Complex Numbers

Let $z = x + iy$

Length of side = 4

$AB = 4$

$|z - \overline z | = 4$

$|2y|\, = 4;$$ \Rightarrow $ $|y|\, = 2$

$BC = 4$

$ \Rightarrow $ $|\overline z - (\overline z - 2{\mathop{\rm Re}\nolimits} (\overline z )|\, = 4$

$ \Rightarrow $ $|2x|\, = 4;\,$$ \Rightarrow $ $|x|\, = 2$

$ \therefore $ $|z|\, = \,\sqrt {{x^2} + {y^2}} = \sqrt {4 + 4} = 2\sqrt 2 $

$\alpha = \omega $ as given $\alpha = {{ - 1 + i\sqrt 3 } \over 2}$

$ \Rightarrow {(2 + \omega )^4} = a + b\omega \,({\omega ^3} = 1)$

$ \Rightarrow {2^4} + {4.2^3}\omega + {6.2^2}{\omega ^3} + 4.2.\,{\omega ^3} + {\omega ^4} = a + b\omega $

$ \Rightarrow 16 + 32\omega + 24{\omega ^2} + 8 + \omega = a + b\omega $

$ \Rightarrow 24 + 24{\omega ^2} + 33\omega = a + b\omega $


$ \Rightarrow - 24\omega + 33\omega = a + b\omega $

$ \Rightarrow a = 0,\,b = 9$

Given, z = x + iy

and $u = {{2z + i} \over {z - ki}}$

$ = {{2(x + iy) + i} \over {(x + iy) - ki}}$

$ = {{2x + i(2y + 1)} \over {x + i(y - k)}} \times {{x - i(y - k)} \over {x - i(y - k)}}$

$ = {{2{x^2} + (2y + 1)(y - k) + i(2xy + x - 2xy + 2kx)} \over {{x^2} + {{(y - k)}^2}}}$

Given,

${\mathop{\rm Re}\nolimits} (u) + {\mathop{\rm Im}\nolimits} (u) = 1$

$ \Rightarrow {{2{x^2} + (2y + 1)(y - k)} \over {{x^2} + {{(y - k)}^2}}} + {{x + 2kx} \over {{x^2} + {{(y - k)}^2}}} = 1$

$ \Rightarrow 2{x^2} + (2y + 1)(y - k) + x + 2kx = {x^2} + {(y - k)^2}$

This curve intersect the y-axis at point P and Q, so at point P and Q x = 0

Putting x = 0 at the above equation,

$ \therefore $ $(2y + 1)(y - k) = {(y - k)^2}$

$ \Rightarrow 2{y^2} + y - 2yk - k = {y^2} + {k^2} - 2ky$

$ \Rightarrow {y^2} + y - (k + {k^2}) = 0$

Let roots of this quadratic equation y₁ and y₂

$ \therefore $ Point P (0, y₁) and Q (0, y₂)

and ${y_{_1}} + {y_2} = 1$ , ${y_1}{y_2} = - k - {k^2}$

$ \therefore $ ${({y_1} - {y_2})^2} = {({y_1} + {y_2})^2} - 4{y_1}{y_2}$

$ = 1 + 4k + 4{k^2}$

$ \Rightarrow |{y_1} - {y_2}| = \sqrt {1 + 4k + 4{k^2}} $

Given, PQ = 5

$ \Rightarrow |{y_1} - {y_2}| = 5$

$ \Rightarrow \sqrt {1 + 4k + 4{k^2}} = 5$

$ \Rightarrow {k^2} + k - 6 = 0$

$ \Rightarrow k = - 3,\,2$

So, k = 2 (Given k > 0)

Let ${z_1} = {x_1} + i{y_1},\,{z_2} = {x_2} + i{y_2}$

Given Re(z₁) = |z₁ – 1|

$ \therefore $ x₁ = |(x₁ - 1) + iy₁|

$ \Rightarrow $ x₁ = $\sqrt {{{\left( {{x_1} - 1} \right)}^2} + y_1^2} $

$ \Rightarrow $ ${x_1}^2 = {({x_1} - 1)^2} + {y_1}^2$

$ \Rightarrow {y_1}^2 - 2{x_1} + 1 = 0$

Also given Re(z₂) = |z₂ – 1|

$ \therefore $ x₂ = |(x₂ - 1) + iy₂|

$ \Rightarrow $ ${x_2}^2 = {({x^2} - 1)^2} + {y_2}^2$

$ \Rightarrow $ $y_2^2 - 2{x_2} - 1 = 0$

Performing equation (1) - (2),

$({y_1}^2 - {y_2}^2) + 2({x_2} - {x_1}) = 0$

$ \Rightarrow $ $({y_1} + {y_2})({y_1} - {y_2}) = 2({x_1} - {x_2})$

$ \Rightarrow $ ${y_1} + {y_2} = 2\left( {{{{x_1} - {x_2}} \over {{y_1} - {y_2}}}} \right)$

Now given, $\arg ({z_1} - {z_2}) = {\pi \over 6}$

$ \Rightarrow $ ${\tan ^{ - 1}}\left( {{{{y_1} - {y_2}} \over {{x_1} - {x_2}}}} \right) = {\pi \over 6}$

$ \Rightarrow {{{y_1} - {y_2}} \over {{x_1} - {x_2}}} = {1 \over {\sqrt 3 }}$

$ \therefore $ ${y_1} + {y_2} = 2\sqrt 3 $

$3 + 2\sqrt { - 54} $

$ = 9 - 6 + 2\sqrt { - 54} $

$ = 9 + {\left( {\sqrt 6 i} \right)^2} + 2.3.\sqrt 6 i$

$ = {3^2} + {\left( {\sqrt 6 i} \right)^2} + 2.3.\sqrt 6 i$

$ = {\left( {3 + \sqrt 6 i} \right)^2}$

Similarly, $\left( {3 - 2\sqrt { - 54} } \right) = {\left( {3 - \sqrt 6 i} \right)^2}$

$ \therefore {\left( {3 + 2\sqrt { - 54} } \right)^{{1 \over 2}}} - {\left( {3 - 2\sqrt { - 54} } \right)^{{1 \over 2}}}$

$ = \pm \left( {3 + \sqrt 6 i} \right) - \left[ { \pm \left( {3 - \sqrt 6 i} \right)} \right]$

$ = 6, - 6,2\sqrt 6 i, - 2\sqrt 6 i$

$ \therefore $ Possible imaginary parts are $2\sqrt 6 i, - 2\sqrt 6 i$

${\left( {{{1 + \sin {{2\pi } \over 9} + i\cos {{2\pi } \over 9}} \over {1 + \sin {{2\pi } \over 9} - i\cos {{2\pi } \over 9}}}} \right)^3}$

