Question

(1-w+w2)(1-w2+w4)(1-w4+w2)...to 2n factors​

Answer 1

SOLUTION

TO DETERMINE

$\sf{(1 - \omega + {\omega}^{2}) (1 - {\omega }^{2} + {\omega}^{4}) (1 - {\omega }^{4} + {\omega}^{2})}.....2n \: factors$

FORMULA TO BE IMPLEMENTED

$\sf{Since \: \omega \: is \: a \: cube \: root \: of \: Unity }$

So

$1. \: \: \sf{{\omega}^{3}} = 1$

$2. \: \: \sf{(1 + \omega + {\omega}^{2}) = 1}$

EVALUATION

$\sf{(1 - \omega + {\omega}^{2}) (1 - {\omega }^{2} + {\omega}^{4}) (1 - {\omega }^{4} + {\omega}^{2})}.....2n \: factors$

$= \sf{(1 - \omega + {\omega}^{2}) (1 - {\omega }^{2} + {\omega}^{3}. \omega\: ) (1 - {\omega }^{3} .\omega+ {\omega}^{2})}.....2n \: factors$

$= \sf{(1 - \omega + {\omega}^{2}) (1 - {\omega }^{2} + \omega\: ) (1 - \omega+ {\omega}^{2})}.....2n \: factors$

$= \sf{(1 + {\omega}^{2} - \omega) (1 + \omega - {\omega }^{2} \: ) (1+ {\omega}^{2} - \omega)}.....2n \: factors$

$= \sf{( -2 \omega) ( - 2{\omega }^{2} \: ) ( - 2 \omega)}.....2n \: factors \: \: (using \: formula \: 2)$

$\sf{ = {2}^{2n}. {( - 1)}^{2n} . {( \omega)}^{n} . { ({ \omega}^{2} )}^{n} \: }$

$\sf{ = {2}^{2n}. { (\omega.{ \omega}^{2} )}^{n} \: }$

$\sf{ = {2}^{2n}. { ({ \omega}^{3} )}^{n} \: }$

$\sf{ = {( {2}^{2}) }^{n}. {(1)}^{n} \: }$

$\sf{ = {4 }^{n}}$

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LEARN MORE FROM BRAINLY

Prove that

z1/z2 whole bar is equal to z1 bar/z2 bar.

Bar here means conjugate

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