Complex Numbers

Both P and Q are true.

$\because$ Length of direct ditance $\le$ length of arc

i.e. | z₂ $-$ z₁ | = length of line AB $\le$ length of arc AB.


| z₃ $-$ z₂ | = length of line BC $\le$ length of arc BC.

$\therefore$ Sum of length of these 10 lines $\le$ sum of length of arcs (i.e. 2$\pi$) (because $\theta$₁ + $\theta$₂ + $\theta$₃ + .... + $\theta$₁₀ = 2$\pi$ (given)

$\therefore$ | z₂ $-$ z₁ | + | z₃ $-$ z₂ | + ..... + | z₁ $-$ z₁₀ | $\le$ 2$\pi$ $\to$ P is true.

And | z$_k^2$ $-$ z$_{k - 1}^2$ | = | z_k $-$ z_{k $-$ 1} | | z_k + z_{k $-$ 1} |

As we know that,

| z_k + z_{k $-$ 1} | $\le$ | z_k | + | z_{k $-$ 1} | $\le$ 2

$\therefore$ | z$_2^2$ $-$ z$_1^2$ | + | z$_3^2$ $-$ z$_2^2$ | + .... + | z$_1^2$ $-$ z$_{10}^2$ | $\le$ 2 ( | z₂ $-$ z₁ | + | z₃ $-$ z₂ | + .... + | z₁ $-$ z₁₀ | )

$\le$ 2(2$\pi$)

$\le$ 4$\pi$ $\to$ Q is true.

Circle ${x^2} + {y^2} + 5x - 3y + 4 = 0$ cuts the real axis (X-axis) at ($-$4, 0), ($-$1, 0).


$\arg \left( {{{z + \alpha } \over {z + \beta }}} \right) = {\pi \over 4}$ implies z is on arc and ($-$ $\alpha$, 0) and ($-$ $\beta$, 0) subtend ${\pi \over 4}$ on z.

So, $\alpha$ = 1 and $\beta$ = 4

Hence, $\alpha$$\beta$ = 1 $\times$ 4 = 4 and $\beta$ = 4

$\begin{array}{left} \left|\begin{array}{aligned} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|=a\left(b c-a^2\right)-b\left(b^2-a c\right)+c\left(a b-c^2\right) \\ \Rightarrow a b c-a^3-b^3+a b c+a b c-c^3=0 \\ \Rightarrow 3 a b c-a^3-b^3-c^3=0 \\ \Rightarrow a^3+b^3+c^3-3 a b c=0 \\ \Rightarrow (a+b+c)=0 \end{array}$

Since, $a=\left|Z_1\right|, b=\left|Z_2\right|, c=\left|Z_3\right|$

$\therefore \quad\left|Z_1\right|+\left|Z_2\right|+\left|Z_3\right|=0$

$\left|z_1+z_2\right|^2=\left|z_1\right|^2+\left|z_2\right|^2$ $\quad$ (given) ..... (i)

Let $z_1=x+i y, z_2=a+i b$.

Then

$\begin{aligned} & \frac{z_1}{z_2}=\frac{x+i y}{a+i b}=\frac{x+i y}{a+i b} \times \frac{a-i b}{a-i b} \\ & \frac{z_1}{z_2}=\frac{(a x+y b)+i(a y-x b)}{a^2+b^2} \quad \text{...... (ii)} \end{aligned}$

From Eq. (i), we get

$\begin{aligned} & \text { LHS }=\left|z_1+z_2\right|^2=|x+a+i(y+b)|^2 \\ & =(x+a)^2+(y+b)^2 \\ & =x^2+a^2+2 a x+y^2+b^2+2 b y \\ & \text { RHS }=|x+i y|^2+|a+i b|^2 \\ & =x^2+y^2+a^2+b^2 \\ & \because \text { LHS }=\text { RHS } \\ & \Rightarrow x^2+y^2+a^2+b^2+2(a x+b y) \\ & =x^2+y^2+a^2+b^2 \\ & \Rightarrow \quad 2(a x+b y)=0 \Rightarrow a x+b y=0 \\ \end{aligned}$

Put $a x+b y=0$ in Eq. (ii),

$\frac{z_1}{z_2}=\frac{i(a y-x b)}{a^2+b^2}$, which is purely imaginary.

$\begin{aligned} & \text { LHS }=\frac{3-4 i x}{3+4 i x} \times \frac{3-4 i x}{3-4 i x} \\ & =\frac{9+16 i^2 x^2-24 i x}{9-\left(16 i^2 x^2\right)}=\frac{9-16 x^2-24 i x}{9+16 x^2} \\ & =\frac{9-16 x^2}{9+16 x^2}-\frac{24 i x}{9+16 x^2}=\alpha-i \beta=\text { RHS } \\ & \therefore \quad \alpha=\frac{9-16 x^2}{9+16 x^2} \\ & \beta=\frac{24 x}{9+16 x^2} \\ & \Rightarrow \quad a^2+\beta^2=\frac{\left(9-16 x^2\right)^2+(24 x)^2}{\left(9+16 x^2\right)^2} \\ & =\frac{\left(9+16 x^2\right)^2}{\left(9+16 x^2\right)^2}=1 \\ & \therefore \quad \alpha^2+\beta^2=1 \\ & \end{aligned}$