= ${\left( {{{1 + \sin \left( {{\pi \over 2} - {{5\pi } \over {18}}} \right) + i\cos \left( {{\pi \over 2} - {{5\pi } \over {18}}} \right)} \over {1 + \sin \left( {{\pi \over 2} - {{5\pi } \over {18}}} \right) - i\cos \left( {{\pi \over 2} - {{5\pi } \over {18}}} \right)}}} \right)^3}$

= ${\left( {{{1 + \cos \left( {{{5\pi } \over {18}}} \right) + i\sin \left( {{{5\pi } \over {18}}} \right)} \over {1 + \cos \left( {{{5\pi } \over {18}}} \right) - i\sin \left( {{{5\pi } \over {18}}} \right)}}} \right)^3}$

= ${\left( {{{2{{\cos }^2}\left( {{{5\pi } \over {36}}} \right) + 2i\sin \left( {{{5\pi } \over {36}}} \right)\cos \left( {{{5\pi } \over {36}}} \right)} \over {2{{\cos }^2}\left( {{{5\pi } \over {36}}} \right) - 2i\sin \left( {{{5\pi } \over {36}}} \right)\cos \left( {{{5\pi } \over {36}}} \right)}}} \right)^3}$

= ${\left( {{{\cos \left( {{{5\pi } \over {36}}} \right) + i\sin \left( {{{5\pi } \over {36}}} \right)} \over {\cos \left( {{{5\pi } \over {36}}} \right) - i\sin \left( {{{5\pi } \over {36}}} \right)}}} \right)^3}$

= ${\left( {{{{e^{i{{5\pi } \over {36}}}}} \over {{e^{ - i{{5\pi } \over {36}}}}}}} \right)^3}$

= ${\left( {{e^{i{{5\pi } \over {18}}}}} \right)^3}$

= ${{e^{i{{5\pi } \over {18}} \times 3}}}$ = ${{e^{i{{5\pi } \over 6}}}}$

= $\cos {{5\pi } \over 6} + i\sin {{5\pi } \over 6}$

= $ - {{\sqrt 3 } \over 2} + {i \over 2}$

Let z = x + iy

given that |Re(z)| + |Im(z)| = 4

$ \therefore $ |x| + |y| = 4
Maximum value of |z| = 4

Minimum value of |z| = perpendicular distance of line AB from (0, 0) = $2\sqrt 2 $

$ \therefore $ |z| $ \in $ $\left[ {2\sqrt 2 ,4} \right]$

$ \therefore $ |z| cannot be $\sqrt {7} $.

Given $\left| {{{z - i} \over {z + 2i}}} \right| = 1$

|z – i| = |z + 2i|

(let z = x + iy)

$ \Rightarrow $ x² + (y – 1)² = x² + (y + 2)²

$ \Rightarrow $ y = $ - {1 \over 2}$

Also given |z| = ${5 \over 2}$

$ \Rightarrow $ x² + y² = ${{25} \over 4}$

$ \Rightarrow $ x² = 6

$ \therefore $ z = $ \pm \sqrt 6 $ - $ - {1 \over 2}i$

|z + 3i| = $\sqrt {6 + {{25} \over 4}} $ = ${7 \over 2}$

x² + bx = 45 = 0 (b $ \in $ R)
has roots $\alpha $ + i$\beta $, $\alpha $ – i$\beta $
sum of roots = – b = 2$\alpha $
product of roots = 45 = $\alpha $² + $\beta $²

Let z = x + iy

$ \therefore $ |x + iy +1| = 2$\sqrt {10} $

${\left| {x + iy + 1} \right|^2} = {\left( {2\sqrt {10} } \right)^2}$

$ \Rightarrow $ (x + 1)² + y² = 40

$ \Rightarrow $ ($\alpha $ + 1)² + $\beta $² = 40

[putting real part $\alpha $ in place of x and imaginary part $\beta $ in place of y]

$ \Rightarrow $ $\alpha $² + 2$\alpha $ + 1 + $\beta $² = 40

$ \Rightarrow $ 45 + 2$\alpha $ + 1 = 40

$ \Rightarrow $ $\alpha $ = -3

$ \therefore $ -b = 2$\alpha $ = 2$ \times $(-3) = -6

$ \Rightarrow $ b = 6

By checking options we found b² – b = 30.

Let z = ${{3 + i\sin \theta } \over {4 - i\cos \theta }}$

= ${{3 + i\sin \theta } \over {4 - i\cos \theta }} \times {{\left( {4 + i\cos \theta } \right)} \over {\left( {4 + i\cos \theta } \right)}}$

= ${{\left( {12 - \sin \theta \cos \theta } \right) + i\left( {4\sin \theta + 3\cos \theta } \right)} \over {16 + {{\cos }^2}\theta }}$

As Z is purely real

$ \therefore $ 4sin$\theta $ + 3cos$\theta $ = 0

$ \Rightarrow $ tan $\theta $ = $ - {3 \over 4}$

$ \therefore $ $\theta $ lies in the 2^nd quadrant then

arg(sinθ + icosθ) = $\pi $ + ${\tan ^{ - 1}}\left( {{{\cos \theta } \over {\sin \theta }}} \right)$

= $\pi - {\tan ^{ - 1}}\left( {{4 \over 3}} \right)$

${\mathop{\rm Re}\nolimits} \left( {{{z - 1} \over {2z + i}}} \right) = 1$

Put z = x + iy

$ \therefore $ ${\mathop{\rm Re}\nolimits} \left( {{{\left( {x + iy} \right) - 1} \over {2\left( {x + iy} \right) + i}}} \right) = 1$

$ \Rightarrow $ ${\mathop{\rm Re}\nolimits} \left( {\left( {{{\left( {x - 1} \right) + iy} \over {2x + i\left( {2y + 1} \right)}}} \right)\left( {{{2x - i\left( {2y + 1} \right)} \over {2x - i\left( {2y + 1} \right)}}} \right)} \right) = 1$

$ \Rightarrow $ ${\mathop{\rm Re}\nolimits} \left( {{{\left\{ {\left( {x - 1} \right) + iy} \right\}\left\{ {2x - i\left( {2y + 1} \right)} \right\}} \over {4{x^2} + {{\left( {2y + 1} \right)}^2}}}} \right) = 1$

Real part of this equation is = 1

$ \therefore $ ${{2x\left( {x - 1} \right) + y\left( {2y + 1} \right)} \over {4{x^2} + {{\left( {2y + 1} \right)}^2}}}$ = 1

$ \Rightarrow $ 2x² + 2y² +2x + 3y + 1 = 0

$ \Rightarrow $ x² + y² +x + ${3 \over 2}$y + ${1 \over 2}$ = 0

This is an equation of circle.

