Need help solving an quartic equation.

[Student4]

Hi.

Homework Statement

Find the four roots of:
https://www.physicsforums.com/attachment.php?attachmentid=39371&stc=1&d=1317233718

Homework Equations

Calculus: p.6 Polynomials and Rational functions

The Attempt at a Solution

Pretty much stuck on starting it. I can't really find a way to start factoring it.

 

[spamiam]

Try using a substitution to turn it into a quadratic.

 

[Student4]

hi spamiam, thanks for fast reply. Could you just show me fast how to get this one down to quadratic, only if you have time ofc. Have done similar equations with substitution, but this spesific one i can't get my head around.

 

[dynamicsolo]

Concerning spamiam's suggestion, keep in mind that z⁴ can be written as ( z² )² ...

 

[SammyS]

Let u = z² or u = z² - 1

 

[Student4]

so;

(z²)²-2z²+4=0 and let u=z²

u²-2u+4=0

u= 1+i√3 and u= 1-i√3

And i find z

z²= 1+i√3 and z²= 1-i√3

z= √(1+i√3), z= -√(1+i√3), z= √(1-i√3), z=√(1-i√3)

is this correct? i think the answer shud be more simplified

 

[dynamicsolo]

If you write the result in Cartesian form (a + ib), you won't be able to "simplify" this much. Do you know how to write complex numbers in polar form? You could then get "simpler" expressions using DeMoivre's Theorem.

 

[spamiam]

dynamicsolo is right, polar form is much cleaner, especially if you note that $\frac{1\pm i\sqrt{3}}{2}$ are primitive sixth roots of unity and can be written $e^{\frac{2 \pi i}{6}}$ and $e^{2 \pi i\frac{5}{6}}$.

 

[Student4]

Hi, dynamicsolo, yes shud know it, but just started my course so I am abit new to polar form. I need to find modulus and argument of the complex number first right?

 

[spamiam]

Complex numbers can be depicted as points in the plane. Just draw yourself a triangle: the real axis is horizontal, and the imaginary is vertical. This gives you a triangle with what side lengths? Then the modulus is just the length of the hypotenuse, and the argument is the angle between the hypotenuse and the horizontal axis.

 

[epenguin]

But for this problem #2 and #4 have as near told you what to do as anyone could without writing out your work for you.

 

[Student4]

Still abit confused

First root z= √(1+i√3)
simplifed with polarform and DeMoivre's Theorem:

modulus: √(1^2 + √3^2 ) = 2
argument: tan = √3/1 = pi/3

and polarform w=( 2*(cos(pi/3) + i sin(pi/3) )^1/2 ,and then with the theorem i can write: √2 (cos(1/2*pi/3) + i sin(1/2*pi/3))

And then i get: √6/2 + i*√2/2 as my first root.

but for next root z= -√(1+i√3) i have a problem. What do i do when i have -√?

either i plug that in at the end and get same as first root just -(root 1)

or i switch operator z= -√(1+i√3) -> z= √(1-i√3), but then i get same root as my 3th root z= √(1-i√3)Any tip what to do