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If $z$ is a complex number where $1+z+z^2=0$, then the value of $(z+\large\frac{1}{z}$$)^2+(z^2+\large\frac{1}{z^2}$$)^2+(z^3+\large\frac{1}{z^3}$$)^2+.......................(z^6+\large\frac{1}{z^6}$$)^2$ = ?

(A) 18

(B) 54

(C) 6

(D) 12

Can you answer this question?
 
 

1 Answer

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Toolbox:
  • $1+\omega+\omega^2=0$ where $\omega$ is cube root of 1
  • $\omega^3=\omega^6=\omega^9=............=1$
  • $\omega=\omega^4=\omega^7=..................$
  • $\omega^2=\omega^5=\omega^8=.....................$
Since $1+z+z^2=0,$ $z=\omega$
Dividing both the sides by $z$
$\Rightarrow\:z+\large\frac{1}{z}$$=-1$
$z^2+\large\frac{1}{z^2}$$=(z+\large\frac{1}{z}$$)^2-2=1-2=-1$
$z^3+\large\frac{1}{z^3}$$=1+1=2$
$z^4+\large\frac{1}{z^4}$$=z+\large\frac{1}{z}$$=-1$
and so on
$\Rightarrow\:(z+\large\frac{1}{z}$$)^2+(z^2+\large\frac{1}{z^2}$$)^2+................(z^6+\large\frac{1}{z^6}$$)^2$
$=(-1)^2+(-1^2)+2^2+(-1)^2+(-1)^2+2^2$
$=12$
answered Jul 5, 2013 by rvidyagovindarajan_1
 

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