$ \therefore $ Locus is a circle whose

center is $\left( { - {1 \over 2}, - {3 \over 4}} \right)$ and radius ${{\sqrt 5 } \over 4}$

$ \therefore $ Diameter = 2 $ \times $ ${{\sqrt 5 } \over 4}$ = ${{\sqrt 5 } \over 2}$

It is given that the complex number satisfying

$|{z^2} + z + 1| = 1 \Rightarrow \left| {{{\left( {z + {1 \over 2}} \right)}^2} + {3 \over 4}} \right| = 1$

$ \because $ $|{z_1} - {z_2}|\, \ge \,||{z_1}| - |{z_2}||$

$ \therefore $ $\left| {{{\left( {z + {1 \over 2}} \right)}^2} - \left( { - {3 \over 4}} \right)} \right|\, \ge \,\left| {{{\left| {z + {1 \over 2}} \right|}^2} - \left| { - {3 \over 4}} \right|} \right|$

$ \Rightarrow \left| {{{\left| {z + {1 \over 2}} \right|}^2} - {3 \over 4}} \right|\, \le \,1$

$ \Rightarrow - 1 \le \,{\left| {z + {1 \over 2}} \right|^2} - {3 \over 4}\, \le \,1$

$ \Rightarrow - {1 \over 4}\, \le \,{\left| {z + {1 \over 2}} \right|^2}\, \le \,{7 \over 4}$

$ \Rightarrow 0 \le \,{\left| {z + {1 \over 2}} \right|^2}\, \le \,{7 \over 4}$ {$ \because $ |z| $ \ge $ 0}

$ \Rightarrow 0 \le \,{\left| {z + {1 \over 2}} \right|^2}\, \le \,{{\sqrt 7 } \over 2}$ ....(i)

$ \because $ $ \Rightarrow \,\left| {z + {1 \over 2}} \right|\, \le \,{{\sqrt 7 } \over 2}$

$ \because $ $|{z_1} + {z_2}|\, \ge \,||{z_1}| - |{z_2}||$

$ \therefore $ $\left| {z + {1 \over 2}} \right|\, \ge \,\left| {|z| - {1 \over 2}} \right|$

$ \Rightarrow \left| {|z| - {1 \over 2}} \right|\, \le \,\left| {z + {1 \over 2}} \right| \le \,{{\sqrt 7 } \over 2}$

$ \Rightarrow - {{\sqrt 7 } \over 2}\, \le \,|z| - {1 \over 2} \le {{\sqrt 7 } \over 2}$

$ \Rightarrow {{1 - \sqrt 7 } \over 2} \le \,|z|\, \le {{\sqrt 7 + 1} \over 2}$

$ \Rightarrow |z|\, \le \,{{1 + \sqrt 7 } \over 2}$, $ \therefore $ |z| $ \le $ 2

$ \because $ $\left| {{{\left( {z + {1 \over 2}} \right)}^2} + {3 \over 4}} \right|\, \le \,\left| {{{\left| {z + {1 \over 2}} \right|}^2} + {3 \over 4}} \right|$

$ \Rightarrow 1 \le \left| {{{\left| {z + {1 \over 2}} \right|}^2} + {3 \over 4}} \right| \Rightarrow {\left| {z + {1 \over 2}} \right|^2} + {3 \over 4} \ge 1$

$ \Rightarrow {\left| {z + {1 \over 2}} \right|^2} \ge {1 \over 4} \Rightarrow \left| {z + {1 \over 2}} \right|\, \ge {1 \over 2}$ ....(ii)

from Eqs. (i) and (ii), we get

${1 \over 2} \le \left| {z + {1 \over 2}} \right|\, \le {{\sqrt 7 } \over 2}$

Given $\operatorname{Re}\left(\frac{z-2}{z-4 i}\right)=0$ and

$ \begin{aligned} & m\left(\frac{z-2}{z-4 i}\right)=1 \\ & \Rightarrow \quad \frac{z-2}{z-4 i}=i \Rightarrow z-2=i(z-4 i) \\ & \Rightarrow \quad z-2=i z+4 \Rightarrow z(1-i)=6 \\ & \Rightarrow \quad z=\frac{6}{1-i} \Rightarrow z=3(1+i) \end{aligned} $

∴ Only one $z$ in Argand plane.

Let $z=\sqrt{3}+i \Rightarrow|z|=\sqrt{3+1}=2$ and $\arg (z)=\tan ^{-1} \frac{1}{\sqrt{3}}=\frac{\pi}{6}$

∴ Polar form of $z=2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)$

$ \begin{aligned} & \Rightarrow \quad z^{10}=2^{10}\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)^{10} \\ & \Rightarrow \quad z^{10}=2^{10}\left(\cos \frac{10 \pi}{6}+i \sin \frac{10 \pi}{6}\right) \\ & \Rightarrow \quad(\sqrt{3}+i)^{10}=2^{10}\left(\frac{1}{2}-i \frac{\sqrt{3}}{2}\right) \\ & \Rightarrow \quad(\sqrt{3}+i)^{10}=512-i 512 \sqrt{3}=a+i b \\ & \Rightarrow \quad a=512 \text { and } b=-512 \sqrt{3} \end{aligned} $

$ \begin{aligned} &\text { Given, } z^2+z+1=0\\ &\begin{aligned} & \therefore \quad z=\frac{-1 \pm \sqrt{1-4}}{2}=\frac{-1 \pm \sqrt{3} i}{2} \\ & \therefore z=\omega, \omega^2 \\ & \text { Now } z+\frac{1}{Z}=\omega+\frac{1}{\omega}=\omega+\omega^2=-1 \\ & \quad z^2+\frac{1}{z^2}=\omega^2+\frac{1}{\omega^2}=\omega^2+\omega=-1 \\ & \therefore\left(z+\frac{1}{z}\right)^3+\left(z^2+\frac{1}{z^2}\right)^3+ \\ & \quad \ldots+\left(z^{2020}+\frac{1}{z^{2020}}\right)^3 \\ & \Rightarrow \quad(-1)^3+(-1)^3+\ldots+(-1)^3=-2020 \end{aligned} \end{aligned} $

Given $n_1, n_2, n_3$ are roots of the equation

$ \begin{array}{r} E_1: x^3+x^2+l x+n=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \therefore \text { Sum of roots } x_1+x_2+x_3=-1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

Now, $E_2: x^3+a x^2+b x+c=0$ is reciprocal equation of class one $\Rightarrow c=1$ and $a=b$

$ \therefore \quad E_2: x^3+a x^2+a x+1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Given, Eq. (iii) have roots

