# Binomial formula (HELP)

1. Nov 23, 2005

### Bob19

Hello

I'm suppose to show

Given z^4 + z^3 + z^2 + z + 1 = 0 where

$$z = cos(\frac{2 \pi}{5}) + i sin(\frac{2 \pi}{5})$$

by using the binomial product formula.

r^n - s^n

Is that then

if r,s = z then z^4 - z^4 = 0 ?????

Sincerely and Best Regards

Bob

2. Nov 23, 2005

### matt grime

i don't see any binomial product formula in your post, i just see r^n-s^n, surely that should equal something...?

3. Nov 23, 2005

### Bob19

Theorem

$$r^n - s^n = (r-s) \sum_{j=0} ^{n-1} r^{j} s^{n-j-1}$$

where r,s are numbers in the Ring R and n is a natural number.

I'm presented with the following polymial

z^4 + z^3 + z^2 + z +1 = 0

where $$z = cos(\frac{2 \pi}{5}) + i sin(\frac{2 \pi}{5})$$

To show the above I'm told to use the theorem:

Do this then imply that

z^4 - z^4 = (z-1)(z^4 + z^3 + z^2 + z +1) = 0

???

/Bob

4. Nov 23, 2005

### HallsofIvy

Staff Emeritus
Yes, it is certainly the case that z^4- z^4= 0 no matter what z is!

But, also certainly, that is not the question you intended to ask.

It is true that r^n- s^n= (r- s)(r^(n-1)+ r^(n-2)s+ ...+ rs^(n-1)+ s^(n-1). (Although that is not waht I would call the "binomial product formula.)

So if z^4 + z^3 + z^2 + z + 1 = 0, multiplying both sides by z-1,
z^5- 1= 0 ro z^5= 1 and so z is a "root of unity". Was that what you intended?

5. Nov 23, 2005

### matt grime

good, now how can you get z^4+z^3+z^2+z+1 into a formula like that? setting r=z will help, now what about a choice for r?

[/quote]where $$z = cos(\frac{2 \pi}{5}) + i sin(\frac{2 \pi}{5})$$
To show the above I'm told to use the theorem:
Do this then imply that
z^4 - z^4 = (z-1)(z^4 + z^3 + z^2 + z +1) = 0
???
/Bob[/QUOTE]

the fact that z^4=z^4 implies that z^4-z^4=0.

your last line of maths is not an example of the binomial product formula

6. Nov 23, 2005

### Bob19

Hello Hall,

it is that thank Yoy

But there is a second equation which has caused me some trouble too:

z^8 + z ^ 6+ z^4 + z^2 +1 = 0

z is the same as in the first one, but how I get i simular result as in the first one? If I multiply left and right by (z-1). I don't get an equation on the form z^n = 1.

Any idears on how I solve this one?

Sincerley
Bob

7. Nov 23, 2005

### matt grime

all of the powers of z are even, aren't they,so it really is an equation in z^2,m isn't it?

8. Nov 23, 2005

### Bob19

yes but if I divide the equation with z^2 I get

z^4 + z ^ 3 + z ^2 + z + 1/z^2 = 0

how do You get z^ 2 from that?

Sincerely
Bob

9. Nov 23, 2005

### matt grime

Firstly dividing by z^2 won't help, and secondly, even if you did divide by z^2 that isn't the result.

10. Nov 23, 2005

### Bob19

Okay I misunderstood

I have now multiplied it by (z^(2) -1) and got

z^16 + z ^12 - 1 = 0 or z^16 + z^12 = 1

Am I on the right track now ?

Sincerley
Bob

11. Nov 23, 2005

### matt grime

why are you making it more complicated? when i said it was a polynomial in z^2 what did I mean by that?

Also, you're starting from the answer. you should not equate z^8+z^6+z^4+z^2+1 to zero to begin with since that is what you are trying to show.

let's use some different letters.

if i write

a^4+a^3+a^2+a+1

and

b^8+b^6+b^4+b^2+1

can you see a simple way to transform the second quantity into the first?

12. Nov 23, 2005

### Bob19

You meant that z^2 is a root for the polynomial ?

thereby (z^2)^4 + (z^2)^3 + (z^2)^2 +1 = 0

/Bob

Last edited: Nov 23, 2005
13. Nov 23, 2005

### matt grime

z^2 is not a root of the polynomial, though two things are being conflated there. but you managed to get the correct notion about making it a polynomial in z^2

14. Nov 23, 2005

### Bob19

If I don't divide or multiply by z^2 and z^2 is not a root of the polynomial, how in God's name is it transformed into z^2 = 0 ??

The only way I can see that is

z^8 + z^6 + z^4 + z^2 +1 = z^8 + z^ 6 + z^4 + 1

and then get z^2 = 0

Sincerely

Bob

Last edited: Nov 23, 2005
15. Nov 23, 2005

### matt grime

who said it was transformed into that?

you know that

w^4+w^3+w^2+w+1= (w^5-1)/(w-1)

that was the first thing you proved.

and you know that w=z^2 gives what you want to examine. and you know that z is a fifth root of unity.

incidentally, you do realize these are just geometric progressions?

Last edited: Nov 23, 2005
16. Nov 23, 2005

### Bob19

Hi Matt and thank You,

Sorry that I have been slow to understand your surgestions.

Following the first equation,

then the second eqaution:

z^8 + z^6 + z^4 + z^2 +1 = (z^10 + z^8 + z^6 + z^4 + z^2)/(z^2)

Am I on the right path now?

Sincerely and Best Regards,
Bob

Last edited: Nov 23, 2005