# 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:
• erucibon and Ian Ch