Imaginary roots.

Just remember math for polar coordinates:

To multiply two polar numbers, multiply the magnitude and add the angle.

So, let''s find the two roots of 2.
Obviously the magnitude has to be sqrt(2). Now, what angles can we add up to get zero, mod 360
Well, that''s just zero and 180.
Giving us (sqrt(2),0) and (sqrt(2),180), which are sqrt(2), and -sqrt(2).

To do this for higher powers, you want to find
n*angle = 0 (mod 360), where n is the power. That gives you the angle, and the magnitude is trivial to calculate.

look up cyclotomic polynomials and complex roots of unity. You''ll find what you''re looking for.

x^3 - 8 is a difference of two cubes and as such, is easily factored, like so:

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

And even if it weren''t, anytime you have a real root r of a cubic polynomial, you can just divide the polynomial by (x - r). That will give you a quadratic polynomial, which you can solve using the quadratic formula to find the other two roots.

For a analytic approach to how to find that first real root, when things like factoring aren''t possible, look up "Cubic Equation" on mathworld.wolfram.com; they''ve got the whole algorithm spelled out.

Listen to sjelkjd. He presents the general case solution in its practical form.

Oztan

quote:

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Original post by sjelkjd
So, let''s find the two roots of 2.
Obviously the magnitude has to be sqrt(2). Now, what angles can we add up to get zero, mod 360
Well, that''s just zero and 180.
Giving us (sqrt(2),0) and (sqrt(2),180), which are sqrt(2), and -sqrt(2).

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I''m confused; what are your final answers? Obviously, sqrt(2) ^ 3 != 8...

I found this link on Google.

Cédric

sjelkjd sought the square roots of 2...
to get the third roots of 8:
the real third root of the absolute value of 8 is 2; so the absolute value of all other third root is also 2.
in polar coordinates (r,phi), 2 can be written as
(2,0) = (2,360°) = (2,720°)
divide the angles of the n representations by n for the n n-th roots.
solution 0: (2,0)
solution 1: (2,120°)
solution 3: (2,240°)

you can convert this to the a + b*i representation:
x0 = 2
x1 = -sqrt(3)/2 + i/2
x2 = -sqrt(3)/2 - i/2


btw: an nth degree polynominal dosn''t necessarily have n solutions, it has got n solutions at most.
the polynominal can be represented by its solutions:
(x - x0)^e0*(x - x1)^e1*...*(x - xm)^em
where the sum e0 + e1 + ... + em = n

quote:

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Original post by ga
sjelkjd sought the square roots of 2...
to get the third roots of 8:
the real third root of the absolute value of 8 is 2; so the absolute value of all other third root is also 2.
in polar coordinates (r,phi), 2 can be written as
(2,0) = (2,360°) = (2,720°)
divide the angles of the n representations by n for the n n-th roots.
solution 0: (2,0)
solution 1: (2,120°)
solution 3: (2,240°)

you can convert this to the a + b*i representation:
x0 = 2
x1 = -sqrt(3)/2 + i/2
x2 = -sqrt(3)/2 - i/2

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Ah, ok... But your x1 and x2 have a problem. Their magnitude is not 2. For anyone interested,

x1 = 2 * cos(120) + 2 * sin(120) * i
x2 = 2 * cos(240) + 2 * sin(240) * i

quote:

---

btw: an nth degree polynominal dosn''t necessarily have n solutions, it has got n solutions at most.
the polynominal can be represented by its solutions:
(x - x0)^e0*(x - x1)^e1*...*(x - xm)^em
where the sum e0 + e1 + ... + em = n

---

If the coefficients are real, a nth degree polynomial does have n complex solutions. It''s the fundamental theorem of algebra.

Cédric

quote:

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Original post by cedricl
x1 = 2 * cos(120) + 2 * sin(120) * i
x2 = 2 * cos(240) + 2 * sin(240) * i

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Or:
x1 = 2*e^(i*radians(120))
x2 = 2*e^(i*radians(240))

and in general:
xi = x*e^(i*radians(i*360/n))

Man, polar coordinates/complex numbers are so cool =)

cedricl: sorry I forgot to write 2*...

I'll give you a simple polynominal of grade 2 which has got only 1 complex solution:
(x-1)^2
or x^2 - 2x + 1


in the fundamental theorem of algebra such solutions are called double solutions, i. e. they are counted twice.
the fundamental theorem of algebra further says you can represent the polynomial as
(x-x0)*(x-x1)*...*(x-xn)
but the x0, x1, ... don't need to be different from each other.

[edited by - ga on March 20, 2003 6:13:30 PM]