For Those Who Do Not Know Omega
z 3 = 1
What is the set of all z that satisfy the equation above?
Note: ω = 2 − 1 + 3 i where i = − 1 .
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10 solutions
I solved it this way, but as it was given that omega = -1±√-3/2,I considered both roots as omega.According to set theory same element is not repeated. Thus {1,omega }is also correct. Please tell me if I am wrong.
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Yes, you are correct. The text states that omega is the pair of complex roots and so the correct answer is also (1, omega). Incidentally omega and omega^2 are the same pair of roots.
In the text you call \omega Two different Numbers. The text is confusing
It's nice answer ,, Try to solve it by De Moivre's theorem.
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I think it is easy by De Moivre's Theorem but I want to make understand everyone the solution.So I used this method.
How u able to solve my no of class equation u r v small how u know that much
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I just do love math !
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@Abdulrahman El Shafei – Wow......by the way....how old are you @A K
That's what I did.
How did you discover that the third answer is actually w^2, i'm missing a step here..
I solved with roots of unity, but this is really nice too.
I got the right answer, but I solved the equation using a different method that takes longer but works just as good...
I think the question should have specified that the solution set is in complex numbers
z 3 = 1
By taking 1 as complex number, it could be put in trignometric form: z 3 = cos 0 + i sin 0
Therefore z = 3 cos 0 + i sin 0 .
By applying Demovire's theorem:
z = cos ( 3 0 + 3 6 0 n ) + i sin ( 3 0 + 3 6 0 n )
where n ϵ {0, 1, 2} giving us a toal of 3 answers.
When n = 0 :
z = cos 0 + i sin 0 = 1
When n = 1 :
z = cos 1 2 0 + i sin 1 2 0 = 2 − 1 + 2 3 i which can be denoted to by ω
When n = 2 :
z = cos 2 4 0 + i sin 2 4 0 = 2 − 1 − 2 3 i which is the square of ω .
The three roots we will get are { 1 , ω , ω 2 } .
I was actually tempted to use cis notation at first, but then I just solved with roots of unity because it was simpler for me. This is a really nice solution though.
ha! you're just joking! when ω = 0. z = 1. so. if x² is sinⁿ. then. x² = ω².
Using the root of unity theorem, we know that the 3 roots are: 1 , e 3 2 π , e 3 4 π . □
There should be an i in the exponent, yeah?
That's just because a cubic equation always have 3 solutions.
But one can have the same solution twice for example, x^3=0's solutions are 0,0,0
You don't even need to calculate this, because in the domain of complex number an ecuation of 3rd grade has 3 roots and there is only one answer with 3 roots
Good observation! (+1)
Since it was cubic equation, I chose the solution with three values.
it has 3 roots. pick the only answer with 3 possibilities.
We know that the group of the n th roots of unity is cyclic of order n , in this case it's a group of the form { 1 , x , x 2 } for some x . Thus the only option for answer is { 1 , ω , ω 2 }
Yep, we can also use the Fundamental theorem of algebra, but we need to make sure there's no multiplicity before we choose the only option with three solutions
z 3 = 1 has three solutions. There is only 1 answer with 3 solutions.
The given equation is z 3 = 1 ⇒ ( z − 1 ) ( z 2 + z + 1 ) = 0 . Either z = 1 or z 2 + z + 1 = 0
Then z = 2 . 1 − 1 ± − 1 2 − 4 . 1 . 1 ⇒ z = 2 − 1 ± − 3
We got the one root of the equation, callled ω , as 2 − 1 + 3 . i .
If we take ω 2 , we will discover it is another root of the equation. The root is 2 − 1 − 3 . i .
The three roots we will get are { 1 , ω , ω 2 } .