$\begin{aligned}(1-i \sqrt{3})^9 & =2^9\left(\frac{1}{2}-i \frac{\sqrt{3}}{2}\right)^9 \\ & =2^9\left(\cos \frac{\pi}{3}-i \sin \frac{\pi}{3}\right)^9=2^9\left(e^{-i \pi / 3}\right)^9 \\ & =2^9\left(e^{-i 3 \pi}\right)=2^9(\cos 3 \pi-i \sin 3 \pi) \\ & =2^9\left((-1)^3-0\right)=-2^9\end{aligned}$

${\left( {\sqrt 2 {{(\sqrt 3 - 1)} \over {2\,.\,2}} + i\sqrt 2 {{(\sqrt 3 + 1)} \over {2\,.\,2}}} \right)^{2020}}$

$ = {\left( {{{\left( {{{\sqrt 3 } \over 2} - {1 \over 2}} \right)} \over {\sqrt 2 }} + i{{\left( {{{\sqrt 3 } \over 2} + {1 \over 2}} \right)} \over {\sqrt 2 }}} \right)^{2020}}$

$ = {\left( {{{{1 \over {\sqrt 2 }}\left( {{{\sqrt 3 } \over 2} - {1 \over 2}} \right)} \over 1} + {{i{1 \over {\sqrt 2 }}\left( {{{\sqrt 3 } \over 2} + {1 \over 2}} \right)} \over 1}} \right)^{2020}}$

$ = {\left( {\left( {{1 \over {\sqrt 2 }}\,.\,{{\sqrt 3 } \over 2} - {1 \over {\sqrt 2 }}\,.\,{1 \over 2}} \right) + i\left( {{1 \over {\sqrt 2 }}\,.\,{{\sqrt 3 } \over 2} + \left( {{1 \over {\sqrt 2 }}\,.\,{1 \over 2}} \right)} \right)} \right)^{2020}}$

$ = (\cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ ) + i(\sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ ){)^{2020}}$

$ = {(\cos 75^\circ + i\sin 75^\circ )^{2020}}$

$ = {\left( {\cos {{5\pi } \over {12}} + i\sin {{5\pi } \over {12}}} \right)^{2020}}$

$ = \left( {\cos {{5\pi } \over {12}} \times 2020 + i\sin {{5\pi } \over {12}} \times 2020} \right)$ [De-moivre]

$ = \left( {\cos {{2525\pi } \over 3} + i\sin {{2525\pi } \over 3}} \right)$

$ = (\cos (2525 \times 60)^\circ + i\sin (2525 \times 60)^\circ $

$ = (\cos (n\pi - 60)^\circ + i\sin (n\pi - 60)^\circ )$, where n = 842

$ = ({( - 1)^n}\cos 60^\circ + i( - \sin 60^\circ ))$

$ = (\cos 60^\circ - i\sin 60^\circ )$ [$\because$ n = even]

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

$\begin{aligned} & \text { } z_1=2+3 i, z_2=3+2 i \\ & \Rightarrow\left[\begin{array}{cc}z_1 & z_2 \\ -\bar{z}_2 & \bar{z}_1\end{array}\right]\left[\begin{array}{cc}\bar{z}_1 & -z_2 \\ \bar{z}_2 & z_1\end{array}\right] \\ & =\left[\begin{array}{cc}\left|z_1\right|^2+\left|z_2\right|^2 & 0 \\ 0 & \left|z_1\right|^2+\left|z_2\right|^2\end{array}\right] \\ & \Rightarrow \quad\left|z_1\right|^2=4+9=13 \\ & \Rightarrow \quad\left|z_2\right|^2=9+4=13=\left[\begin{array}{cc}26 & 0 \\ 0 & 26\end{array}\right]=26 I \\ & \end{aligned}$

$\begin{array}{rlrl} & \text { } & (1+i)(1+3 i)(1+7 i) & =x+i y \\ \Rightarrow & & (-2+4 i)(1+7 i) & =x+i y \\ \Rightarrow & & -30-10 i & =x+i y \\ \Rightarrow & & |x+i y| & =\sqrt{30^2+10^2} \\ & & =\sqrt{1000}=10 \sqrt{10}\end{array}$

$\because 1, \alpha_1, \alpha_2, \alpha_3, \alpha_4$ are the roots of $z^5-1=0$

$\begin{gathered} \Rightarrow \quad(z-1)\left(z-\alpha_1\right)\left(z-\alpha_2\right)\left(z-\alpha_3\right)\left(x-\alpha_4\right)=z^5-1 \\ \Rightarrow \quad(\omega-1)\left(\omega-\alpha_1\right)\left(\omega-\alpha_2\right)\left(\omega-d_3\right)\left(\omega-\alpha_4\right)=\omega^5-1 \\ (\omega-1)\left(\omega-\alpha_1\right)\left(\omega-\alpha_2\right)\left(\omega-\alpha_3\right)\left(\omega-\alpha_4\right)+\omega \\ =\left(\omega^5-1\right)+\omega=\omega^2-1+\omega \\ =-1-1=-2 \end{gathered}$