$ \begin{aligned} & \frac{x_1-1}{2}, \frac{x_2-1}{2}, \frac{x_3-1}{2} \\ \Rightarrow & \frac{x_1-1}{2}+\frac{x_2-1}{2}+\frac{x_3-1}{2}=-a \\ \Rightarrow & \frac{x_1+x_2+x_3}{2}-\frac{3}{2}=-a \end{aligned} $

From Eq. (ii),

$ \begin{aligned} & \quad x_1+x_2+x_3=-1 \\ & \therefore \quad \frac{-1}{2}-\frac{3}{2}=-a \Rightarrow a=2 \\ & \therefore \text { Eq. }(\text { iii }), x^3+2 x^2+2 x+1=0 \\ & (x+1)\left(x^2+x+1\right)=0 \quad \therefore \quad x=-1, \\ & x^2+x+1=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad x=\frac{-1 \pm \sqrt{3} i}{2} \\ & \therefore \text { Given, } \frac{x_1-1}{2}=-1 \Rightarrow x_1=-1 \\ & \quad \frac{x_2-1}{2}=\frac{-1+\sqrt{3} i}{2} \\ & \Rightarrow \quad x_2=\sqrt{3} i \\ & \quad \frac{x_3-1}{2}=\frac{-1-\sqrt{3} i}{2} \\ & \Rightarrow \quad x_3=-\sqrt{3} i \end{aligned} $

∴ Roots of these two equation excluding common roots are

$ \sqrt{3} i,-\sqrt{3} i, \frac{-1+\sqrt{3} i}{2}, \frac{-1-\sqrt{3} i}{2} $

Given $\alpha, \beta, \gamma, \delta$ are roots of

$ \begin{aligned} & x^4+x^2+1=0 \\ & \therefore \quad x^2=\frac{-1 \pm \sqrt{1-4}}{2}=\frac{-1 \pm \sqrt{3} i}{2} \\ & \therefore \quad x= \pm \sqrt{\frac{1+\sqrt{3} i}{2}}, x= \pm \sqrt{\frac{-1-\sqrt{3} i}{2}} \\ & \text { Let } \alpha=\sqrt{\frac{1+\sqrt{3} i}{2}}, \beta=-\sqrt{\frac{1+\sqrt{3} i}{2}}, \\ & \quad \gamma=\sqrt{\frac{-1-\sqrt{3} i}{2}} \\ & \quad \delta=-\sqrt{\frac{-1-\sqrt{3} i}{2}} \\ & \Rightarrow \quad \alpha^3+\beta^3+\gamma^3+\delta^3=0 \end{aligned} $

∴ Given expression,

$ \frac{\alpha^3+\beta^3+\gamma^3+\delta^3}{\alpha^6+\beta^6+\gamma^6+\delta^6}=0 $

We have, $|z|-z=2+i$

$ \begin{aligned} & \text { Let } \quad z=x+i y \\ & \therefore \sqrt{x^2+y^2}-x-i y=2+i \end{aligned} $

Equating real part and imaginary part, we get

$ \begin{aligned} & & \sqrt{x^2+y^2}-x & =2 \text { and } y=-1 \\ & \therefore & \sqrt{x^2+1} & =x+2 \end{aligned} $

$ \begin{array}{rlrl} \Rightarrow & x^2+1 =x^2+4 x+4 \\ \Rightarrow & x =-\frac{3}{4} \\ \Rightarrow & z =-\frac{3}{4}-i \\ \therefore & |z| =\sqrt{\frac{9}{16}+1} \\ & =\sqrt{\frac{25}{16}}=\frac{5}{4} \end{array} $

Given, $\arg (z-2-3 i)=\frac{\pi}{4}$

$ \begin{aligned} \Rightarrow \quad z & =x+i y \\ z-2-3 i & =x+i y-2-3 i \\ & =(x-2)+(y-3) i \\ \arg (z-2-3 i) & =\tan ^{-1}\left(\frac{y-3}{x-2}\right)=\frac{\pi}{4} \\ \Rightarrow \quad \frac{y-3}{x-2} & =\tan \frac{\pi}{4} \\\quad y-3 & =x-2 \\ \Rightarrow \quad x-y+1 & =0 \end{aligned} $

∴ Locus of $z$ is $x-y+1=0$.

Given, $z_1, z_2, z_3, \ldots, z_n$ are roots of equation

$ \begin{aligned} (z+1)^n & =z^n \Rightarrow\left(\frac{z+1}{z}\right)=(1)^{1 / n} \\ \frac{z+1}{z} & =\left(\cos \frac{2 k \pi}{n}+i \sin \frac{2 k \pi}{n}\right) \\ z+1 & =Z\left(\cos \frac{2 k \pi}{n}+i \sin \frac{2 k \pi}{n}\right) \\ z & =\frac{1}{\cos \frac{2 k \pi}{n}+i \sin \frac{2 k \pi}{n}-1} \\ z & =\frac{1}{2}\left(1-i \cot \frac{k \pi}{n}\right) \\ \operatorname{Re} z & =\frac{1}{2^{\prime}} \operatorname{Im}(z)=\left(\frac{-1}{2} \cot \frac{k \pi}{n}\right) \\ 2 \operatorname{Re}(z i) & =1, \cot ^{-1}[2|\operatorname{Im} z i|]=\frac{k \pi}{n} \end{aligned} $

$ \begin{aligned} & \sum_{i=1}^v \frac{\cot ^{-1}(2(\operatorname{Im} z i)-1}{2 \operatorname{Re}(z i)}=\sum_{i=1}^v\left(\frac{k \pi}{n}-1\right) \\ & \quad=\frac{\pi}{n}(1+2+3+\ldots . . n)-n \\ & \quad=\pi \frac{(n)(n+1)}{2(n)}-n \\ & \quad=\pi \frac{(n+1)}{2}-n \\ & \quad=\frac{1}{2}[n \pi+\pi-2 n] \\ & \quad=\frac{1}{2}[\pi-(\pi-2) n] \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} z_1 & =x_1+i y_1 \\ z_2 & =x_2+i y_2 \\ z_3 & =x_1+\frac{i x_2}{2} \\ z_4 & =2 y_1+i y_2 \\ \left|z_1\right| & =1,\left|z_2\right|=2 \text { and } \operatorname{Re} \\ \left(z_1 z_2\right)=0 & \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Let } z_1=e^{i \alpha} \text { and } z_2=2 e^{i \beta}\\ &z_1 z_2=2 e^{i(\alpha+\beta)} \end{aligned} $

$ \begin{aligned} & z_1 z_2=2(\cos (\alpha+\beta)+i \sin (\alpha+\beta)) \\ & \operatorname{Re}\left(z_1+z_2\right)=2 \cos (\alpha+\beta)=0 \\ & \therefore \quad \alpha+\beta=\frac{\pi}{2} \\ & \quad \beta=\frac{\pi}{2}-\alpha \end{aligned} $

