Finding the cube root of 1 using Euler's formula

lo2

1. The problem statement, all variables and given/known data

I have found this video where there is this problem:

Find the cube root of 1

2. Relevant equations

I have found this video:

http://www.youtube.com/watch?v=sn3orkHWqUQ
Where from 8:02 to the end she solves this problem.


3. The attempt at a solution

My question is why is she making such a big fuss about this? Is the cube root of 1 not just simply only 1? And if not could someone please explain what she is trying to explain?

Mentallic

so we're looking for the value of z where [itex]z=1^{1/3}[/itex]
if we cube both sides of this equality, we get [itex]z^3=1[/itex], which is equivalent to [itex]z^3-1=0[/itex]
Can you factorize that expression?

voko

It is not simply only 1 because it is not simply only 1. There are three different complex cubic roots of 1, and that's what is shown in the video.

AGNuke

You are true. Each polynomial function have a degree, and equal number of roots, whether real or imaginary. Cubic expressions always have 3 roots. So, there are 3 cube roots of 1. They are 1, ω, ω². It can be found by simple quadratic as follows. I hope you know the identity of factorizing (a-b)³
[tex]x^3-1=0\Rightarrow (x-1)(x^2+x+1)=0[/tex]

Solutions of the quadratic polynomial factor.
[tex]x=\omega = \frac{-1-i\sqrt{3}}{2};x=\omega ^2=\frac{-1+i\sqrt{3}}{2}[/tex]

This can also be done by Euler's formula,
[tex]z^3=e^{i2n\pi}\Rightarrow z=e^{\frac{i2n\pi}{3}}[/tex]

By putting n = 0, 1, 2 (3 will yield same result as 0, 4 as 1...)

[tex]z_1=e^{\frac{i0\pi}{3}};z_2=e^{\frac{i2\pi}{3}};z_3=e^{\frac{i4\pi}{3}}[/tex]

By the identity, [itex]e^{i\theta}=cos\theta+isin\theta[/itex], you get z₁, z₂ and z₃

lo2

Ok. I can follow you on the first way of finding the roots of the cubic root!

I am not sure I can follow why these two things are equal to each other could you please elaborate?

eumyang

AGNuke took the cube root of both sides, or raised both sides to the 1/3 power. However, there is an i missing in the result, so the equation after the right arrow should be:
[tex]z = e^{i2n\pi/3}[/tex]
This is not quite right, because the middle root written above is a cube root of -1, not 1. The three roots should be
[itex]z_1 = e^{i \cdot 0\pi/3}[/itex]
[itex]z_2 = e^{i \cdot 2\pi/3}[/itex]
[itex]z_3 = e^{i \cdot 4\pi/3}[/itex]

AGNuke

Corrected!
What you can't follow, that [itex]e^{i\theta}=\cos\theta+i\sin\theta[/itex]?