# Roots of a fourth degree polynomial

1. Oct 29, 2012

### V0ODO0CH1LD

1. The problem statement, all variables and given/known data

z^4 - z^2 + 1 = 0 is an equation in ℂ.

Which of the following alternatives is the sum of two roots of this equation:

(i) 2√3; (ii) -(√3)/2; (iii) (√3)/2; (iv) -i; (v) i/2

2. Relevant equations

3. The attempt at a solution

All I know is that the sum of all roots should equal 0, because that's the coefficient in front of the second highest degree term. And that the sum of the roots squared should equal 2, because that is the coefficient of the second highest degree term plus -2 times the third highest degree term.

2. Oct 29, 2012

### Staff: Mentor

The equation is quadratic in form. Let x = z2 and use the quadratic formula to find values of x (= z2).
No, the coefficient of the z2 term is -1, not 0.

3. Oct 29, 2012

### ehild

It is a quadratic equation for z2. Solve it, then take the square roots of the solutions.

ehild

4. Oct 29, 2012

### V0ODO0CH1LD

Okay, so I said x = z^2 in z^4 - z^2 + 1 = 0 and solved it as if it were a quadratic:

x^2 - x + 1 = 0 <=> (x - (1/2))^2 + 3/4 = 0

(x - (1/2))^2 = -3/4 <=> x - 1/2 = ±√(-3/4)

x = ±√(3)i/2 + 1/2

Now I can just square the roots of the equation for x and I get the roots of the equations for z?

Wait, ehild said I should take the square root of the roots. Why!?

Last edited: Oct 29, 2012
5. Oct 29, 2012

### Staff: Mentor

Think about what you're doing. If you square x, you'll get z4. You want z.

6. Oct 29, 2012

### V0ODO0CH1LD

Wow, thanks! I feel kinda of stupid now.. Anyway, I got to z = √(±√(3)i/2 + 1/2). Which means the sum of two roots of the equation could equal √(√(3)i/2 + 1/2) + √(-√(3)i/2 + 1/2). Any tips on how to simplify this to a point where it looks like one of the answers in the original question?

7. Oct 29, 2012

### Staff: Mentor

I don't know what you did to get the above.
I think the best way to go is to work with x = 1/2 ± i (√3/2), and write this in polar form, or r(cosθ + i sinθ). Here r = 1, so that simplifies things a bit, and the angles should be pretty easy to calculate. Getting a square root is pretty easy - you just take the square root of the magnitude (r) and divide the angle by 2.

For example, consider the complex number i, which in polar form is 1(cos $\pi/2$ + i sin $\pi/2$). √i = 1(cos $\pi/4$ + i sin $\pi/4$) = (√2/2) + (√2/2)i. This is one of the square roots of i. You can verify this by multiplying (√2/2) + (√2/2)i by itself.

8. Oct 29, 2012

### Curious3141

Rather than actually calculating the roots, a much quicker way to do this would be to use the "sum of roots" and "product of roots" formulae for a quadratic. Let two roots of the quartic be $z_1$ and $z_2$.

Then you know that:

$$z_1^2 + z_2^2 = 1$$
$$z_1^2z_2^2 = 1$$

Now use those equations and figure out what values ${(z_1 + z_2)}^2$ can take, and take the square root of that. There's only one choice that matches.

9. Oct 29, 2012

### SammyS

Staff Emeritus
The quartic, z4 - z2 + 1, can be expressed as the product of two quadratics. Because of symmetry we can suppose that they are of the following form, where a and b are real coefficients.

$\displaystyle z^4-z^2+1$
$\displaystyle =(z^2+az+1)(z^2+bz+1)$

$\displaystyle =z^4+(a+b)z^3+(2+ab)z^2+(a+b)z+1$​

Equating coefficients of powers of z gives a+b=0 and 2+ab=-1.

These can be solved for a & b.

10. Oct 30, 2012

### haruspex

The existing posts, noting that you don't actually need to find all the roots, are the right way to go. But fwiw, the easiest way to spot the simplification is to look at it geometrically. Where in the plane is 1 + i√3? What is that in (r, θ)?

11. Oct 30, 2012

### V0ODO0CH1LD

If I solve for a and b on those I would get +√3 and -√3 which add up to 0 which is not one of the answers.

I don't get the last thing. You mean to find the roots of (z^2)^2 - z^2 + 1 = 0 as if they were the roots of a quadratic and then take the square root of those?

EDIT: sorry, I meant find the roots of an equation like x^2 - x + 1 = 0 and take the square root of those!

12. Oct 30, 2012

### SammyS

Staff Emeritus
$\displaystyle \sqrt{3}\ \ \text{ and }\ -\sqrt{3}\$ are not the roots of anything in this problem. They're the coefficients of z in the respective quadratics which factor the given quartic in this problem.

So, find the roots of each of those quadratics:
$\displaystyle (z^2+\sqrt{3}z+1)\ \ \text{ and }\ (z^2-\sqrt{3}z+1)\ .$​
This ia not as elegant a solution as those offered by others, but it does work.

Once you get the answer this way, go back and see how the suggestions of others also lead you to the answer.

13. Oct 31, 2012

### ehild

It is correct, the roots are the square roots of either $1/2+i\sqrt{3}/2$ or $1/2-i\sqrt{3}/2$ . You know that a complex number can be written in the form r(cos(θ)+isin(θ)). What are r and theta?
You have learnt something about the roots of a complex number. What was it?

ehild

14. Oct 31, 2012

### V0ODO0CH1LD

Yeah, the roots are:

+ cos (π/6) + sin(π/6)i;
+ cos (π/6) – sin(π/6)i;
– cos (π/6) + sin(π/6)i;
– cos (π/6) – sin(π/6)i;

Which when added give ±√3 or ±i.

But according to what the others have said, I didn't need to actually find the roots in the first place. I can see how the other ways (Curious3141's, SammyS', Mark44's) to get the roots would work. I just don't see how I could have used those properties of quadratics to find the sums.

15. Oct 31, 2012

### ehild

Curious' method gives the sum of roots directly.

From the quadratic equation for z2 you get that z12+z22=1 and z12z22=1 (Vieta's formulas). The second equation means that z1z2=±1.
(z1+z2)2=z12+z22+2z1z2=1±2.

ehild

16. Oct 31, 2012

### SammyS

Staff Emeritus
Yes, this is correct. (Well, the sum can also be zero.)

As for my method, it did involve finding all of the roots. As follows:

$\displaystyle z^4-z^2+1$
$\displaystyle =(z^2+\sqrt{3}z+1)(z^2-\sqrt{3}z+1)=0$​
So that $\displaystyle \ z^2+\sqrt{3}z+1=0\ \ \text{ or }\ z^2-\sqrt{3}z+1=0\ .$

Using the quadratic formula to solve those gives:
$\displaystyle z=\frac{-\sqrt{3}\pm i }{2}\ , \ \ z=\frac{\sqrt{3}\pm i }{2}\ .$​