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Mathematics Question on Complex Numbers and Quadratic Equations

If z2+z+1=0z^2 + z + 1 = 0 where zz is a complex number, then the value of (z+1z)2+(z2+1z2)2+(z3+1z3)2\left(z+ \frac{1}{z}\right)^{2} + \left(z^{2} + \frac{1}{z^{2}}\right)^{2} + \left(z^{3} + \frac{1}{z^{3}}\right)^{2} equals

A

4

B

5

C

6

D

7

Answer

6

Explanation

Solution

We have,
z2+z+1=0z^{2}+z+1=0
⇒z=−1±1−42\Rightarrow z=\frac{-1 \pm \sqrt{1-4}}{2}
=−1±3i2=\frac{-1 \pm \sqrt{3} i}{2}
∴z=w\therefore z=w or w2w^{2}
Let z=wz=w, then
(z+1z)2+(z2+1z2)2+(z3+1z3)2\left(z+\frac{1}{z}\right)^{2}+\left(z^{2}+\frac{1}{z^{2}}\right)^{2}+\left(z^{3}+\frac{1}{z^{3}}\right)^{2}
=(ω+1ω)2+(ω2+1ω2)2+(ω3+1ω3)2=\left(\omega+\frac{1}{\omega}\right)^{2}+\left(\omega^{2}+\frac{1}{\omega^{2}}\right)^{2}+\left(\omega^{3}+\frac{1}{\omega^{3}}\right)^{2}
=(ω+ω2)2+(ω2+ω)2+(ω3+1)2 [∵ω3=1]=\left(\omega+\omega^{2}\right)^{2}+\left(\omega^{2}+\omega\right)^{2}+\left(\omega^{3}+1\right)^{2}\,\left[\because \omega^{3}=1\right]
=(−1)2+(−1)2+(1+1)2=(-1)^{2}+(-1)^{2}+(1+1)^{2}
[∵1+ω+ω2=0]\left[\because 1+\omega+\omega^{2}=0\right]
=1+1+4=6=1+1+4 = 6
The value will be same when z=ω2z=\omega^{2}.