# Complex polynomial and zeroes

1. Aug 22, 2009

### Mentallic

1. The problem statement, all variables and given/known data
http://img198.imageshack.us/img198/5517/polynomial1.jpg [Broken]

3. The attempt at a solution

I was able to answer (i) by solving the quadratic in $$z^2$$ and showing it had complex roots, thus solving for z will also give complex roots.

For (ii) I would have gone about it by taking the long tedious process of finding $$z^6$$, however, by the marking criteria this is only a 1 mark question, so there must be an easier way to answer this.

For (iii) I'm unsure.

Thanks for any help

Last edited by a moderator: May 4, 2017
2. Aug 22, 2009

### CompuChip

Re: Polynomial

For iii) just plug in $z = \alpha^2$ and use that $1 + \alpha^2 + \alpha^4 = 0$ (because it is a root).

ii) is a little trickier, but kinda nice (if I say so myself).
Try multiplying $1 + \alpha^2 + \alpha^4 = 0$ by $\alpha^2$.

3. Aug 22, 2009

### Дьявол

Re: Polynomial

(ii) Because you've found the solutions of z:

$$1+z^2+z^4=0$$

$$y=z^2$$

$$y_{1,2}=\frac{-1 \pm i\sqrt{3}}{2}$$

$$z_{1,2,3,4}=\pm\sqrt{\frac{-1 \pm i\sqrt{3}}{2}}$$

So you need $\alpha ^ 6$ i.e $z^6$

Choose one of the 4 roots of z and do this:

$$z^6= \frac{(-1+i\sqrt{3})^3}{2^3}$$

Solve it, and you will see that it turns out 1.

For (iii), you already got y, just plug it in the equation and you'll see that it turn out 0.

Regards.

4. Aug 22, 2009

### CompuChip

Re: Polynomial

Dyavol, there is a solution to (ii) that doesn't require using the quadratic formula. Just follow the hint in my post ;)

5. Aug 22, 2009

### Дьявол

Re: Polynomial

Yep, your proof is much easier.

I hope Mentallic will spot it.

6. Aug 22, 2009

### Mentallic

Re: Polynomial

CompuChip I have to agree with you there, that IS tricky, but really nice too
I think I'm going to have to keep that idea in mind to see if i can apply it to any other polynomial questions. Thanks for that.

Дьявол, for your solution to (ii) I would've done the same thing, which is quickly noticeable that it's too much work for a 1 mark question.
However, it isn't as bad as having to solve $$(-1\pm i\sqrt{3})^3$$ because $$-1\pm i\sqrt{3}$$ is quickly and easily converted into mod-arg form: $$2cis(\pm \frac{2\pi}{3})$$

Thus, $$(-1\pm i\sqrt{3})^3=(2cis(\pm \frac{2\pi}{3}))^3=2^3cis(\pm 2\pi)=2^3$$

But of course, it could've been worse.

Sorry I still don't understand (iii) very well. Am I finding the root $$a^2$$ and plugging it into the original equation? This also seems like quite a bit of work for 1 mark. Maybe there is an easier way? If not, that's cool.

7. Aug 22, 2009

### Dick

Re: Polynomial

It's not hard. You know a^6=1. (a=alpha). You want to check a^2 is a root, so you want to show 1+a^4+a^8=0. You also know a^2+a^4=(-1), right? Square both sides of that equation. You really don't actually have to find the complex roots to solve any of these questions. That's why it's 1 mark.

Last edited: Aug 22, 2009
8. Aug 22, 2009

### Mentallic

Re: Polynomial

Ahh I see, so $$(a^2+a^4)^2=a^4+2a^6+a^8=1$$

Hence, $$1+a^4+a^8=0$$

Cool thanks for all the help!

9. Aug 22, 2009

### VietDao29

Re: Polynomial

Since you've already solved the problem, I'll give you another way. It is to use the fact that $$\alpha ^ 6 = 1$$, but in a little bit different manner:

$$\alpha$$ is a zero of p(z), so:

$$1 + \alpha ^ 2 + \alpha ^ 4 = 0$$.

We want to prove that $$\alpha ^ 2$$ is also a zero of p(z), i.e, we want to prove:

$$1 + \alpha ^ 4 + \alpha ^ 8 = 0$$.

It goes like this:

$$1 + \alpha ^ 4 + \alpha ^ 8 = \alpha ^ 6 + \alpha ^ 4 + \alpha ^ 8 = \alpha ^ 4 (1 + \alpha ^ 2 + \alpha ^ 4) = \alpha ^ 4 * 0 = 0$$. (Q.E.D)

10. Aug 22, 2009

### Mentallic

Re: Polynomial

Ahh that's very clever!

11. Aug 23, 2009

### CompuChip

Re: Polynomial

It's what I had in mind, somewhat similar to ii), where you write

$$0 = 1 + \alpha^2 + \alpha^4 = \alpha^2(1 + \alpha^2 + \alpha^4) = \alpha^2 + \alpha^4 + \alpha^6 = \underbrace{(1 + \alpha^2 + \alpha^4)}_{{} = 0} + (\alpha^6 - 1)$$

12. Aug 23, 2009

### Mentallic

Re: Polynomial

This too is very smart. Now, if I can just spot stuff like this as efficiently as you guys have done, I'll be set