Let z = 1 - i 3 2 , i = - 1 . Then the value of 21 + (z + (1)/(z))...

Given:

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

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

$⇒ z = - \omega$

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

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

${\omega}^{3} = 1$

Now, let

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

$⇒ A = {\left(- \omega - \dfrac{1}{\omega}\right)}^{3} + {\left({\omega}^{2} + \dfrac{1}{{\omega}^{2}}\right)}^{3} + {\left(- {\omega}^{3} - \dfrac{1}{{\omega}^{3}}\right)}^{3} + \ldots . + {\left(- {\omega}^{21} - \dfrac{1}{{\omega}^{21}}\right)}^{3}$

$⇒ A = - {\left(\dfrac{{\omega}^{2} + 1}{\omega}\right)}^{3} + {\left(\dfrac{{\omega}^{4} + 1}{{\omega}^{2}}\right)}^{3} + {\left(- 1 - \dfrac{1}{1}\right)}^{3} + \ldots . + {\left(- {\omega}^{21} - \dfrac{1}{{\omega}^{21}}\right)}^{3}$

$⇒ A = - {\left(\dfrac{- \omega}{\omega}\right)}^{3} + {\left(\dfrac{- {\omega}^{2}}{{\omega}^{2}}\right)}^{3} + {\left(- 1 - 1\right)}^{3} + \ldots . + {\left(- {\omega}^{21} - \dfrac{1}{{\omega}^{21}}\right)}^{3}$

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

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

$⇒ A = - 8$

Then,

$21 + A = 21 - 8$

$= 13$