Need help finding roots for a complex number using angles

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fishspawned
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so i am starting with the equation x3 = √(3) - i

first : change to a vector

magnitude = √[ (√(3))2 + 12] = 2
and angle = tan-1( 1/√(3) ) = 30 degrees
(in fourth quadrant)

so i have a vector of 2 ∠ - 30

so i plot the vector on the graph and consider that :

1. the fundamental theorum of algebra tells me i must have three roots.
2. multiplying two complex roots written as vectors will result in the magnitudes multiplied and the angles added together

so to get the root, i simply find the cube root of the magnitude and divide the angle by three

this gets me two initial answers:

3√(2) ∠ - 10
and
3√(2) ∠ 110

the question now comes up with the third root, because I am assuming there should be only one more. But I'm getting MANY more. Since the magnitude remains as 3√(2) then i simply need to find an angle that can be divided into three to get -30 (or 330) degrees.

well : if i take (360 + 330)/3 = 130, it works
if i take (-30 - 360)/3 = -110 , it works.
in fact any addition of 360 works, so I can only guess that these are not really roots. But they do produce entirely different complex numbers - all of them.

I have been assuming there are only three possible roots. Where is my logic going wrong here? Please help. I would greatly appreciate a little enlightenment
 
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fishspawned said:
so i am starting with the equation x3 = √(3) - i

first : change to a vector

magnitude = √[ (√(3))2 + 12] = 2
and angle = tan-1( 1/√(3) ) = 30 degrees
(in fourth quadrant)

so i have a vector of 2 ∠ - 30

so i plot the vector on the graph and consider that :

1. the fundamental theorum of algebra tells me i must have three roots.
2. multiplying two complex roots written as vectors will result in the magnitudes multiplied and the angles added together

so to get the root, i simply find the cube root of the magnitude and divide the angle by three

this gets me two initial answers:

3√(2) ∠ - 10
and
3√(2) ∠ 110

the question now comes up with the third root, because I am assuming there should be only one more. But I'm getting MANY more. Since the magnitude remains as 3√(2) then i simply need to find an angle that can be divided into three to get -30 (or 330) degrees.

well : if i take (360 + 330)/3 = 130, it works
(360 + 330)/3 = 230. It is the angle of the third root.
fishspawned said:
if i take (-30 - 360)/3 = -110 , it works.
(-30 - 360)/3 = -130, which is equivalent to 230, the previous result.
fishspawned said:
in fact any addition of 360 works, so I can only guess that these are not really roots. But they do produce entirely different complex numbers - all of them.

I have been assuming there are only three possible roots. Where is my logic going wrong here? Please help. I would greatly appreciate a little enlightenment

If you add any times 360° to the angle of the vector, you get the same vector. So you have only three different roots, and usually they are chosen form the interval [0, 360°) .
 
ha! a simple addition mistake threw me completely off. thank you so much. everything makes sense now and i can walk away from this .content