# Finding the cube root of 1 using Euler's formula

1. Aug 11, 2012

### 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:

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?

Last edited by a moderator: Sep 25, 2014
2. Aug 11, 2012

### Mentallic

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

3. Aug 11, 2012

### 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.

4. Aug 11, 2012

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

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

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

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

$$z_1=e^{\frac{i0\pi}{3}};z_2=e^{\frac{i2\pi}{3}};z_3=e^{\frac{i4\pi}{3}}$$

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

Last edited: Aug 11, 2012
5. Aug 11, 2012

### 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?

6. Aug 11, 2012

### 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:
$$z = e^{i2n\pi/3}$$
This is not quite right, because the middle root written above is a cube root of -1, not 1. The three roots should be
$z_1 = e^{i \cdot 0\pi/3}$
$z_2 = e^{i \cdot 2\pi/3}$
$z_3 = e^{i \cdot 4\pi/3}$

Last edited: Aug 11, 2012
7. Aug 11, 2012

### AGNuke

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