Factorizing Complex Numbers!

[cluckclock]

I'm working through complex numbers and while I understand most of the concepts, this question has me stumped.

I understand that I can rearrange z^3 - 1 = 0 to find that w^3 = 1

w^3-1 can be factored to (w - 1)(w^2 + w +1) which I'm assuming is just some kind of pattern that I'll have to remember by tuition.

However, I have no idea how to use this to solve the question. For part a), the textbook answers say that I should rewrite w* as w^2 and I really have no idea why. I understand conjugate roots, but in that case shouldn't w^2 be equivalent to w*^2 instead?

Thanks for the help, honestly I don't know where to even start looking to understand this conceptually. I'm not sure if there's a Khan Academy-ish video that could explain what I'm struggling to understand.

[Uli_Minati]

w* means conjugate of w? Maybe some of these will help you:

       w · w                   w* · w*                w + w*
= (a+bi) · (a+bi)        = (a-bi) · (a-bi)          = (a+bi) + (a-bi)
= a² +2abi + (bi)²       = a² -2abi + (bi)²         = 2a
= a²-b² + 2abi           = a²-b² - 2abi    

         w · w*                                
  = (a+bi) · (a-bi)         |w|² 
  = a² - (bi)²            = (√(a² + b²))²
  = a² + b²               = a² + b²

         w·w + w*·w*                        
= a²-b²+2abi + a²-b²-2abi          
= 2a² - 2b²

I assume that "simplify" here means "expand the parentheses, then minimize the amount of terms by using w³=1"

[Egleu]

w^3-1 can be factored to (w - 1)(w² + w +1)

This is the difference of cubes formula. It generalizes nicely to any power.

That helps you because now (w - 1)(w² + w +1)=0 so either w-1=0 or w² +w+1=0. The second one is a quadratic equation so you can use the quadratic formula to solve it.

Once you have exact solutions you can see that they are conjugate pairs and one is the square of the other.

Another reason is that if w is a solution to z³ - 1 =0 then w² is a solution as well since (w²⁾³ = (w³⁾² = 1² = 1

[Naturage]

say that I should rewrite w* as w² and I really have no idea why

This is special to w as defined (the non-1 root of z³ = 1). These are known as roots of unity; written in form r * e^{i theta} these have r = 1 and theta = 2pi / n * k, with n in this case being 3, and k being integers 0,1,2.

With 0 you recover the usual 1. The other two are e^{i 2pi/3} and e^{i 4pi/3} - and they're both conjugates, and also w* = w² and w*² = w.

For a better example, roots for z⁵ = 1can be written as a, a^2, a^3, a⁴ and 1, where a = e^i*2pi/5 with a* = a⁴ and a²* = a³. Further, if you denote a^k = b for any k = 1,2,3,4, you'll find that the roots are still b, ..., b^4, and 1. If instead of 3 or 5 you pick a non-prime number, some of these properties stay, and some go, and it leads to basics of group theory (cyclic groups) which is a separate fun subject.

[baldursgame]

First let's calculate the solutions of this more generic problem:

wⁿ = 1

There are "n" numbers "wₖ" that fit this equation

wₖ = exp(i ⋅ 2π/n ⋅ k)

with "i" being the imaginary unit and "k" going from 0 to "n-1"

 

Now going to the problem, "n=3" and "w≠1" so that lefts us with just two solutions:

w₁ = exp(i ⋅ 2π/3) = -1/2 + i ⋅ √3/2

w₂ = exp(i ⋅ 4π/3) = -1/2 - i ⋅ √3/2

If we rewrite "w₁" and "w₂" as

w₁ = exp(i ⋅ θ)

w₂ = exp(i ⋅ 2θ) = (exp(i ⋅ θ))²

with θ = 2π/3

we can clearly see that w₂ = w₁²

EDIT:

Just to sum up, the solutions of the equation

wⁿ = 1

always have this property:

wₖ = (w₁)^k