Question

If w=e^i(2π/2n+1) ,n>=1
And z=1/2+w+w^2+......w^n then
Prove that
(2z+1)^2n+1 +(2z-1)^2n+1=0

Answer 1

hmm... thinking
\[2z-1 = 2 \sum_{i=1}^{n} w^i \\ 2z+1= 2+2 \sum_{i=1}^{n} w^i \\ ... \]
please let me know anything else that you think might help or any thing you are thinking of trying but failed on

Answer 2

I tried it a li'l bit like this.
I got z=1+w/1-w
And then i tried to solve for it .
I got z =icotπ/5 then i replaced this value of z in the expression been asked.But couldn't get it next.

Answer 3

I even tried to apply binomial expansion but all in vain.

Answer 4

\[z+\frac{1}{2}=1+w+w^2 + \cdots +w^n \\ (z+\frac{1}{2})(w-1)=w^{n+1}-1 \\ z+\frac{1}{2}=\frac{w^{n+1}-1}{w-1} \\ z+\frac{1}{2}=\frac{(e^{ i \frac{2 \pi}{2n+1}})^{n+1}-1}{e^{i \frac{2 \pi}{2n+1}}-1} \\\]

...

Answer 5

\[z+\frac{1}{2}=\frac{(e^{ i \frac{2 \pi}{2n+1}})^{2n+1-n}-1}{e^{i \frac{2 \pi}{2n+1}}-1} \\ z+\frac{1}{2}=\frac{e^{i 2\pi} (e^{i \frac{ 2 \pi}{2n+1}(-n)})-1}{e^{i \frac{2 \pi}{2n+1}}-1} \\ z+\frac{1}{2}= \frac{e^{ i \frac{-2 n \pi}{2n+1}}-1}{e^{i \frac{2 \pi}{2n+1}}-1}\]

Answer 6

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

Answer 7

so the only think I figure out from this is that
\[w^{-n}-1=w^{n+1}-1\]
unless I did something wrong
or that
\[w^{-n}=w^{n+1}\]

Answer 8

I don't find a mistake in your way though.

Answer 9

@ganeshie8 any ideas here?

Answer 10

I know we can multiply 2 on both sides and get 2z+1 in terms of w
but the thing that is in terms of w is ugly

Answer 11

\[w^{-n}=e^{i \frac{-n 2\pi}{2n+1}} =e^{-i \pi (\frac{2n+1-1}{2n+1})}=e^{- i \pi(1-\frac{1}{2n+1})} =-1 e^{ i \frac{\pi}{2n+1}}=-1 w^{\frac{1}{2}}=-w^{\frac{1}{2}}\]

Answer 12

Just some stuff for reference:\[1, \omega, \omega^2, \cdots , \omega^{2n}\]These are the solutions to the equation\[z^{2n+1}-1 = 0\]Obviously, they also satisfy\[\sum \omega^i =0\]\[\prod \omega^i = 1\]Aaand another thing to consider is that \((\omega^i )^* = \omega^{2n+1-i}\).
So that's really all the omega-math.

Answer 13

\[2z+1 = 2 (1 + \omega + \omega^2 + \cdots + \omega^n)\]\[2z-1=2(\omega + \omega^2 + \cdots + \omega^n) \]So really let's just get into it by plugging both into the expression.\[2^{n+1} (1+\omega + \cdots + \omega^n)^{2n+1} + 2^{n+1}(\omega + \omega^2 + \cdots + \omega^n)^{2n+1}=0 \]Which means we have to prove\[(1+\omega + \cdots + \omega^n)^{2n+1}+(1+\omega+\cdots + \omega^{n-1})^{2n+1}=0\]

Answer 14

I found something interesting about my way above...
\[z+\frac{1}{2}=\frac{-w^{\frac{1}{2}}-1}{w-1} \\ z+\frac{1}{2}=-\frac{w^\frac{1}{2}+1}{w-1} \\ z+\frac{1}{2}=-\frac{1}{w^{\frac{1}{2}}-1} \\ -(z+\frac{1}{2})=\frac{1}{w^{\frac{1}{2}}-1} \\ -(2z+1)=\frac{2}{w^{\frac{1}{2}}-1} \\ \frac{-1}{2z+1}=\frac{w^{\frac{1}{2}}-1}{2} \\ \frac{-2}{2z+1}=w^{\frac{1}{2}}-1 \\ \frac{-2}{2z+1}+1=w^{\frac{1}{2}} \\ \frac{-2}{2z+1}+\frac{2z+1}{2z+1}=w^{\frac{1}{2}} \\ \frac{2z-1}{2z+1}=w^{\frac{1}{2}}\]

Answer 15

Thanks,i got it to be 0 .

Answer 16

CAN we prove this ?
Re(z^{2k+1})=0