$ \begin{aligned} \therefore \quad z_1 & =\cos \alpha+i \sin \alpha \\ z_2 & =2(\cos \beta+i \sin \beta) \\ z_2 & =2(\sin \alpha+i \cos \alpha) \\ \therefore \quad z_3 & =\cos \alpha+i\left(\frac{2 \sin \alpha}{2}\right) \\ z_3 & =\cos \alpha+i \sin \alpha \\ \left|z_3\right| & =1 \\ z_4 & =2(\sin \alpha+i \cos \alpha) \\ \left|z_4\right| & =2 \\ z_3 z_4 & =2(\cos \alpha+i \sin \alpha)(\sin \alpha+i \cos \alpha) \\ z_3 z_4 & =2[(\cos \alpha \sin \alpha-\cos \alpha \sin \alpha) \\ & \left.\quad+i\left(\sin ^2 \alpha+\cos ^2 \alpha\right)\right] \end{aligned} $

$ \begin{gathered} z_3 z_4=2 i \\ \operatorname{Re}\left(z_3 z_4\right)=0 \end{gathered} $

We have, $|z| \geq 3$

Then, $\left|z+\frac{3}{z}\right| \leq|z|+\left|\frac{3}{z}\right|$

$ =\left|z+\frac{3}{z}\right| \leq 3+\left|\frac{3}{z}\right| $

∴ Least value of $\left|z+\frac{3}{z}\right|$ is not 1

And $\left|z_1-z_2\right| \leq\left|z_1\right|+\left|z_2\right|$ which is true.

$ \begin{aligned} & \text { At } \theta=\frac{\pi}{2} \\ & \left(\frac{\cos \theta+i \sin \theta}{\sin \theta+i \cos \theta}\right)^{2020}+\left(\frac{1+\cos \theta+i \sin \theta}{1-\cos \theta+i \sin \theta}\right)^{2021} \end{aligned} $

$ \begin{aligned} & =\left(\frac{0+i \cdot 1}{1+i \cdot 0}\right)^{2020}+\left(\frac{1+0+i}{1-0+i}\right)^{2021} \\ & =(i)^{2020}+\left(\frac{1+i}{1+i}\right)^{2021} \\ & =1+(1)^{2021}=1+1 \\ & =2=x+i y \Rightarrow x=2, y=0 \\ \therefore \quad x & +y=2+0=2 \end{aligned} $

As, $\omega$ is complex root of unity, then

$ \begin{gathered} 1+\omega+\omega^2=0, \omega^3=1 \\ \therefore \quad(\omega x+2)\left(\omega^2 x+2\right)-3 \\ =\omega^3 x^2+2 \omega x+2 \omega^2 x+4-3 \\ =x^2+2 x\left(\omega+\omega^2\right)+1 \quad\left(\because \omega^3=1\right) \\ =x^2+2 x(-1)+1=x^2-2 x+1 \\ \therefore \quad \sum_{x=1}^{10}\left((\omega x+2)\left(\omega^2 x+2\right)-3\right) \\ =\sum_{x=1}^{10}\left(x^2-2 x+1\right)=\sum_{x=1}^{10} x^2-2 \sum_{x=1}^{10} x+\sum_{x=1}^{10} 1 \\ =\frac{10(10+1)(2 \times 10+1)}{6}-\frac{2 \times 10(10+1)}{2}+10 \\ =\frac{10 \times 11 \times 21}{6}-10 \times 11+10 \\ =385-100=285 \end{gathered} $

We have, $|z| \leq 2$

$ \Rightarrow \sqrt{x^2+y^2} \leq 2 \Rightarrow x^2+y^2 \leq 4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Again, $(1-i) z+(1+i) \bar{z} \geq 4$

$ \begin{aligned} & \Rightarrow \quad(1-i)(x+i y)+(1+i)(x-i y) \geq 4 \\ & \Rightarrow \quad x+i y-i x+y+x-i y+i x+y \geq 4 \\ & \Rightarrow \quad 2 x+2 y \geq 4 \Rightarrow x+y \geq 2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

So, $A \cap B=\left\{x^2+y^2 \leq 4\right\} \cap\{x+y \geq 2\}$

$ \therefore \quad z=\sqrt{3}+\frac{1}{2} i $

Given equation, $z^2\left(1-z^2\right)=16, z \in \mathbf{C}$

Now, let $z^2=w=r(\cos \theta+i \sin \theta)$

where $r>0$.

$ \begin{aligned} & \therefore \quad 1-z^2=\frac{16}{z^2} \Rightarrow z^2+\frac{16}{z^2}=1 \\ & \Rightarrow \quad\left(r+\frac{16}{r}\right) \cos \theta+i\left(r-\frac{16}{r}\right) \sin \theta=1 \end{aligned} $

On comparing real and imaginary parts, we get

$ \left(r+\frac{16}{r}\right) \cos \theta=1 \text { and }\left(r-\frac{16}{r}\right) \sin \theta=0 $

∴ Either $\sin \theta=0$

$ \Rightarrow \quad \cos \theta=1 \Rightarrow r+\frac{16}{r}=1 $

(not possible)

or $\quad r-\frac{16}{r}=0 \Rightarrow r^2=16$

$ \begin{array}{lc} \Rightarrow & r=4 \\ \Rightarrow & |z|^2=4 \Rightarrow|z|=2 \end{array} $

Let $z=x+i y$

Then, vertices of rectangle are $(x, y)$, $(x,-y),(-x,-y),(-x, y)$.

Now, area of rectangle $=(2 x)(2 y)=4 x y$

It is given that,

$ \begin{array}{rlrl} & & 4 x y & =2 \sqrt{3} \Rightarrow 2 x y=\sqrt{3} \\ & \therefore & & x =\frac{1}{2}, y=\sqrt{3} \\ & \therefore & & z=\frac{1}{2}+\sqrt{3} i \end{array} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \left(\frac{\cos \theta+i \sin \theta}{\sin \theta+i \cos \theta}\right)^8+\left(\frac{1+\cos \theta-i \sin \theta}{1+\cos \theta+i \sin \theta}\right)^{16} \\ & =\left(\frac{\cos \theta+i \sin \theta}{i(\cos \theta-i \sin \theta)}\right)^8 +\left(\frac{2 \cos ^2 \frac{\theta}{2}-i 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}+i 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)^{16} \end{aligned} \end{aligned} $

