Help with cube roots of 1 please

[Guest]

Can you help to explain this question please?

Factorise \(\displaystyle \L x^3 - 1\), find the three cube roots of 1 and illustrate them on an Argand diagram.

Factorised: \(\displaystyle \L (x - 1)(x^2 + x + 1)\)

To solve, I equated it to zero:
\(\displaystyle \L (x - 1)(x^2 + x + 1) = 0\)

I worked out two of the roots using the formula:

\(\displaystyle \L x = \frac{{ - 1 \pm \sqrt {1 - 4} }}{2} = - \frac{1}{2} \pm \frac{{i\sqrt 3 }}{2}\)


The third root I thought would be 1, looking at (x-1) in the expression, but I'm told it is 0. Can you explain to me why that is?

Also, how would you work out the angles in order to draw them on an argand diagram?

Thank you

 

[pka]

No, you are correct! The nunber 1 is a cube root of unity.
To diagram, use the ordinary xy-axis but label the x-axis as Re and y-axis as Im.
The number 1 is the point (1,0).

 

[Guest]

Thanks for your help

 

[soroban]

Hello, Amber!

Factor \(\displaystyle \, x^3\,-\,1\) and find the three cube roots of 1
Illustrate them on an Argand diagram.

Factorised: \(\displaystyle \,(x\,-\,1)(x^2\,+\,x\,+\,1)\)

To solve, I equated it to zero: \(\displaystyle \, (x\,-\, 1)(x^2\,+\,x\,+\,1)\:=\:0\)

I worked out two of the roots using the formula: \(\displaystyle \, x\:=\:\frac{-1\,\pm\,i\sqrt {3}}{2}\)

You're correct . . . the other root is \(\displaystyle 1.\)

On an Argand diagram, the three points are: \(\displaystyle \,(1,\ 0),\;\left(-\frac{1}{2},\ \frac{\sqrt{3}}{2}\right),\;\left(-\frac{1}{2},\ -\frac{\sqrt{3}}{2}\right)\)

You will find that they are evenly spaced . . . \(\displaystyle 120^o\) apart.


The four 4^th roots are spaced \(\displaystyle 90^o\) apart, the five 5^th roots are spaced \(\displaystyle 72^o\) apart, . . . and so on.