Need help with a proof

[saint_n]

Hey!I have a tut question and I am 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

 

[Tide]

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

 

[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 I am really bad at writing this out.I will try to learn some latex..thanks

 

[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

 

[HallsofIvy]

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

 

[saint_n]

I don't 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 don't see where to proceed from here then...

 

[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! ;-)

 

[saint_n]

Do you know the name of it because of never heard of it.I wouldn't know where to start.If i have the name i can probably do some research on it and continue from there maybe

 

[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

 

[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

 

[marlon]

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

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

 

[saint_n]

I figured that we have to use complex numbers but we weren't 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??

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! ;-)

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

 

[Tide]

I figured that we have to use complex numbers but we weren't 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 I am having difficulty to find as i mostly get information on curvature of surfaces.

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!

 

[saint_n]

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