The area of the triangle vertices A(z), B(iz) and C(z + iz) 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

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

->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

$S=21+{\displaystyle \sum _{k=1}^{21}{\left(z^k+\frac{1}{z^k}\right)}^3}=\sum {\left(z^k+{\overline{z}}^k\right)}^3=8\sum {\left(Re\left(z^k\right)\right)}^3$

$\left(As\left|z\right|=1\Rightarrow \frac{1}{z}=\overline{z}\right)$

$S=2{\displaystyle \sum _{k=1}^{21}cosk\pi +6}{\displaystyle \sum _{k=1}^{21}cos\frac{k\pi }{3}+21}$

= $2\left(-1\right)+6Re\left(\alpha +{\alpha }^2+{\alpha }^3\right)+21$

Where $\alpha =e^{i\left(2\pi /6\right)}$

$=2\left(-1\right)+6\left(\frac{1}{2}-\frac{1}{2}-1\right)+21$

$=-8+21=13$

From properties of nth root of unity

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

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