# Homework Help: Algebra-should be simple, but isn't

1. Feb 8, 2005

### cepheid

Staff Emeritus
Algebra--should be simple, but isn't

I'm having trouble factoring the polynomial:

p(z) = 1 + z + z2 + z3 + z4

I don't know how to factor a quartic except by trial and error. Our prof assured us that we probably wouldn't want to make use of the formula for finding the zeros, even if we had it. He encouraged us to just try out various things like 1, -1, i, -i, etc to see if we could find a zero by inspection. I'm not having any luck with that method. What's the strategy here?

2. Feb 8, 2005

### Sirus

If p(z)=0, then z is a root, and synthetic division will yield a cubic equation, which you can work with in the same manner. So substitute integral values for z until you find one that makes the equation equal to zero; then use that term. For example, if 1 works substituted for z, then divide synthetically by (z - 1).

3. Feb 8, 2005

### primarygun

I would guess the factor is not a real integer.
Since p(2)=31. I've tried 1 and -1. And for the real , I guess wildly.

4. Feb 8, 2005

### Kelvin

$$p\left( z \right) = 1 + z + z^2 + z^3 + z^4 \hbox{ and we have } p\left( 1 \right) = 1 + 1 + 1 + 1 + 1 = 5 \ne 0$$

$$\hbox{Let}$$

$$p\left( z \right) = 0$$

$$\frac{{z^5 - 1}}{{z - 1}} = 0$$

$$z^5 = 1 = \cos 0 + i\sin 0$$

$$z = \cos \frac{{2n\pi }}{5} + i\sin \frac{{2n\pi }}{5},\ n\ =\ 1,\ 2,\ 3,\ 4$$

so we have five roots and we can factorize p(z)

5. Feb 8, 2005

### Sirus

Ah, yes, I forgot: this function does not have integral roots.

6. Feb 8, 2005

### cepheid

Staff Emeritus
Sorry, I can't seem to understand what's going on here.

7. Feb 8, 2005

### Kelvin

I think you have some problems about the last 2 steps only, right?

$$z^5 = 1 = \cos 0 + i\sin 0$$

this should be trivial. I just express 1 in complex number form. you know, any number can be expressed in complex form.

$$\cos 0 = 1$$

$$\sin 0 = 0$$

$$i = \sqrt{-1}$$

so $$1 = \cos 0 + i\sin 0$$

$$z = \cos \frac{{2n\pi }}{5} + i\sin \frac{{2n\pi }}{5},\ n\ =\ 1,\ 2,\ 3,\ 4$$

I have used de Moivre's theorem here which states

$$\left( {\cos \theta + i\sin \theta } \right)^n = \cos \left( {n\theta } \right) + i\sin \left( {n\theta } \right)$$
for any rational number n

so we have five roots and we can factorize p(z)

8. Feb 9, 2005

### cepheid

Staff Emeritus
No, I understand all of that. It's where (z^5 - 1)/ (z-1) came from in the first place that I cannot see.

9. Feb 9, 2005

### cepheid

Staff Emeritus
Hmmm....He seems to have moved on. Can anyone else help?

10. Feb 9, 2005

### s_a

Sum of a GP (geometric progression).

11. Feb 9, 2005

### Galileo

Calculate $(1+z+z^2+z^3+z^4)(z-1)$.

I think it's called a cyclotomic polynomial.

12. Feb 9, 2005

### cepheid

Staff Emeritus
Ok okay! Very cool. But I think Kelvin forgot to stress that z $\ne$ 1 because 1 is not in the domain of the function p(z). As a result, n = 0 is not allowed, and we have FOUR roots (just as expected), not five (as Kelvin typed, although I think he knew there were only four because he said n = 1, 2, 3, 4, he just typed "five roots" by accident).

Thanks for all of your help!

Edit: OKAY! This is super weird. If you express p(z) in its original form, then p(1) = 5, but if you express it in the form that comes from taking the sum of the geometric series, then p(1) is undefined!!! What's going on? lim (z -> 1) of that expression is 5, but p(1) itself as expressed in that form is undefined! How can it be a valid expression of that sum then?

Last edited: Feb 9, 2005
13. Feb 9, 2005

### ehild

You can replace p(z) by (z^5 - 1)/ (z-1) if z is not equal to 1, and by p(z)=5 if z=1. If z is not 1,

$$p(z)=\frac{z^5-1}{z-1}=\frac{(z-1)(z-e^{i2\pi/5})(z-e^{i4\pi/5})(z-e^{i6\pi/5})(z-e^{i8\pi/5})}{z-1}= (z-e^{i2\pi/5})(z-e^{i4\pi/5})(z-e^{i6\pi/5})(z-e^{i8\pi/5})$$

but the value of this product is 5 at z=1, so p(z) is identical with it.

ehild

14. Feb 9, 2005

### cepheid

Staff Emeritus
Of course! (*slaps forehead*) -- I should have understood right away. Thanks for clarifying.

