Complex roots questions (1 Viewer)

erucibon

Active Member
Joined
Jul 2, 2018
Messages
210
Gender
Male
HSC
N/A
1) Find the roots of z^6 + 2z^3 +2 =0

I have tried letting z^3 = w to turn it into a quadratic and solved for z^3 but I'm not sure what to do next.

2) Factorise the polynomial z^6 + 64 into real quadratic factors

I tried to use the difference of cubes to turn it into (z^2+4)(z^4-4z^2+16), and then solve for z^2 using the quadratic equation for the second term. Not sure what to do next and if this is an approach that works.

3) solve the complex equation z^5 = z(bar), expressing in a +ib

I have succesfully gotten 6 of the solutions but what to write for the solution 0.

Any help on any of the questions is appreciated! Thanks!
 

fan96

617 pages
Joined
May 25, 2017
Messages
543
Location
NSW
Gender
Male
HSC
2018
Uni Grad
2024
1. You should have .

Convert to polar form: .

Consider the case .

Let so that

.

We get .

The rest is very similar to the process of finding the roots of unity. We want to solve



for some integer . Re-arrange to get



We're looking for cube roots, so we should expect to find three distinct answers. Set in the above to get your solutions.

Now do the same for the other case .

2. The (relatively) obvious method is to find all the roots of and in this way factorise the polynomial into linear factors, and then multiply the linear factors back together in a way that creates real quadratic factors.

Here's a cleaner way to do this.

First make the substitution . The polynomial then becomes




We just need to factorise .

Now, I propose that you can write this polynomial in the form

.

Why this choice? Notice the coefficient of is , so we would expect the coefficients of in each factor to also be . The constant term is also , so we should expect the constant terms in each factor to be as well.

You can expand the RHS and equate coefficients on each side to find . Remember to substitute back for when you're done.

(As an exercise, you can try the same thing with the polynomial you obtained, .)

3. Just write . It's already in the form , where .
 
Last edited:

4u Boy

Member
Joined
Oct 25, 2019
Messages
48
Gender
Male
HSC
2019
Using cube root of unity where w is a root

SHow: (1+w)(1+2w)(1+3w)(1+5w) = 21


PLZZ HELP
 

fan96

617 pages
Joined
May 25, 2017
Messages
543
Location
NSW
Gender
Male
HSC
2018
Uni Grad
2024
Very Hard complex roots q
Plz help
This is not ideal but one method is to use the difference of squares identity to continuously factorise the LHS of the equation into a product of seven polynomial factors of degree one.

You can rewrite a sum of squares as a difference of squares by using .

e.g.

Then you end up with the real root and six other roots in the form



And with a bit of algebra you can convert this to the form the question asks for.

I'm sure there's a much nicer way to do this though.
 

Drdusk

Moderator
Moderator
Joined
Feb 24, 2017
Messages
2,025
Location
a VM
Gender
Male
HSC
2018
Uni Grad
2023
This is not ideal but one method is to use the difference of squares identity to continuously factorise the LHS of the equation into a product of seven polynomial factors of degree one.

You can rewrite a sum of squares as a difference of squares by using .

e.g.

Then you end up with the real root and six other roots in the form



And with a bit of algebra you can convert this to the form the question asks for.

I'm sure there's a much nicer way to do this though.
Isn't it to the power of 8, not 4.
 

Trebla

Administrator
Administrator
Joined
Feb 16, 2005
Messages
8,113
Gender
Male
HSC
2006
For part a), let y = 1 + 1/z and the equation reduces to a roots of unity problem (but with one less solution as y cannot equal 1).

Use some double angle manipulations when you return it in terms of z and you should get the answer.
 

fan96

617 pages
Joined
May 25, 2017
Messages
543
Location
NSW
Gender
Male
HSC
2018
Uni Grad
2024
Isn't it to the power of 8, not 4.
The first factorisation, , is straightforward. Realising that you can turn a sum of squares into a difference of squares is not so trivial - that's the part I showed.

(actually, this method generalises to any positive even power...)
 

4u Boy

Member
Joined
Oct 25, 2019
Messages
48
Gender
Male
HSC
2019
For part a), let y = 1 + 1/z and the equation reduces to a roots of unity problem (but with one less solution as y cannot equal 1).

Use some double angle manipulations when you return it in terms of z and you should get the answer.
This method worked i got the solution
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top