Question

A complex number problem
specially nth roots of unity is the topic:)

Answer 1

@KingGeorge Please help:)

Answer 2

I'm getting the correct value when \(n=4\). Both the RHS and LHS are 156. As for proving it, that's got to wait until tomorrow since I'm tired.

Answer 3

@KingGeorge plz help now:)

Answer 4

KingGeorge is a mod ... congratulations!!

Answer 5

Thanks experimentX!

Answer 6

would u plz show me @KingGeorge that how did u gt
l.h.s=r.h.s. :)

Answer 7

For the \(n=4\) I just used Wolfram to compute the product. http://www.wolframalpha.com/input/?i=%285-%28e%5E%28i+pi%2F2%29%29%29%285-%28e%5E%28i+pi%29%29%29%285-%28e%5E%283i+pi%2F2%29%29%29

I then compared to \[\frac{5^n-1}{4}\]

Answer 8

for the proof, my idea would be use the fact that the sum of n nth roots of unity is equal 0.

Answer 9

My idea is a bit different. Notice that \[\frac{5^n-1}{4}=\frac{5^n-1}{5-1}=1+5+25+125+...+5^{n-1}\]So all we need to show is that the sum of \(m\) different \(n\)-th roots of unity is either 1 or -1 in certain specific cases.

Answer 10

\[w+w ^{2}+...+w ^{n-1}=-1\]

Answer 11

becouse total sum is 0, :)

Answer 12

and 1 is also root

Answer 13

may i use omega cube =1 here?

Answer 14

sorry my net is slow:(

Answer 15

You can only \(\omega^3=1\) if \(\omega\) is a third root of unity.

Answer 16

only if omega i 3th root

Answer 17

oh ok i see now that wt was my fault when i was taking n=4.
thanx a lot @KingGeorge & @myko :)

Answer 18

A slightly more fleshed out idea of my idea is as follows. Suppose we have multiplied out all of the binomials. Then we get a polynomial in \(\omega\) that looks like \[5^{n-1}-5^{n-2}(\omega+\omega^2+\omega^3+...)+5^{n-3}(\omega\omega^2+\omega\omega^3+...+\omega^{n-2}\omega^{n-1})-...\]\[...\pm5(\omega\omega^2...\omega^{n-2}+\omega\omega^2...\omega^{n-3}\omega^{n-1}+...\omega^2\omega^3...\omega^{n-1})\mp \]\[\mp (\omega\omega^2\omega^3...\omega^{n-1})\]

Answer 19

So if we can show when the sum of those products of \(n\)-th roots of unity is 1 or -1, we're done.

Answer 20

doesn't it depend on n ??

Answer 21

ok.
i gt it now fully @KingGeorge thanx:)

Answer 22

Yes, and that's the tricky part. It should depend on if \(n\) is even or odd.