How to find the third root of z^3=1?

[hamad12a]

Homework Statement

in a given activity: solve for z in C the equation: z^3=1

Homework Equations

prove that the roots are 1, i, and i^2

The Attempt at a Solution

using z^3-1=0 <=> Z^3-1^3 == a^3-b^3=(a-b)(a^2+2ab+b^2)
it's clear the solution are 1 and i^2=-1 but i didn't find "i" as a solution by using this method
i need your help please

 

[jedishrfu]

Are you sure the equation is right or the roots are right?

1 is a root but i and i^2 aren't

When you plug them into z^3 = 1 you should get 1 each time but i gives -i = 1 and i^2 gives -1 = 1 right?

 

[SammyS]

Homework Statement

in a given activity: solve for z in C the equation: z^3=1

Homework Equations

prove that the roots are 1, i, and i^2

The Attempt at a Solution

using z^3-1=0 <=> Z^3-1^3 == a^3-b^3=(a-b)(a^2+2ab+b^2)
it's clear the solution are 1 and i^2=-1 but i didn't find "i" as a solution by using this method
i need your help please

Hello hamad12a. Welcome to PF!

You will see the following request in many threads here at PF, especially in those started by relative new-comers.

Please state the entire problem as it was given to you.

 

[hamad12a]

Are you sure the equation is right or the roots are right?

1 is a root but i and i^2 aren't

When you plug them into z^3 = 1 you should get 1 each time but i gives -i = 1 and i^2 gives -1 = 1 right?

the mistake was in applying the formulae a^3-b^3 which equals to :

• 
  a3 – b3 = (a – b)(a2 + ab + b2)

i wrote 2ab instead of ab only without myltiplying it by 2
so, again the roots are: -1/2+sqrt(3)/2*i and -1/2-sqrt(3)/2

by using the D=b^2-4ac for the quadratic one and a=b for the first term which is between brackets

 

[jedishrfu]

Another way of finding these roots in this case is DeMoivre's formula:

https://en.wikipedia.org/wiki/De_Moivre's_formula

and polar symmetry of them being spaced 120 degrees apart with the same modulus of 1.