$ \begin{aligned} = & \frac{1}{i^8}\left(\frac{\cos \theta+i \sin \theta}{\cos \theta-i \sin \theta}\right)^8+\left(\frac{\cos \frac{\theta}{2}-i \sin \frac{\theta}{2}}{\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}}\right)^{16} \\ = & \frac{\cos 8 \theta+i \sin 8 \theta}{\cos 8 \theta-i \sin 8 \theta}+\frac{\cos 8 \theta-i \sin 8 \theta}{\cos 8 \theta+i \sin 8 \theta} \\ = & (\cos 8 \theta+i \sin 8 \theta)(\cos 8 \theta-i \sin 8 \theta)^{-1} \\ & \quad+\left(\cos 8 \theta-i \sin 8 \theta(\cos 8 \theta+i \sin 8 \theta)^{-1}\right. \\ = & (\cos 8 \theta+i \sin 8 \theta)(\cos 8 \theta+i(\sin 8 \theta)+ \\ & \quad(\cos 8 \theta-i \sin 8 \theta)(\cos 8 \theta-i \sin 8 \theta) \\ = & (\cos 8 \theta+i \sin 8 \theta)^2+(\cos 8 \theta-i \sin 8 \theta)^2 \\ = & \cos ^2 8 \theta+i^2 \sin ^2 8 \theta+2 i \cos 8 \theta \sin 8 \theta \\ & \quad+\cos ^2 8 \theta+i^2 \sin ^2 8 \theta-2 i \cos 8 \theta \sin 8 \theta \\ = & 2\left(\cos ^2 8 \theta-\sin ^2 8 \theta\right)=2 \cos 16 \theta \end{aligned} $

We have, $Z = {{7 + i} \over {3 + 4i}}$

$ \Rightarrow Z = \left( {{{7 + i} \over {3 + 4i}}} \right)\left( {{{3 - 4i} \over {3 - 4i}}} \right) = {{21 + 4 - 28i + 3i} \over {9 + 16}} = 1 - i$

$\therefore$ ${Z^{14}} = {(1 - i)^{14}} = {[{(1 - i)^2}]^7} = {(1 - 1 - 2i)^7}$

$ = {( - 2)^7}{(i)^7} = {2^7}i$

The given equation is

$2(1 + i){x^2} - 4(2 - i)x - 5 - 3i = 0$

$ \Rightarrow x = {{4(2 - i) \pm \sqrt {16{{(2 - i)}^2} + 8(1 + i)(5 + 3i)} } \over {4(1 + i)}}$

$ = - {i \over {1 + i}}$ or ${{4 - i} \over {1 + i}} = {{ - 1 - i} \over 2}$ or, ${{3 - 5i} \over 2}$

Now, $\left| {{{ - 1 - i} \over 1}} \right| = \sqrt {{1 \over 4} + {1 \over 4}} = \sqrt {{1 \over 2}} $

and $\left| {{{3 - 5i} \over 2}} \right| = \sqrt {{9 \over 4} + {{25} \over 4}} = \sqrt {{{17} \over 2}} $

Also, $\sqrt {{{17} \over 2}} > \sqrt {{1 \over 2}} $

Hence, required root is ${{3 - 5i} \over 2}$.

We have,

$Z = r{e^{i\theta }}$

$Z = r(\cos \theta + i\sin \theta )$

${e^{iz}} = {e^{ir(\cos \theta + i\sin \theta )}}$

$ \Rightarrow {e^{iz}} = {e^{ir\cos \theta + {i^2}r\sin \theta }}$

$ \Rightarrow {e^{iz}} = {e^{ir\cos \theta - r\sin \theta }}$

$ \Rightarrow {e^{iz}} = {e^{ - r\sin \theta }}\,.\,{e^{ir\cos \theta }}$

$\therefore$ $\arg ({e^{iz}}) = r\cos \theta $

Let Re (z) = x, then

${{2(x + 10i) - n} \over {2(x + 10i) + n}} = 2i - 1$

$ \Rightarrow \left( {2x - n} \right) + 20i = - \left( {2x + n} \right) - 40 - 20i + 2ni$

$ \Rightarrow 2x - n = 2x - n - 40$

& 20 = -20 + 2n $ \Rightarrow $ x = -10 & n = 20

Let the complex number z = x + iy

Now given | (x + iy) - 1 | = | (x+iy) - i |

$ \Rightarrow $ (x - 1)² + y² = x² + (y - 1)²

$ \Rightarrow $ x = y

$ \therefore $ the correct answer is option (D)

$\left| {zw} \right| = 1$

$ \Rightarrow $ $\left| z \right|\left| w \right| = 1$

Let $w = {1 \over r}{e^{i\theta }}$

then z = $r{e^{i\left( {\theta + {\pi \over 2}} \right)}}$

$\overline z w = {e^{ - i\left( {\theta + {\pi \over 2}} \right)}}.{e^{i\theta }} = {e^{ - i(\pi /2)}} = - i$

$z\overline w = {e^{i\left( {\theta + {\pi \over 2}} \right)}}.{e^{ - i\theta }} = {e^{i\pi /2}} = i$

$z = {{{{\left( {1 + i} \right)}^2}} \over {a - i}} \times {{a + i} \over {a + i}}$

$ \Rightarrow z = {{\left( {1 - 1 + 2i} \right)\left( {a + i} \right)} \over {{a^2} + 1}} = {{2ai - 2} \over {{a^2} + 1}}$

$ \Rightarrow \left| z \right| = \sqrt {{{\left( {{{ - 2} \over {{a^2} + 1}}} \right)}^2} + {{\left( {{{2a} \over {{a^2} + 1}}} \right)}^2}} = \sqrt {{{4 + 4{a^2}} \over {{{({a^2} + 1)}^2}}}} $

$ \Rightarrow \sqrt {{{4(1 + {a^2})} \over {{{({a^2} + 1)}^2}}}} = {2 \over {\sqrt {{a^2} + 1} }}$.... (i)

Now given $\left| z \right| = \sqrt {{2 \over 5}} $

so $\sqrt {{2 \over 5}} = {2 \over {\sqrt {1 + {a^2}} }}$ from equation (i)

By squaring both sides

$ \Rightarrow {2 \over 5} = {4 \over {1 + {a^2}}}$

$ \Rightarrow 1 + {a^2} = 10$

a² = 9

$ \Rightarrow $ a = $ \pm $ 3

$ \because $ (a > 0) then a = 3

Now $\overline z = {{{{\left( {1 - i} \right)}^2}} \over {3 + i}}$

= ${{\left( {1 + {i^2} - 2i} \right)} \over {3 + i}}$

= ${{\left( {1 + {i^2} - 2i} \right)\left( {3 - i} \right)} \over {\left( {3 + i} \right)\left( {3 - i} \right)}}$

= ${{\left( { - 2i} \right)\left( {3 - i} \right)} \over {\left( {9 - {i^2}} \right)}}$

