\frac{2b}{a-b} = -\frac{b}{a}$. Then I get that $3a=b$. Plugging into (2) I get

(7) $a(a^2 - (3a)^2)=1 = a(a^2 -9a^2) = -8a^3$. So that $a= \frac{-1}{2}$. Now I get that $b= \frac{3}{2}$, which would give me $\frac{1}{2} + \frac{3}{2}i$. But on Wolfram, the imaginary component is close to $.9$. Where am I going wrong?

Answer 1

Hint:

The simpler way is to factorize: $$ x^3-1=(x-1)(x^2+x+1) $$

can you find all the roots?

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Anyway, your algebra is wrong because:

$$ (a+ib)^3=a^3+3a^2(ib)+3a(ib)^2+(ib)^3=a^3-3ab^2+i(3a^2b-b^3) $$

Answer 2

You can just use difference of cubes, which gets the answer much more quickly:

$$a^3-b^3 = (a-b)(a^2+ab+b^2)$$

Applying it here, you get

$$x^3-1 = (x-1)(x^2+x+1) = 0$$

The first factor obviously gives the real root of $x = 1$, so solve for the second factor. It should be pretty straightforward.

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As for your error, you have expanded incorrectly in $(1)$. Recall that

$$(a+b)^3 = a^3+3a^2b+3ab^2+b^3$$

and you apply it to

$$(a+bi)^3$$