Need help finding roots for a complex number using angles

In summary, the conversation discusses finding the roots of the equation x3 = √(3) - i by converting it to a vector and using the fundamental theorem of algebra. The speaker initially finds two possible roots and then realizes there should be one more. They discover that any addition of 360° to the angle of the vector produces the same vector, leading to the conclusion that there are only three possible roots. The conversation ends with the speaker gaining clarity on their mistake and being able to solve the problem.
  • #1
fishspawned
66
16
Member warned that the homework template is required
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
 
Physics news on Phys.org
  • #2
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°) .
 
  • #3
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
 

FAQ: Need help finding roots for a complex number using angles

What is a complex number?

A complex number is a number that has two parts: a real part and an imaginary part. It is represented in the form a + bi, where a and b are real numbers and i is the imaginary unit equal to the square root of -1.

How do you find the roots of a complex number using angles?

To find the roots of a complex number using angles, you can use the polar form of the complex number. You can convert the complex number to its polar form by finding its magnitude (distance from the origin) and angle (direction from the positive real axis). The roots can then be found by dividing the angle by the number of desired roots.

Can you find the roots of a complex number using the rectangular form?

Yes, you can also find the roots of a complex number using the rectangular form. To do this, you can use the formula: zn = r1/n(cos((θ+2πk)/n) + i sin((θ+2πk)/n)), where n is the number of roots desired and k is the root number (k=0,1,2,...,n-1).

How do you represent the roots of a complex number on a complex plane?

The roots of a complex number can be represented on a complex plane by plotting the points that correspond to the roots. Each root will have a unique angle and distance from the origin, which can be used to plot the point on the complex plane.

Are there any other methods for finding the roots of a complex number?

Yes, there are other methods for finding the roots of a complex number, such as using De Moivre's formula or using the quadratic formula for special cases. However, using angles is a commonly used and efficient method for finding the roots of a complex number.

Similar threads

Replies
3
Views
2K
Replies
5
Views
2K
Replies
3
Views
2K
Replies
4
Views
2K
Replies
1
Views
1K
Back
Top