# complex numbers

• Aug 13th 2009, 03:38 PM
MariaJJ
complex numbers
not sure if this is the right forum so sorry if i got it wrong. i'm having alot of trouble with this question. the numbers in brackets in front of the w's are meant to be subscripts

if $w_n = \exp\left(2\pi i/n\right)$ show $w_n^n = 1$ and that

$1 + w_n +w_n^2 + w_n^3 + \dots + w_n^{n-1} = 0$

and show

$\left(x + y\cdot w_3 + z\cdot w_3^2\right)\left(x + y\cdot w_3^2 + z\cdot w_3\right) = x^2 + y^2 +z^2 - xy -yz -zx$

sorry for the lack of tex in it. thanks so much for any help given!!

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Edit by Chris L T521: Reformatted question with $\text{\LaTeX}$. Please inform me if I have misinterpreted anything.
• Aug 13th 2009, 05:18 PM
pankaj
$w^k_{n}=e^{\frac{2k\pi}{n}}=\left(e^{\frac{2\pi}{n }}\right)^k$

If $k=n$,then, $w^n_{n}=\cos 2\pi+i\sin 2\pi=1+i.0=1$

$1+w_{n}+w^2_{n}+w^3_{n}+.....+w^{n-1}_{n}=\frac{1-w^n_{n}}{1-w_{n}}=\frac{1-1}{1-w_{n}}=0$

For, $n=3$,anove results are, $w^3_{3}=1$ and $1+w_{3}+w^2_{3}=0$

$x^2+y^2w^3_{3}+z^2_{3}+xy(w_{3}+w^2_{3})+yz(w^2_{3 }+w^4_{3})+zx((w_{3}+w^2_{3})$

$=x^2+y^2+z^2+xy(-1)+yz(w^2_{3}+w_{3})+zx(-1)$

$=x^2+y^2+z^2-xy-yz-zx$