Roots of a fourth degree polynomial

[V0ODO0CH1LD]

Homework Statement

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

Homework Equations

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.

 

[Mark44]

Homework Statement

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

The equation is quadratic in form. Let x = z² and use the quadratic formula to find values of x (= z²).

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

Homework Equations

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.

No, the coefficient of the z² term is -1, not 0.

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.

 

[ehild]

Homework Statement

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

Homework Equations

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.

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


ehild

 

[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!?

 

[Mark44]

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?

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

 

[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?

 

[Mark44]

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).

I don't know what you did to get the above.

Any tips on how to simplify this to a point where it looks like one of the answers in the original question?

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.

 

[Curious3141]

Homework Statement

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

Homework Equations

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.

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.

 

[SammyS]

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.

 

[haruspex]

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?

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)?

 

[V0ODO0CH1LD]

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.

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.

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.

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!

 

[SammyS]

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.

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.

\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.

 

[ehild]

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?

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 learned something about the roots of a complex number. What was it?

ehild

 

[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.

 

[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

 

[SammyS]

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.

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}\ .​