Let $z=\frac{1-i\sqrt[]{3}}{2},i=\sqrt[]{-1}.$  Then the value of

$21+{\left(z+\frac{1}{z}\right)}^3+{\left(z^2+\frac{1}{z^2}\right)}^3+{\left(z^3+\frac{1}{z^3}\right)}^3+.....+{\left(z^{21}+\frac{1}{z^{21}}\right)}^{3}is\_\_\_\_\_\_\_\_\_.$

(2 – i) z = (2 + i) $\overline{z}$ $\stackrel{}{}$ , put z = x + iy

$y=\frac{x}{2}........\left(i\right)$

(ii) $\left(2+i\right)z+\left(i-2\right)\overline{z}-4i=0$ $\stackrel{}{}$

x + 2y = 2

(iii) $iz+\overline{z}+1+i=0$ $\stackrel{}{}$

Equation of tangent x – y + 1 = 0

Solving (i) and (ii)

$x=1.y=\frac{1}{2}\Rightarrow centre\left(1,\frac{1}{2}\right)$

Perpendicular distance of point $\left(1,\frac{1}{2}\right)$ $\frac{}{}$ from x – y + 1 = 0 is p = r $\Rightarrow \left|\frac{1-\frac{1}{2}+1}{\sqrt[]{2}}\right|=r$ $\frac{\frac{}{}}{\text{exception 25:}}$

$r=\frac{3}{2\sqrt[]{2}}$

f (x) = λ (x-2)²
⇒ 12 = λ (2)² ⇒ λ = 3
f (x) = 3 (x-2)² f (6) = 3 × 4² = 48

Kindly consider the following figure

$Given\left|z+5\right|\le 4..........\left(i\right)$

->Represent a circle ${\left(x+5\right)}^2+y^2\le 16$

$=z\left(1+i\right)+\overline{z}\left(1-i\right)\ge -10..........\left(ii\right)$

->Represent a line X – y $\ge -5$

So max |z + 1|² = AQ²

$={\left(-4-2\sqrt[]{2}\right)}^2+8$

$\begin{array}{l}=32+16\sqrt[]{2}\\ =\alpha +\beta \sqrt[]{2}\end{array}$  

Hence α + β) = 48

From properties of nth root of unity

$1^{2021}+{\alpha }^{2021}+{\beta }^{2021}+{\gamma }^{2021}+{\delta }^{2021}=0$

$\Rightarrow {\alpha }^{2021}+{\beta }^{2021}+{\gamma }^{2021}+{\delta }^{2021}=-1$