Chapter 9.3, Problem 62AYU

Find the six complex sixth roots of unity (1) and plot them.

To determine

To find: The six complex fourth roots of unity $\left(1\right)$ and plot them.

Answer to Problem 62AYU

Solution:

$1, -1, i, -i$

Explanation of Solution

Let $x$ be the sixth root of unity $x=\sqrt[6]{1}={\left(1\right)}^{\frac{1}{6}}$.

$x={\left(1\right)}^{\frac{1}{6}}$, taking fourth power of equation,

$x^6={\left(1^6\right)}^{\frac{1}{6}}$

$x^6=1$

$x^6-1=0$

$x^6-1^6=0$

Put $x^3=t$; $t^2-1^2=0$.

$\left(t+1\right)\left(t-1\right)=0$ by $a^2-b^2=\left(a+b\right)\left(a-b\right)$

$\left(x^3+1^3\right)\left(x^3-1^3\right)=0$

$x^3-1^3=0$

$\left(x-1\right)\left(x^2+x+1\right)=0$

by $a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)$

$x=1$; $x^2+x+1\text{ = 0}$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-1\pm \sqrt{1-4}}{2}$

$x=\frac{-1\pm \sqrt{3}i}{2}$

$x^3+1^3=0$

$\left(x+1\right)\left(x^2-x+1\right)=0$

by $a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)$,

$x=-1$; $x^2-x+1\text{ = 0}$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{1\pm \sqrt{1-4}}{2}$

$x=\frac{1\pm \sqrt{3}i}{2}$

Therefore the roots are $1, -1, \frac{-1\pm \sqrt{3}i}{2}, \frac{1\pm \sqrt{3}i}{2}$.