$\begin{aligned} & \text { } a > 0, z=x+i y \\ & \log _{\cos ^2{ }_\theta}|z-a| > \log _{\cos ^2{ }_\theta}|z-a i| \\ & 0 < \cos ^2 \theta < 1 \\ & \text { So, }|z-a| < |z-a i| \\ & \Rightarrow \quad(x-a)^2+y^2 < x^2+(y-a)^2 \\ & \Rightarrow x^2+a^2-2 a x+y^2 < x^2+y^2+a^2-2 a y \\ & \Rightarrow-2 a x < -2 a y \\ & \Rightarrow \quad x > y\end{aligned}$

$i x^2-2(i+1) x+(2-i)=0$ ..... (i)

Let the other roots be $\alpha$.

$\begin{aligned} & {[x-(2-i)][x-\alpha]=0} \\ & \Rightarrow x^2-(2-i+\alpha) x-(\alpha(2-i))=0 \\ & \text { \{comparing with Eq. (i) } \\ & \Rightarrow i x^2-(2 i+1+\alpha i) x-[\alpha(2 i+1)]=0 \\ & \Rightarrow \quad 2 i+1+\alpha i=2 i+2 \\ & \Rightarrow \\ & \alpha i=1 \\ & \Rightarrow \quad \alpha=\frac{1}{i}=-i \\ & \end{aligned}$

$\because|z-2|=|z-1|$

Let $z=x+i y$, then

$\begin{array}{ll} \Rightarrow & |x+i y-2|^2=|x+i y-1|^2 \\ \Rightarrow & (x-2)^2+y^2=(x-1)^2+y^2 \\ \Rightarrow & x^2-4 x+4=x^2-2 x+1 \\ \Rightarrow & -2 x+4 x=4-1 \Rightarrow x=3 / 2 \end{array}$

Which is parallel to $Y$-axis.

${\left( {{{1 + i} \over {1 - i}}} \right)^m} = 1 \Rightarrow {\left[ {{{(1 + i)(1 + i)} \over {(1 - i)(1 + i)}}} \right]^m} = 1$

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

$\Rightarrow i^m=1$

$\Rightarrow m$ is multiple of $4\left[\because i^{4 n}=1, \forall n \in N\right]$

Option (a) is not a multiple of 4 .

$\therefore \quad m \neq 1934$

$(\sin \theta-i \cos \theta)^3=(-i)^3(\cos \theta+i \sin \theta)^3$

$=(-i)^3(\cos 3 \theta+i \sin 3 \theta)$ [by De-Moivre theorem]

Real part of $(\cos 4+i \sin 4+1)^{2020}=$ Real part of $\left(2 \cos ^2 2+i 2 \sin 2 \cos 2\right)^{2020}$

$\begin{aligned} & =\text { Real part of } 2^{2020} \cos ^{2020}(2)(\cos 2+i \sin 2)^{2020} \\ & =\text { Real part of } 2^{2020} \cos ^{2020}(2)(\cos 4040+i \sin 4040) \\ & =2^{2020} \cos ^{2020} 2 \cdot \cos 4040 \quad[z+\bar{z}=2 \operatorname{Re}(z)] \end{aligned}$

Given z = x + iy

and ${z^2} = i{\left| z \right|^2}$

$ \Rightarrow $ (x + iy)² = i(x² + y²)

$ \Rightarrow $ x² - y² + 2ixy = i(x² + y²) + 0

Comparing both side we get,

x² - y² = 0

$ \Rightarrow $ x² = y²

and 2xy = (x² + y²)

$ \Rightarrow $ (x - y)² = 0

$ \Rightarrow $ x = y

$ \therefore $ z lies on line x = y

Given z = x + iy

|z| – Re(z) $ \le $ 1

$ \Rightarrow $ $\sqrt {{x^2} + {y^2}} $ - x $ \le $ 1

$ \Rightarrow $ $\sqrt {{x^2} + {y^2}} $ $ \le $ 1 + x

$ \Rightarrow $ x² + y² $ \le $ 1 + 2x + x²

$ \Rightarrow $ y² $ \le $ 2x + 1

$ \Rightarrow $ y² $ \le $ 2$\left( {x + {1 \over 2}} \right)$

${\left( {{{ - 1 + i\sqrt 3 } \over {1 - i}}} \right)^{30}}$

= ${\left( {{{2\omega } \over {1 - i}}} \right)^{30}}$

= ${{{2^{30}}.{\omega ^{30}}} \over {{{\left( {{{\left( {1 - i} \right)}^2}} \right)}^{15}}}}$

= ${{{2^{30}}.1} \over {{{\left( {1 + {i^{^2}} - 2i} \right)}^{15}}}}$

= ${{{2^{30}}.1} \over { - {2^{15}}.{i^{15}}}}$

= –2¹⁵i

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} $