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Home  >>  TN XII Math  >>  Complex Numbers
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If $x=a+b$,$y=a\omega+b\omega^2,z=a\omega^2+b\omega$ show that (i) $xyz=a^3+b^3$ where $\omega$ is the complex cube root of unity.

This is the first part of the multi-part question Q2

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1 Answer

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Toolbox:
  • If $\omega$ is a cube root of unity then $\omega^3=1$ and $1+\omega+\omega^2=0$
Step 1:
$x=a+b,y=a\omega+b\omega^2,z=a\omega^2+b\omega$
$x+y+z=a+b+a\omega+b\omega^2+a\omega^2+b\omega$
$\qquad\;\;\;\;\;\;=a(1+\omega+\omega^2)+b(1+\omega+\omega^2)$
We know that $(1+\omega+\omega^2)=0$
$\qquad\;\;\;\;\;\;=0$
Step 2:
$xyz=(a+b)(a\omega+b\omega^2)(a\omega^2+b\omega)$
$\quad\;\;=(a+b)(a^2\omega^3+ab\omega^2+ab\omega^4+b^2\omega^3)$
$\quad\;\;=(a+b)(a^2+ab(\omega^2+\omega)+b^2)$
We know that $(1+\omega+\omega^2)=0$
$\Rightarrow \omega+\omega^2=-1$
$\quad\;\;=(a+b)(a^2-ab+b^2)$
$a^3+b^3=(a+b)(a^2-ab+b^2)$
$\quad\;\;=a^3+b^3$
Hence proved.
answered Jun 12, 2013 by sreemathi.v
 

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