15. Feb 9, 2005

### Kelvin

I didn't try to modify to defintion of p(z)
see the following
$$$\begin{array}{l} \hbox{\rm my approach to this problem is to find the roots of the equation} \\ p\left( z \right) = 0 \\ \hbox{\rm so if the roots of this equation are }z_1 \hbox{\rm , }z_2 \hbox{\rm , }z_3 ,\hbox{\rm }z_4 \hbox{\rm , then} \\ p\left( z \right) = \left( {z - z_1 } \right)\left( {z - z_2 } \right)\left( {z - z_3 } \right)\left( {z - z_4 } \right) \\ \\ \hbox{\rm so we first solve the equation }p\left( z \right) = 0 \\ \hbox{\rm because the degree of }p \hbox{\rm \ is 4, we expect to find 4 roots only} \\ \hbox{However, the equation is not that easy to solve at first glance, so we consider the equation}\\ q\left( z \right) = z^5 - 1 = 0\\ \hbox{\rm the equation }q\left( z \right) = 0\hbox{\rm \ has five roots and four of them are the roots of } \\ p\left( z \right) \\ \\ \hbox{but } q\left( z \right) = 0 \hbox{\rm \ is obviously easier to solve} \\ \hbox{\rm the roots are} \\ z_k = \cos \frac{{2k\pi }}{5} + i\sin \frac{{2k\pi }}{5} \\ \hbox{\rm where }k = 0,\hbox{\rm }1,\hbox{\rm }2,\hbox{\rm }3,\hbox{\rm }4 \\ \\ \hbox{\rm only 4 of them are roots of }p\left( z \right) = 0 \\ \hbox{\rm and the remaining is to check which is not the root of }p\left( z \right) = 0 \\ \\ \\ \end{array}$$$

$$$\begin{array}{l} \hbox{\rm since }p\left( {z_0 } \right) = p\left( 1 \right) = 5 \ne 0 \\ \hbox{\rm so the roots of }p\left( z \right) = 0\hbox{\rm \ are} \\ z_1 \hbox{\rm , }z_2 \hbox{\rm , }z_3 ,\hbox{\rm }z_4 \hbox{\rm } \\ \end{array}$$$

so is it better?

16. Feb 9, 2005

### Kelvin

In my first reply I have mentioned that p(1) = 5 =/= 0
so I assumed z =/= 1 otherwise I can't write

$$\frac{{z^5 - 1}}{{z - 1}} = 0$$

yes, you're right. It shouldn't be "five roots" but "four". :tongue2: coz I was very sleepy
sorry for that

17. Feb 9, 2005

### saltydog

Jesus. Didn't know that. Thanks guys.

So that means for any polynomial of the form:

$$\sum_{n=0}^{k}x^n=0$$

Then can use the value for the sum of a geometric progression:

$$\sum_{n=0}^{k}x^n=\frac{x^{k+1}-1}{x-1}=0$$

Thus:

$$x^{k+1}-1=0$$

Then, using the formula to calculate complex roots:

$$x = \cos \frac{{2n\pi }}{k+1} + i\sin \frac{{2n\pi }}{k+1},\ n\ =\ 1,\ 2,\ 3,\ . ,., ., (k+1)$$

And finally, determine which one is not part of the original solution. I assume that's right? I'll work one out to verify it.

Last edited: Feb 9, 2005
18. Feb 9, 2005

### Kelvin

is it a typo?
$$x = \cos \frac{{2n\pi }}{M} + i\sin \frac{{2n\pi }}{M}$$

$$M$$ is not always 5

19. Feb 9, 2005

### saltydog

Thanks, I corrected it for the benefit of others (and me).

20. Feb 9, 2005

### saltydog

Don't mind me guys, doing this helps me learn it (maybe you too):

For any polynomial of the form:

$$y(x)=\sum_{n=0}^{k}x^n$$

If k is even, all the k roots are determined by the complex values determined by the formula:

$$z = \cos \frac{{2n\pi }}{k+1} + i\sin \frac{{2n\pi }}{k+1}, n=0,1 . . .,k$$

If k is odd, then the equation has a single real root at x=-1 and the remaining k-1 roots are determined by the complex values of the formula above.

Also, all the complex roots occur in the complex plane on the unit circle.

Example:

$$1+x+x^2+x^3+x^4=0$$
$$x_0=(-0.81+0.59i)$$
$$x_1=(-0.81-0.59i)$$
$$x_2=(0.31+0.95i)$$
$$x_3=(0.31-0.951)$$

$$1+x+x^2+x^3+x^4+x^5=0$$
$$x_0=-1$$
$$x_1=(0.5+0.87i)$$
$$x_2=(0.5-0.87i)$$
$$x_3=(-0.5+0.87i)$$
$$x_4=(-0.5-0.87i)$$