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Cube Root of 1

Date: 01/07/98 at 18:48:12
From: Sean
Subject: Cube root of 1

The cube root of 1 has three roots. I know one is +1. Can you show me 
the steps to find the other two?  I'm stuck.

Thanks for any help.

TimeStamp: 01/08/98 at 06:29:38
From: Doctor Anthony
Subject: Re: Cube root of 1

If you are dealing with all possible cube roots of any number, 
positive or negative, you should be familiar with complex numbers and 
DeMoivre's theorem. I shall assume this is the case, but if you don't 
understand the working you will need to study these topics first.

In effect we must find the roots of the equation  z^3 = 1

We could write this           z^3-1 = 0

                 (z-1)(z^2 + z + 1) = 0

and then z = 1 gives one root, and other two are solutions of the 
quadratic z^2+z+1 = 0 (these roots will be complex).

A more general method (applicable to fourth, fifth, ..etc. roots) is 
as follows.

    Using the modulus, argument form of a complex number

               z = r(cos(theta) + i.sin(theta))

If z = 1, then 1 = 1(cos(2k.pi) + i.sin(2k.pi)) where k = any integer.

Take cube roots  (1)^(1/3) = [cos(2k.pi) + i.sin(2k.pi)]^(1/3) and, 
using DeMoivre's theorem,

          (1)^(1/3) =  cos(2k.pi/3) + i.sin(2k.pi/3)

If k=0  z1 =  cos(0) + i.sin(0)            = 1

if k=1  z2 =  cos(2.pi/3) + i.sin(2.pi/3)  = -1/2 + i.sqrt(3)/2   

if k=2  z3 =  cos(4.pi/3) + i.sin(4.pi/3)  = -1/2 - i.sqrt(3)/2

If we take further values of k we shall simply repeat these 
values of z1, z2 and z3.  

If we required fourth roots the method would be identical except 
that we let k = 0, 1, 2, 3, in the expression

      1^(1/4) = cos(2k.pi/4) + i.sin(2k.pi/4)

and for 5th roots

      1^(1/5) = cos(2k.pi/5) + i.sin(2k.pi/5)   

with k ranging from 0 to 4.

-Doctor Anthony,  The Math Forum
 Check out our web site!   

TimeStamp: 01/08/98 at 05:58:47
From: Doctor Joe
Subject: Re: Cube root of 1

Dear Sean,

The general form of the question is called the nth roots of unity. 
In essence, the roots lie on the unit circle of the complex plane, 
centred at the origin. These roots are precisely the vertices of the 
regular n-gon starting with the first vertex at the real number 1 on 
the Argand diagram.

Now, let me show you how you can find the cube roots of 1.

Suppose that z is a cube root of 1, i.e. z^3 = 1

It is easily seen (by taking the modulus on both sides) that 

                 |z^3| = |1|

It follows:      |z|^3 = 1

Since |z| is real, |z| has to be 1.

Thus, in the polar form, z may be expressed as cos A + i sin A where 
A is the principal argument of z (i.e. -pi < A <= pi).

So, we have z^3 = 1 leading to (cos A + i sin A)^3 = 1.

Applying De Moivre's Theorem, we have

    cos 3A + i sin 3A = 1

Comparing the real part of the left and right expressions (and, 
respectively, the imaginary part),

  cos 3A = 1 and sin 3A = 0

It follows that 3A = 0, 2pi, 4pi

Thus,  A = 0, 2pi/3, 4pi/3

Hence, the three distinct complex cube roots of 1 are:

     cos 0 + i sin 0 = 1

     cos 2pi/3 + i sin 2pi/3 = -1/2 + i sqrt(3)/2

     cos 4pi/3 + i sin 4pi/3 = -1/2 - i sqrt(3)/2

These three complex numbers represent the vertices of an equilateral 
triangle circumscribed by the unit circle centred at 0, with the first 
vertex at the real number 1.

I hope that this is clear.

-Doctor Joe,  The Math Forum
 Check out our web site!   
Associated Topics:
High School Imaginary/Complex Numbers

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