# Complex Numbers Question

1. Nov 27, 2008

### thomas49th

1. The problem statement, all variables and given/known data
x³ - 64 = 0
Find the complex roots of this, giving your answers in the form of a + ib where a and b are real
2. Relevant equations

3. The attempt at a solution

well the cube root of 64 is 4, and that's real. I dont see how there can be any imaginery numbers here. 4 x4 x4 = 64. What elses can make x cubed to give 64?

But im wrong arn't i :(

Thanks :)

2. Nov 27, 2008

### 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'...

3. Nov 27, 2008

### 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.

4. Nov 27, 2008

### thomas49th

i dont 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 cant remeber how work out imaginary numbers from this.

Can you walk me through?

Thanks :)

5. Nov 27, 2008

### 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.

6. Nov 27, 2008

### gabbagabbahey

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

7. Nov 27, 2008

### 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 :)

8. Nov 27, 2008

### thomas49th

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

cheers :)

9. Nov 27, 2008

### gabbagabbahey

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.

10. Nov 27, 2008

### 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.

11. Nov 27, 2008

### 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 :)

12. Nov 27, 2008

### Citan Uzuki

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

13. Nov 27, 2008

### 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....

14. Nov 27, 2008

### thomas49th

ahhh im being stupid it's

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

is that right now ;) ?!

Cheers :)

15. Nov 27, 2008

### thomas49th

oh yea you've written it above.

thanks all :)