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Home  >>  TN XII Math  >>  Complex Numbers
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Prove that if $\omega^3=1$,then (iii) $\large\frac{1}{1+2\omega}-\frac{1}{1+\omega}+\frac{1}{2+\omega}$$=0$

This is the third part of the multi-part question Q3

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  • If $\omega$ is a cube root of unity then $\omega^3=1$ and $1+\omega+\omega^2=0$
$\large\frac{1}{1+2\omega}-\frac{1}{1+\omega}+\frac{1}{2+\omega}=\frac{(1+\omega)(2+\omega)-(1+2\omega)(2+\omega)+(1+2\omega)(1+\omega)}{(1+2\omega)(1+\omega)(2+\omega)}$
$\Rightarrow \large\frac{2+\omega+2\omega+\omega^2-2-\omega-4\omega-2\omega^2+1+\omega+2\omega+2\omega^2}{(1+2\omega)(1+\omega)(2+\omega)}$
$\Rightarrow \large\frac{\omega+\omega^2+1}{(1+2\omega)(1+\omega)(2+\omega)}$
$1+\omega+\omega^2=0$
$\Rightarrow 0$
Hence proved.
answered Jun 12, 2013 by sreemathi.v
 

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