= ${{\left( { - 2i} \right)\left( {3 - i} \right)} \over {10}}$

= ${{ - 6i + 2{i^2}} \over {10}}$

= ${{ - 6i - 2} \over {10}}$

= $-{1 \over 5} - {3 \over 5}i$

$\omega = {{5 + 3z} \over {5(1 - z)}}$z

$ \Rightarrow $ $5\omega \left( {1 - z} \right) = 5 + 3z$

$ \Rightarrow $ $5\omega - 5\omega z = 5 + 3z$

$ \Rightarrow $ $5\omega - 5 = 5\omega z + 3z$

$ \Rightarrow $ $z\left( {5\omega + 3} \right) = 5\left( {\omega - 1} \right)$

$ \Rightarrow $ $z = {{5\left( {\omega - 1} \right)} \over {3 + 5\omega }}$

$ \Rightarrow $ $z = {{5\left( {\omega - 1} \right)} \over {5\left( {\omega + {3 \over 5}} \right)}}$

$ \Rightarrow $ $z = {{\left( {\omega - 1} \right)} \over {\left( {\omega + {3 \over 5}} \right)}}$

Given |z| < 1

$ \Rightarrow $ $\left| {{{\omega - 1} \over {\omega + {3 \over 5}}}} \right|$ < 1

$ \Rightarrow $ $\left| {\omega - 1} \right| < \left| {\omega + {3 \over 5}} \right|$

$ \Rightarrow $ $\left| {\omega - 1} \right| < \left| {\omega - \left( { - {3 \over 5}} \right)} \right|$

It says distance of $\omega $ from point 1 is less than distance from point ${\left( { - {3 \over 5}} \right)}$.

x coordinate of perpendicular bisector of points ${\left( { - {3 \over 5},0} \right)}$ and (1, 0),

x = ${{1 - {3 \over 5}} \over 2}$ = ${1 \over 5}$

As $\omega $ is closer to point (1, 0) so $\omega $ should present in the right side of perpendicular bisector.

$ \therefore $ ${\mathop{\rm Re}\nolimits} (\omega ) > {1 \over 5}$

$ \Rightarrow $ 5Re( $\omega$) > 1

Other Method :

$\omega = {{5 + 3z} \over {5(1 - z)}}$z

$ \Rightarrow $ $5\omega \left( {1 - z} \right) = 5 + 3z$

$ \Rightarrow $ $5\omega - 5\omega z = 5 + 3z$

$ \Rightarrow $ $5\omega - 5 = 5\omega z + 3z$

$ \Rightarrow $ $z\left( {5\omega + 3} \right) = 5\left( {\omega - 1} \right)$

$ \Rightarrow $ $z = {{5\left( {\omega - 1} \right)} \over {3 + 5\omega }}$

Given |z| < 1

$ \Rightarrow $ $\left| {{{5\left( {\omega - 1} \right)} \over {5\omega + 3}}} \right|$ < 1

$ \Rightarrow $ $5\left| {\omega - 1} \right| < \left| {5\omega + 3} \right|$

$ \Rightarrow $ $25\left( {{\omega ^2} - 2\omega + 1} \right) < 25{\omega ^2} + 30\omega + 9$

$ \Rightarrow $ $25\left( {\omega \overline \omega - \omega - \overline \omega + 1} \right)$ <

$25\omega \overline \omega + 15\omega + 15\overline \omega + 9$

(Using ${\left| z \right|^2} = z\overline z $)

$ \Rightarrow $ $16 < 40\omega + 40\overline \omega $

$ \Rightarrow $ $\omega + \overline \omega > {2 \over 5}$

$ \Rightarrow $ $2{\mathop{\rm Re}\nolimits} \left( \omega \right) > {2 \over 5}$

$ \Rightarrow $ 5Re( $\omega$) > 1

Let h + ik = ${{\alpha + i} \over {\alpha - i}}$

= ${{\left( {\alpha + i} \right)\left( {\alpha + i} \right)} \over {\left( {\alpha - i} \right)\left( {\alpha + i} \right)}}$

= ${{\left( {{\alpha ^2} - 1} \right) + 2i\alpha } \over {{\alpha ^2} + 1}}$

$ \therefore $ h = ${{{\alpha ^2} - 1} \over {{\alpha ^2} + 1}}$ and k = ${{2\alpha } \over {{\alpha ^2} + 1}}$

By squaring and adding we get

h² + k² = 1

$ \therefore $ This is circle whose radius is 1.

$z = {{\sqrt 3 } \over 2} + {i \over 2}$

$ \Rightarrow $ z = $\cos {\pi \over 6}$ + i $\sin {\pi \over 6}$

$ \Rightarrow $ z = ${e^{i{\pi \over 6}}}$

x² – 2x + 2 = 0

$ \therefore $ x = ${{2 \pm \sqrt { - 4} } \over 2} = 1 \pm i$

Now, ${\alpha \over \beta } = {{1 + i} \over {1 - i}} = {{{{\left( {1 + i} \right)}^2}} \over {1 - {i^2}}} = i$

or ${\alpha \over \beta } = {{1 - i} \over {1 + i}} = {{{{\left( {1 - i} \right)}^2}} \over {1 + {i^2}}} = - i$

$ \therefore $ ${\left( { i} \right)^n}$ = 1

and ${\left( { - i} \right)^n}$ = 1

So n must be a multiple of 4

$ \therefore $ Minimum value of n = 4

$\left| {{z_1}} \right| = 9,\,\,\left| {{z_2} - \left( {3 + 4i} \right)} \right| = 4$

${C_1},(0,0)$ radius r₁ = 9

C₂ (3, 4), radius r₂ = 4

C₁C₂ = $\left| {{r_1} - {r_2}} \right| = 5$

$ \therefore $  Circle touches internally

$ \therefore $  ${\left| {{z_1} - {z_2}} \right|_{\min }} = 0$

${{z - \alpha } \over {z + \alpha }} + {{\overline z - \alpha } \over {\overline z + \alpha }} = 0$

$z\overline z + z\alpha - \alpha \overline z - {\alpha ^2} + z\overline z - z\alpha + \overline z \alpha - {\alpha ^2} = 0$

${\left| z \right|^2} = {\alpha ^2},$  $a = \pm 2$

$\left| z \right| + z = 3 + i$

$z = 3 - \left| z \right| + i$

Let  $3 - \left| z \right| = a \Rightarrow \left| z \right| = \left( {3 - a} \right)$

$ \Rightarrow z = a + i \Rightarrow \left| z \right| = \sqrt {{a^2} + 1} $

$ \Rightarrow 9 + {a^2} - 6a = {a^2} + 1 \Rightarrow a = {8 \over 6} = {4 \over 3}$

