# 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 $\bg_white z^3 = -1 \pm i$.

Convert to polar form: $\bg_white z^3 = \sqrt 2 \exp\left(\pm {3\pi}/{4}\cdot i\right)$.

Consider the case $\bg_white z^3 = \sqrt 2 \exp\left( {3\pi}/{4}\cdot i\right)$.

Let $\bg_white z = r \exp (i\theta)$ so that

$\bg_white r^3 \exp(i3\theta)= \sqrt 2 \exp\left( \frac{3\pi}{4}i\right)$.

We get $\bg_white r^3 = \sqrt 2$.

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

$\bg_white 3\theta = \frac{3\pi}{4} + 2k\pi$

for some integer $\bg_white k$. Re-arrange to get

$\bg_white \theta = \frac{\pi(3 + 8k)}{12}.$

We're looking for cube roots, so we should expect to find three distinct answers. Set $\bg_white k = 0, 1, 2$ in the above to get your solutions.

Now do the same for the other case $\bg_white z^3 = \sqrt 2 \exp\left( -{3\pi}/{4}\cdot i\right)$.

2. The (relatively) obvious method is to find all the roots of $\bg_white z^4 -4z^2+16$ 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 $\bg_white z = 2w$. The polynomial then becomes

$\bg_white 64(w^6+1)$
$\bg_white = 64(w^2+1)(w^4-w^2+1).$

We just need to factorise $\bg_white w^4-w^2+1$.

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

$\bg_white w^4-w^2+1=(w^2+aw+1)(w^2+bw+1)$.

Why this choice? Notice the coefficient of $\bg_white w^4$ is $\bg_white 1$, so we would expect the coefficients of $\bg_white w^2$ in each factor to also be $\bg_white 1$. The constant term is also $\bg_white 1$, so we should expect the constant terms in each factor to be $\bg_white 1$ as well.

You can expand the RHS and equate coefficients on each side to find $\bg_white a, b$. Remember to substitute back for $\bg_white z$ when you're done.

(As an exercise, you can try the same thing with the polynomial you obtained, $\bg_white z^4 -4z^2+16$.)

3. Just write $\bg_white 0$. It's already in the form $\bg_white a+ib$, where $\bg_white a = b = 0$.

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#### 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
$\bg_white (1+ \omega) = -\omega^2 \hspace{2mm}\text{as}\hspace{2mm}(1 + \omega + \omega^2 = 0)$

$\bg_white \therefore \text{LHS} = -\omega^2(1 + 2\omega)(1 + 3\omega)(1 + 5\omega)$

$\bg_white = -\omega^2(1 + 5\omega + 6\omega^2)(1 + 5\omega)\hspace{2mm}(\text{Expanding brackets})$

$\bg_white = -\omega^2(1 + 10\omega + 31\omega^2 + 30\omega^3)\hspace{2mm}(\text{Expanding brackets})$

$\bg_white \text{Now}\hspace{2mm}\omega^3 = 1$

$\bg_white \therefore \text{LHS}\hspace{2mm} = -\omega^2(31 + 10\omega + 31\omega^2)$

$\bg_white = -\omega^2 \bigg(31\underbrace{\big(1 + \omega^2\big)}_{-\omega} + 10\omega\bigg)$

$\bg_white = -\omega^2(-31\omega + 10\omega)$

$\bg_white = 21\omega^3$

$\bg_white = 21$

#### 4u Boy

##### New Member
$\bg_white (1+ \omega) = -\omega^2 \hspace{2mm}\text{as}\hspace{2mm}(1 + \omega + \omega^2 = 0)$

$\bg_white \therefore \text{LHS} = -\omega^2(1 + 2\omega)(1 + 3\omega)(1 + 5\omega)$

$\bg_white = -\omega^2(1 + 5\omega + 6\omega^2)(1 + 5\omega)\hspace{2mm}(\text{Expanding brackets})$

$\bg_white = -\omega^2(1 + 10\omega + 31\omega^2 + 30\omega^3)\hspace{2mm}(\text{Expanding brackets})$

$\bg_white \text{Now}\hspace{2mm}\omega^3 = 1$

$\bg_white \therefore \text{LHS}\hspace{2mm} = -\omega^2(31 + 10\omega + 31\omega^2)$

$\bg_white = -\omega^2 \bigg(31\underbrace{\big(1 + \omega^2\big)}_{-\omega} + 10\omega\bigg)$

$\bg_white = -\omega^2(-31\omega + 10\omega)$

$\bg_white = 21\omega^3$

$\bg_white = 21$
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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Moderator

#### 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 $\bg_white i^2 = -1$.

e.g. $\bg_white (z+1)^4+z^4 = (z+1)^4-i^2z^4 = ((z+1)^2+iz^2) ((z+1)^2-iz^2)=...$

Then you end up with the real root $\bg_white 1/2$ and six other roots in the form

$\bg_white -\frac{1}{1\pm \exp(i \cdot \frac{k\pi}4)}\quad (k = 1, \,2, \,3)$

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 $\bg_white i^2 = -1$.

e.g. $\bg_white (z+1)^4+z^4 = (z+1)^4-i^2z^4 = ((z+1)^2+iz^2) ((z+1)^2-iz^2)=...$

Then you end up with the real root $\bg_white 1/2$ and six other roots in the form

$\bg_white -\frac{1}{1\pm \exp(i \cdot \frac{k\pi}4)}\quad (k = 1, \,2, \,3)$

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

##### 617 pages
Isn't it to the power of 8, not 4.
The first factorisation, $\bg_white (z+1)^8 - z^8 = ((z+1)^4 + z^4)((z+1)^4 - z^4)$, 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

##### 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