Mathematics - Complex Number Question with Solution | TestHub

Given:

$z=\frac{1-i\sqrt{3}}{2}$

$\Rightarrow z=-\left(\frac{-1+i\sqrt{3}}{2}\right)$

$\Rightarrow z=-\omega$

where, $\omega$ is the cube root of unity. So,

$1+\omega +{\omega }^2=0$

${\omega }^3=1$

Now, let

$A={\left(z+\frac{1}{z}\right)}^3+{\left(z^2+\frac{1}{z^2}\right)}^3+{\left(z^3+\frac{1}{z^3}\right)}^3+\dots +{\left(z^{21}+\frac{1}{z^{21}}\right)}^3$

$\Rightarrow A={\left(-\omega -\frac{1}{\omega }\right)}^3+{\left({\omega }^2+\frac{1}{{\omega }^2}\right)}^3+{\left(-{\omega }^3-\frac{1}{{\omega }^3}\right)}^3+\dots .+{\left(-{\omega }^{21}-\frac{1}{{\omega }^{21}}\right)}^3$

$\Rightarrow A=-{\left(\frac{{\omega }^2+1}{\omega }\right)}^3+{\left(\frac{{\omega }^4+1}{{\omega }^2}\right)}^3+{\left(-1-\frac{1}{1}\right)}^3+\dots .+{\left(-{\omega }^{21}-\frac{1}{{\omega }^{21}}\right)}^3$

$\Rightarrow A=-{\left(\frac{-\omega }{\omega }\right)}^3+{\left(\frac{-{\omega }^2}{{\omega }^2}\right)}^3+{\left(-1-1\right)}^3+\dots .+{\left(-{\omega }^{21}-\frac{1}{{\omega }^{21}}\right)}^3$

$\Rightarrow A=1+\left(-1\right)+{\left(-1-1\right)}^3+1+\left(-1\right)+{\left(-1-1\right)}^3\dots .+{\left(-1-1\right)}^3$

$\Rightarrow A=3\left[1+\left(-1\right)\right]+{\left(-1-1\right)}^3$

$\Rightarrow A=-8$

Then,

$21+A=21-8$

$=13$