$ \Rightarrow \left| z \right| = 3 - {4 \over 3} = {5 \over 3}$

${\left( { - 2 - {i \over 3}} \right)^3} = - {{{{\left( {6 + i} \right)}^3}} \over {27}}$

$ = {{ - 198 - 107i} \over {27}} = {{x + iy} \over {27}}$

Hence, $y - x = 198 - 107 = 91$

$z = {\left( {{{\sqrt 3 + i} \over 2}} \right)^5} + {\left( {{{\sqrt 3 - i} \over 2}} \right)^5}$

$z = {\left( {{e^{i\pi /6}}} \right)^5} + {\left( {{e^{ - i\pi /6}}} \right)^5}$

$ = {e^{i5\pi /6}} + {e^{ - i5\pi /6}}$

$ = \cos {{5\pi } \over 6} + i{{\sin 5\pi } \over 6} + \cos \left( {{{ - 5\pi } \over 6}} \right) + i\sin \left( {{{ - 5\pi } \over 6}} \right)$

$ = 2\cos {{5\pi } \over 6} < 0$

${\rm I}(z) = 0$ and ${\mathop{\rm Re}\nolimits} (z) < 0$

Given, $3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|$

$ \Rightarrow $ ${{\left| {{z_1}} \right|} \over {\left| {{z_2}} \right|}} = {4 \over 3}$

$ \Rightarrow $ ${{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}} = {4 \over 3} \times {3 \over 2} = 2$

As we know, for any compled number

${{3{z_1}} \over {2{z_2}}} = {{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}}$(cos$\theta $ + i sin$\theta $)

= 2(cos$\theta $ + i sin$\theta $)

$ \therefore $ ${{2{z_2}} \over {3{z_1}}}$ = ${1 \over {2\left( {\cos \theta + i\sin \theta } \right)}}$

= ${1 \over {2\left( {\cos \theta + i\sin \theta } \right)}} \times {{\left( {\cos \theta - i\sin \theta } \right)} \over {\left( {\cos \theta - i\sin \theta } \right)}}$

= ${\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}$
Now, given

$z = {{3{z_1}} \over {2{z_2}}} + {{2{z_2}} \over {3{z_1}}}$

= 2(cos$\theta $ + i sin$\theta $) + ${\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}$

= ${{5 \over 2}\cos \theta + {3 \over 2}i\sin \theta }$

So, |z| = $\sqrt {{{25} \over 4}{{\cos }^2}\theta + {9 \over 4}{{\sin }^2}\theta } $

= ${1 \over 2}\sqrt {9 + 16{{\cos }^2}\theta } $

z is neither purely real nor purely imaginary and |z| depends on $\theta $.

1 + x + x² = 0

x = ${{ - 1 \pm \sqrt {1 - 4} } \over 2} = {{ - 1 \pm i\sqrt 3 } \over 2}$

z₀ = w, w²

Now  

z = 3 + 6iz$_0^{81}$ $-$ 3iz$_0^{93}$

z = 3 + 6iw⁸¹ $-$ 3iw⁹³      (w⁹³ = w⁸¹ = 1)

$ \Rightarrow $  z = 3 + 3i

then arg(z) = tan^$-$1$\left( {{3 \over 3}} \right)$ = tan^$-$1 (1) = ${\pi \over 4}$

Given complex number,

${{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}$

$ = {{\left( {3 + 2i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$

$ = {{3 + 6i\sin \theta + 2i\sin \theta - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}$

$ = {{\left( {3 - 4{{\sin }^2}\theta } \right) + i\left( {8\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$

As complex number is purely imaginary, So real part of this complex number is zero.

$ \therefore $   ${{3 - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}$ = 0

$ \Rightarrow $   $3 - 4{\sin ^2}\theta = 0$

$ \Rightarrow $   $\sin \theta = \pm {{\sqrt 3 } \over 2}$

as   $\theta $ $ \in $ $\left( { - {\pi \over 2},\pi } \right)$

$ \therefore $   $\theta $ $=$ $-$ ${\pi \over 3},{\pi \over 3},{{2\pi } \over 3}$

$ \therefore $   Sum of those values of A is

$ = - {\pi \over 3} + {\pi \over 3} + {{2\pi } \over 3}$

$ = {{2\pi } \over 3}$

Given equation,

x² + 2x + 2 = 0

$ \therefore $  x = ${{ - 2 \pm \sqrt {4 - 4.1.2} } \over {2.1}}$

x = $-$ 1 $ \pm $ i

$ \therefore $  $\alpha $ = $-$ 1 + i

and $\beta $ = $-$ 1 $-$ i

Note :

x + iy = r (cos$\theta $ + isin$\theta $)

$ \therefore $  (x + iy)ⁿ = rⁿ (cosn$\theta $ + isinn$\theta $)

$ \therefore $  $-$ 1 + i = $\sqrt 2 $ [cos${{3\pi } \over 4}$ + isin${{3\pi } \over 4}$ ]

$ \Rightarrow $  ($-$ 1 + i)¹⁵ = ${\left( {\sqrt 2 } \right)^{15}}$ [cos$\left( {{{15.3\pi } \over 4}} \right) + i\sin \left( {{{15.3\pi } \over 4}} \right)$]

And $-$1 $-$ i = $\sqrt 2 $ $\left[ {\cos \left( { - {{3\pi } \over 4}} \right) + i\sin \left( { - {{3\pi } \over 4}} \right)} \right]$

= $\sqrt 2 \left[ {\cos {{3\pi } \over 4} - \sin {{3\pi } \over 4}} \right]$

$ \therefore $  ($-$1 $-$ i)¹⁵ = ${\left( {\sqrt 2 } \right)^{15}}\left[ {\cos \left( {{{15.3\pi } \over 4}} \right) - i\sin \left( {{{15.3\pi } \over 4}} \right)} \right]$

Now

$\alpha $¹⁵ + $\beta $¹⁵

= ($-$1 + i)¹⁵ + ($-$ 1 $-$ i)¹⁵

=  ${\left( {\sqrt 2 } \right)^{15}}$ $\left[ {2\cos \left( {{{15.3\pi } \over 4}} \right)} \right]$

= ${\left( {\sqrt 2 } \right)^{15}}\left[ {2\cos \left( {11\pi + {\pi \over 4}} \right)} \right]$

= ${\left( {\sqrt 2 } \right)^{15}}\left[ {2\left( { - \cos {\pi \over 4}} \right)} \right]$

= ${\left( {\sqrt 2 } \right)^{15}} \times 2 \times - {1 \over {\sqrt 2 }}$

= $ - {\left( {\sqrt 2 } \right)^{14}}.2$

= $-$ 2⁷ $ \times $ 2

= $-$ 2⁸

= $-$ 256