Can You Find the Complex Roots of x³ - 64 = 0?

[thomas49th]

Homework Statement

x³ - 64 = 0
Find the complex roots of this, giving your answers in the form of a + ib where a and b are real

Homework Equations

The Attempt at a Solution

well the cube root of 64 is 4, and that's real. I don't see how there can be any imaginery numbers here. 4 x4 x4 = 64. What elses can make x cubed to give 64?

But I am wrong arn't i :(

Thanks :)

 

[gabbagabbahey]

If you've studied any complex analysis, you should know that any complex (or real number) can be written in polar form: z=|z|e^{i\phi}...Do that for '64'...

 

[Citan Uzuki]

There are two ways to look at the problem. One is to note that the factors of the polynomial correspond to the roots. You know that one of the factors of x³ - 64 is x-4. What are the others?

The other method is to familiarize yourself with the cube roots of unity, and note that multiplying 4 by any of them will yield another cube root of 64.

 

[thomas49th]

i don't think I've seen that notation before. I think that phi is perhaps the angle on the argand diagram. I know the modulus is sqrt(x² + y²), which comes from pathag, but never seent the e before.

I would of thought the answer to this question would be somthing like 0 + or - 4i but that's only because 4x4x4 is 64. I can't remeber how work out imaginary numbers from this.

Can you walk me through?

Thanks :)

 

[Дьявол]

64=4³
so, x³-4³=0
(x-4)(x²+4x+16)=0
Now you got separate cases
(x-4)=0
(x²+4x+16)=0
Now, I think you can easily find the complex roots of x³-64=0
Regards.

 

[gabbagabbahey]

Have you seen Eulers' formula before? e^{i\phi}=\cos\phi+i\sin\phi?

 

[thomas49th]

ah you factorised the polynomial with a quadratic part. Should I ALWAYS do this?

x = 4
now using completing the square:

x = 2 +12i
x = 2 -12i

is that correct

Thanks :)

 

[thomas49th]

sorry baggagabbhey is havn't seen that formula before. atleast i can't remeber it or find it in the textbook :S

cheers :)

 

[gabbagabbahey]

ah you factorised the polynomial with a quadratic part. Should I ALWAYS do this?

x = 4
now using completing the square:

x = -2 +12i
x = -2 -12i

is that correct

Thanks :)

Factoring the polynomial is one good method; but only useful when you can guess at least one of the roots.

Your two complex roots are incorrect, you should check your algebra again.

I'll show you the polar form method after if you like. It is a much more general method.

 

[Дьявол]

Particularly, this one cubic equation is very easy to solve. If you have another complete cubic equation, and if you can't factor it, the best way to solve it is using Rational Root Theorem, or using Cardano's method.

 

[thomas49th]

gabbagabbahley i would very much like to like to see the polar form method.

did you get my edited version of the post. i might of updates it while you were typing

x = 2 +12i
x = 2 -12i

the 2 is positive?

(x-2)² + 16 - 4 = 0
(x-2)² = -12
x = 2 + 12i and 2 - 12i?

Right?

Thanks :)

 

[Citan Uzuki]

No, the 2 is negative. But more importantly, (12i)² = -144 ≠ -12.

 

[gabbagabbahey]

First you should have (x+2)^2=-12, and second taking the square root of -12 doesn't give you 12i it gives \sqrt{12}i=2\sqrt{3}i

\implies x=-2 \pm 2\sqrt{3}i

To use the polar form method; you really need to know a little about complex exponentials...

 

[thomas49th]

ahhh I am being stupid it's

x = -2 +2sqrt(3)i
x = -2 -2sqrt(3)i

is that right now ;) ?!

Cheers :)

 

[thomas49th]

oh yea you've written it above.

thanks all :)