I know how to directly prove that complex number is one of the cube roots of 1. But the textbook solution also gives another solution as shown below,
However, I don't understand how it jumps to the conclusion that $\frac{-1+\sqrt{3}i}{2}$ is a root of the quadratic equation following the two equations at the top. Thank you!
On
Well maybe it is using Complex conjugate root theorem & Vieta's formula to say that $\alpha+ \bar{\alpha}=-1$ and $\alpha \bar{\alpha}=1$. So $\alpha$ and $\bar \alpha$ are roots of $x^2-(-1)x+1$ or $x^2+x+1$. Hence $\alpha$ divides this which divides $x^3-1$ hence $\alpha$ is a root of $x^3-1$.
Note: $\alpha = \frac{-1+\sqrt3i}2,\;\bar\alpha=\frac{-1-\sqrt3i}2$
On
Do you remember from high school how to find the solutions of $x^2 - 5x + 6 = 0$? What the trick was, was to realize that $x^2 - 5x + 6 = (x - 2)(x - 3)$.
The general form here is that $(x - a)(x - b) = x^2 - (a + b)x + ab$ and hence that $a$ and $b$ are solutions to the equation $x^2 - (a + b)x + ab = 0$.
Here we have the numbers $a = -1/2 + 1/2 \sqrt{3} i$ and $b = \overline{a} = -1/2 - 1/2 \sqrt{3}i$.
The book computes that $a + b = - 1$ so that $-(a+b) = 1$ and $ab = 1$ so that the equation to which $a$ and $b$ are solutions according to the high school trick becomes $x^2 + 1x + 1 = 0$.
Of course, when you were actually in high school you would be given the equation and asked to find solution instead of the other way around, but the reasing is more or less the same.
take $x=\frac{-1+\sqrt{3}i}{2}$ show that $x^3=1$ $$x=\frac12(-1+\sqrt{3}i)\\x^3=\frac{1}{8}((-1)^3+(\sqrt{3}i)^3+3(\sqrt{3}i)(-1)(-1+\sqrt{3}i))=\\ \frac18 (-1-3\sqrt3i+3\sqrt3i-9i^2)=\\\frac18(-1-9i^2)=\\\frac18(-1+9)=\\1$$ other method $$x^3-1=(x-1)(x^2-x+1)=0\\x=1\\x^2-x+1=0\\\to (x-\frac12)^2-(\frac14)+1=0\\(x-\frac12)^2=-(\frac34)\\(x-\frac12)=\pm\frac {\sqrt{-3}}{2}\\ (x-\frac12)=\pm\frac {\sqrt{3}i}{2}\\ x=\frac{1}{2}\pm\frac {\sqrt{3}i}{2}$$