If z+frac{1}{z}=1,then frac{left(z^{20}+1righ | vedclass

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(B) Given $z+\frac{1}{z}=1$,which implies $z^2-z+1=0$.The roots of this quadratic equation are $z = \frac{1 \pm \sqrt{1-4}}{2} = \frac{1 \pm i\sqrt{3}

Vedclass · Accepted Answer

$2$

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(B) Given $z+\frac{1}{z}=1$,which implies $z^2-z+1=0$.The roots of this quadratic equation are $z = \frac{1 \pm \sqrt{1-4}}{2} = \frac{1 \pm i\sqrt{3}}{2} = -\omega$ and $-\omega^2$,where $\omega$ is the complex cube root of unity.Let $E = \frac{(z^{20}+1)(z^{40}+1)(z^{60}+1)}{z^{60}}$.For $z = -\omega$:$z^{20} = (-\omega)^{20} = \omega^{20} = \omega^{18} \cdot \omega^2 = \omega^2$$z^{40} = (-\omega)^{40} = \omega^{40} = \omega^{39} \cdot \omega = \omega$$z^{60} = (-\omega)^{60} = \omega^{60} = 1$Substituting these values into $E$:$E = \frac{(\omega^2+1)(\omega+1)(1+1)}{1} = 2(\omega^3 + \omega^2 + \omega + 1)$.Since $1+\omega+\omega^2=0$,we have $\omega^2+\omega = -1$.$E = 2(1 + (-1)) = 2(0) = 0$. Wait,let's re-evaluate: $E = 2(\omega^3 + \omega^2 + \omega + 1) = 2(1 + \omega^2 + \omega + 1) = 2(1 + 0) = 2$.Thus,the value is $2$.

If $z+\frac{1}{z}=1$,then $\frac{\left(z^{20}+1\right)\left(z^{40}+1\right)\left(z^{60}+1\right)}{z^{60}}=$

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