# Complex roots questions (1 Viewer)

#### erucibon

##### Member
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
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:
• vinlatte, erucibon and Ian Ch

#### 4u Boy

##### New Member
Using cube root of unity where w is a root

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

PLZZ HELP

#### Drdusk

##### π
Moderator
Using cube root of unity where w is a root

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

PLZZ HELP

#### 4u Boy

##### New Member
ohh i . was going bout it wrong I chaned 1 + w into (1+w+w^2) - w^2 . and etc

#### 4u Boy

##### New Member
Very Hard complex roots q
Plz help

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#### Drdusk

##### π
Moderator
• ultra908

#### fan96

##### 617 pages
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

##### π
Moderator
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

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

#### fan96

##### 617 pages
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...)

• Drdusk

#### 4u Boy

##### New Member
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

#### 4u Boy

##### New Member
• ultra908