# Need help with a proof

1. Sep 23, 2004

### saint_n

Hey!I have a tut question and im having problems proving

(z^(n+1))-1 = (z-exp[2*Pi*i*(j+1)/(n+1)]) where 0<=j<=n

I tried doing it by induction which is easy for the 1st case with n = 1,I assume the case n-1 but then i get stuck with the last case.
How will i know that all the z's with a power smaller than n+1 will be zero since i know multiplying all the exp[2*Pi*i*(j+1)/(n+1)] from j=o to j= n+1 should give something like exp[2*Pi*i*(n(n+1))/(2(n+1)) +1] which will be one exp[2*Pi*i*(n(n+1))/(2(n+1)) +1 = cos(2*Pi*(n+2))+isin(2*Pi*(n+2)) and the all the z's will give me z^(n+1)

Can you please tell me what i can do or other ways of doing this so i can solve this

2. Sep 23, 2004

### Tide

I think you're missing something in your original equation - perhaps a product or summation?

3. Sep 23, 2004

### saint_n

sorry!!!yes i was missing something in the original equation.So it suppose to look like this
(z^(n+1))-1 = Product[(z-exp[2*Pi*i*(j+1)/(n+1)]) , 0<=j<=n]
where
Product[(z-exp[2*Pi*i*(j+1)/(n+1)]) , 0<=j<=n] = (z-exp[2*Pi*i*(0+1)/(n+1)]*(z-exp[2*Pi*i*(1+1)/(n+1)]*.......*(z-exp[2*Pi*i*(n+1)/(n+1)]
Sorry im really bad at writing this out.I will try to learn some latex..thanks

4. Sep 23, 2004

### saint_n

After learning a little latex maybe this looks better
$$z^{n+1} - 1 =\Pi (z - e^\frac{2 \pi i(j+1)}{n+1})$$ where the product goes from 0<= j <= n

Last edited: Sep 23, 2004
5. Sep 23, 2004

### HallsofIvy

What are the roots of zn+1= 1? What if you were to convert z to polar form?

6. Sep 24, 2004

### saint_n

I dont see where you going.The roots you can get from the product..do want me to replace z with $$\cos\theta+i\sin\theta$$
hmmm...with the
LHS = $$(\cos\theta+i\sin\theta)^{n+1}-1=(\cos(n+1)\theta+i\sin(n+1)\theta)-1$$
and RHS = $$\Pi(\cos\theta+i\sin\theta-(\cos(2\pi\frac{j+1}{n+1})+i\sin(2\pi\frac{j+1}{n+1})))$$
am i doing this correctly??
I dont see where to proceed from here then...

7. Sep 24, 2004

### Tide

Every polynomial of degree n can be written as a product
$$P_n(z) = A \times \Pi (z-z_j)$$
where A is a constant and the $z_j$ are the zeros of the polynomial. I'll leave the proof to Halls! ;-)

8. Sep 24, 2004

### saint_n

Do you know the name of it because of never heard of it.I wouldnt know where to start.If i have the name i can probably do some research on it and continue from there maybe

9. Sep 24, 2004

### marlon

This is the theorema egregium of Algebra, proved by Gauss. I think you need to realize (just as a start) that you can always find the solutions of any binomial equation (and all other equations) by using the complex numbers. Are you familiar with solving such equation using complex numbers ???

regards
marlon

10. Sep 24, 2004

### saint_n

the highest we ever gone up to is probably degree 5 that i know of..i'll try look up the theorem and see what you people are talking about

11. Sep 24, 2004

### marlon

Yes, but when using complex numbers, you can go as high as you want. That's the nice thing. Remember : you need to use complex numbers here, not real numbers only...

regards
marlon

12. Sep 24, 2004

### saint_n

I figured that we have to use complex numbers but we werent told a method how to factorise complex polynomials with say n degree.Can you tell me a method of how to factorize a complex polynomial??

You mentioned that the quote above is the theorema egregium of Algebra which im having difficulty to find as i mostly get information on curvature of surfaces.

13. Sep 24, 2004

### Tide

The theorem is also known as the Fundamental Theorem of Algebra.

Regarding factoring complex polynomials, I cannot tell you in general how to do it but the particular one you're concerned with can be! Basically, you need to find the values of z that make $z^n - 1 = 0$ which means $z^n = 1$ or $z = 1^{\frac {1}{n}}$. Now, at first this may seem odd to you but we're dealing with possibly complex roots so you can replace the 1 on the right hand side with $e^{2 j \pi i}$ where j is an integer.

So we find that
$$z = e^{2 \pi i \frac {j}{n}}$$
Notice that for j ranging from 1 to n the roots are distinct complex numbers. But if you increase j further than that the complex values repeat themselves! Therefore, there are n distinct complex values of z that make $z^n - 1 = 0$ a true statement.

Finally, call those individual values $z_j$ and form the product
$$\Pi (z - z_j)$$ and your polynomial is factored!

14. Sep 24, 2004

### saint_n

ok,,thanks i also found the Fundamental Theorem of Algebra which im reading up now.
Thank you everyone!!
Cheers

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook