Complex Numbers

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} $

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

The complex number z satisfying $\left| {z - 2 + i} \right| \ge \sqrt 5 $, which represents the region outside the circle (including the circumference) having centre (2, ${ - 1}$) and radius $\sqrt 5 $ units.



Now, for ${{z_0} \in S{1 \over {\left| {{z_0} - 1} \right|}}}$ is maximum.

When ${\left| {{z_0} - 1} \right|}$ is minimum. And for this it is required that ${{z_0} \in S}$, such that z₀ is collinear with the points (2, $ - $1) and (1, 0) and lies on the circumference of the circle $\left| {z - 2 + i} \right|$ = $\sqrt 5 $.

So let z₀ = x + iy, and from the figure 0 < x < 1 and y >0.

So, ${{4 - {z_0} - {{\overline z }_0}} \over {{z_0} - {{\overline z }_0} + 2i}} = {{4 - x - iy - x + iy} \over {x + iy - x + iy + 2i}} = {{2(2 - x)} \over {2i(y + 1)}} = - i\left( {{{2 - x} \over {y + 1}}} \right)$

$ \because $ ${{{2 - x} \over {y + 1}}}$ is a positive real number, so ${{4 - {z_0} - {{\overline z }_0}} \over {{z_0} - {{\overline z }_0} + 2i}}$ is purely negative imaginary number.

$ \Rightarrow $ $\arg \left( {{{4 - {z_0} - {{\overline z }_0}} \over {{z_0} - {{\overline z }_0} + 2i}}} \right) = - {\pi \over 2}$

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

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

We know, $\omega $ = $-$ ${{1 - i\sqrt 3 } \over 2}$

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

$ \Rightarrow $ $\,\,\,\,$ ${\left( {{{ - {\omega ^2}} \over { - \omega }}} \right)^n} = 1$

$ \Rightarrow $ ($\omega $)ⁿ = 1 = $\omega $³

$ \Rightarrow $ $\,\,\,\,$ n = 3

Given equation,

x² $-$ x + 1 = 0

Roots of this equation

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

$\therefore\,\,\,$ $ \propto \, = \,{{1 + \sqrt 3 \,i} \over 2}$

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

We know;

$\omega = {{ - 1 + \sqrt 3 \,i} \over 2} = - \left( {{{1 - \sqrt 3 \,i} \over 2}} \right) = - \beta $

and ${\omega ^2} = {{ - 1 - \sqrt 3 \,i} \over 2} = - \left( {{{1 + \sqrt 3 \,i} \over 2}} \right) = - \propto $

$\therefore\,\,\,$ $ \propto \, = - {\omega ^2}$ and $\beta \, = \, - \omega $

$\therefore\,\,\,$ ${ \propto ^{101}} + {\beta ^{107}}$

$ = {\left( { - {\omega ^2}} \right)^{101}} + {\left( { - \omega } \right)^{107}}$

$ = {\left( { - 1} \right)^{101}}.{\left( {{\omega ^2}} \right)^{101}} + {\left( { - 1} \right)^{107}}.{\left( \omega \right)^{107}}$

$ = - 1.{\left( {{\omega ^2}} \right)^{101}} - {\omega ^{107}}$

$ = - \left( {{\omega ^{202}} + {\omega ^{107}}} \right)$

$ = - \left( {{\omega ^{3.67}}.\omega + {\omega ^{3.35}}.{\omega ^2}} \right)$

$ = - \left( {\omega + {\omega ^2}} \right)\,\,\,$ [ as $\,\,\,$ ${\omega ^{3n}} = 1$]

$ = - \left( { - 1} \right)$ $\,\,\,\,\,\,$ [as $\,\,\,$ $1 + \omega + {\omega ^2} = 0$ ]

$ = 1$

$\left| {z - \left( {3 - 2i} \right)} \right| \le 4$ represents a circle whose center is (3, $-$2) and radius = 4.

$\left| z \right|$ = $\left| z -0\right|$  represents the distance of point 'z' from origin (0, 0)



Suppose RS is the normal of the circle passing through origin 'O' and G is its center (3, $-$2).

Here, OR is the least distance

and OS is in the greatest distance

OR = RG $-$ OG and OS = OG + GS . . . . .(1)

As, RG = GS = 4

OG = $\sqrt {{3^2} + \left( { - 2{)^2}} \right)} $ = $\sqrt {9 + 4} $ = $\sqrt {13} $

From (1), OR = 4 $-$ $\sqrt {13} $ and OS = 4 + $\sqrt {13} $

So, required difference = $\left( {4 + \sqrt {13} } \right)$ $-$ $\left( {4 - \sqrt {13} } \right)$

= $\sqrt {13} + \sqrt {13} $ = $2\sqrt {13} $

As w = ${{1 + \left( {1 - 8\alpha } \right)z} \over {1 - z}}$, w is purely imaginary

$ \therefore w$ + $\bar w$ = 0

$ \Rightarrow $ ${{1 + \left( {1 - 8\alpha } \right)z} \over {1 - z}}$ + ${{1 + \left( {1 - 8\alpha } \right)\bar z} \over {1 - \bar z}}$ = 0

$ \Rightarrow $ [1 + (1 - 8$\alpha $)][1 - $\bar z$] + [1 + ( 1 - 8$\alpha $)][1 - z] = 0

$ \Rightarrow $ 2 - (z + $\bar z$) + (1 - 8$\alpha $)(z + $\bar z$) - 2(1 - 8$\alpha $) = 0

$ \Rightarrow $ 2 - (z + $\bar z$) + (z + $\bar z$) - 8$\alpha $(z + $\bar z$) - 2 + 16$\alpha $ = 0

$ \Rightarrow $ 16$\alpha $ = 8$\alpha $(z + $\bar z)$

z + $\bar z$ = 2 or $\alpha $ = 0

but z + $\bar z$ = 2 is not possible as Re(Z) $ \ne $ 1

$ \therefore $ $\alpha $ = 0

$ \therefore $ $\alpha $ $ \in $ {0}

We have,

$sz + t\overline z + r = 0$ ...(i)

On taking conjugate,

$\overline {sz} + \overline t z + \overline r = 0$ ... (ii)

On solving Eqs. (i) and (ii), we get

$z = {{\overline r t - r\overline s } \over {|s{|^2} - |t{|^2}}}$

(a) For unique solutions of z

$|s{|^2} - |t{|^2} \ne 0 \Rightarrow |s|\, \ne \,|t|$

It is true

(b) If |s| = |t|, then ${\overline r t - r\overline s }$ may or may not be zero. So, z may have no solutions.

$ \therefore $ L may be an empty set.

It is false.

(c) If elements of set L represents line, then this line and given circle intersect at maximum two point. Hence, it is true.

(d) In this case locus of z is a line, so L has infinite elements. Hence, it is true.

(a) Let z= $-$1$-$i and arg(z) = $\theta $

Now, $\tan \theta = \left| {{{im(z)} \over {{\mathop{\rm Re}\nolimits} (z)}}} \right| = \left| {{{ - 1} \over { - 1}}} \right| = 1$

$ \Rightarrow $ $\theta = {\pi \over 4}$

Since, x < 0, y < 0

$ \therefore $ $\arg (z) = - \left( {\pi - {\pi \over 4}} \right) = - {{3\pi } \over 4}$

(b) We have, f(t) = arg($-$1 + it)

$\arg ( - 1 + it) = \left\{ \matrix{ \pi - {\tan ^{ - 1}}t,\,t \ge 0 \hfill \cr - (\pi + {\tan ^{ - 1}}t),\,t < 0 \hfill \cr} \right.$

This function is discontinuous at t = 0.

(c) We have,

$\arg \left( {{{{z_1}} \over {{z_2}}}} \right) = arg({z_1}) - arg({z_2}) + 2n\pi $

$ \therefore $ $\arg \left( {{{{z_1}} \over {{z_2}}}} \right) - arg({z_1}) + arg({z_2})$

= arg(z₁) $-$ arg(z₂) + 2n$\pi $ $-$ arg(z₁) + arg(z₂) = 2n$\pi $

So, given expression is multiple of 2$\pi $.

(d) We have, $\arg \left( {{{(z - {z_1})({z_2} - {z_3})} \over {(z - {z_3})({z_2} - {z_1})}}} \right) = \pi $


Now, $\arg \left( {{{{z_1} - z} \over {{z_3} - z}}} \right) = \theta $

and $\arg \left( {{{{z_3} - {z_2}} \over {{z_1} - {z_2}}}} \right) = \pi - \theta $

Therefore, $\arg \left( {{{{z_1} - z} \over {{z_3} - z}}} \right) + \arg \left( {{{{z_3} - {z_2}} \over {{z_1} - {z_2}}}} \right) = \theta + \pi - \theta $

That is,

$\arg \left( {{{{z_1} - z} \over {{z_3} - z}}.{{{z_3} - {z_2}} \over {{z_1} - {z_2}}}} \right) = \arg \left( {{{\left( {{z_1} - z} \right)\left( {{z_3} - {z_2}} \right)} \over {\left( {{z_3} - z} \right)\left( {{z_1} - {z_2}} \right)}}} \right) = \pi $

This implies that for any three given distinct complex numbers z₁ , z₂ and z₃ , the locus of the point z satisfying the condition
$\arg \left( {{{\left( {{z_1} - z} \right)\left( {{z_3} - {z_2}} \right)} \over {\left( {{z_3} - z} \right)\left( {{z_1} - {z_2}} \right)}}} \right) = \pi $, lies on a circle.

$ \therefore $ (a), (b), (d) are false statement.

Hence, option (a), (b), (d) are correct answer.

Let z = x + iy

Then,

Im $\left( {{{iz - 2} \over {z - i}}} \right)$ + 1 = 0

$ \Rightarrow $ ${\mathop{\rm Im}\nolimits} \left[ {\left( {{{i\left( {x + iy} \right) - 2} \over {x + iy - i}}} \right)} \right] + 1 = 0$

$ \Rightarrow $${\mathop{\rm Im}\nolimits} \left[ {\left( {{{ix - y - 2} \over {x + i\left( {y - 1} \right)}}} \right)} \right] + 1 = 0$

$ \Rightarrow $${\mathop{\rm Im}\nolimits} \left[ {\left( {{{ix - y - 2} \over {x + i\left( {y - 1} \right)}}} \right)\left( {{{x - i\left( {y - 1} \right)} \over {x - i\left( {y - 1} \right)}}} \right)} \right] + 1 = 0$

$ \Rightarrow $${\mathop{\rm Im}\nolimits} \left[ {\left( {{\matrix{ i{x^2} - {i^2}x\left( {y - 1} \right) - xy + \hfill \cr \,\,\,\,iy\left( {y - 1} \right) - 2x + i2\left( {y - 1} \right) \hfill \cr} \over {{x^2} - {i^2}{{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0$

$ \Rightarrow $ ${\mathop{\rm Im}\nolimits} \left[ {\left( {{\matrix{ i{x^2} + x\left( {y - 1} \right) - xy \hfill \cr \,\,\, - 2x + i\left( {y - 1} \right)\left( {y + 2} \right) \hfill \cr} \over {{x^2} + {{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0$

$ \Rightarrow $${\mathop{\rm Im}\nolimits} \left[ {\left( {{\matrix{ x\left( {y - 1} \right) - xy\, - 2x \hfill \cr \,\, + i\left[ {\left( {y - 1} \right)\left( {y + 2} \right) + {x^2}} \right] \hfill \cr} \over {{x^2} + {{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0$

$ \Rightarrow $${{\left( {y - 1} \right)\left( {y + 2} \right) + {x^2}} \over {{x^2} + {{\left( {y - 1} \right)}^2}}} + 1 = 0$

$ \Rightarrow $2x² + 2y² - y - 1 = 0

$ \Rightarrow $x² + y² - $\left( {{1 \over 2}} \right)$y - $\left( {{1 \over 2}} \right)$ = 0

$ \therefore $ Center of the circle is $\left( {0,{1 \over 4}} \right)$

$ \therefore $ Radius = $\sqrt {{0^2} + {{\left( {{1 \over 4}} \right)}^2} + {1 \over 2}} $

= $\sqrt {{1 \over {16}} + {1 \over 2}} $

= $\sqrt {{9 \over {16}}} $

= ${{3 \over 4}}$

Given,

2 $\,\left| \, \right.$z + 3i$\,\left| \, \right.$ = $\,\left| \, \right.$z $-$i$\,\left| \, \right.$

Let z = x + iy

$ \Rightarrow $$\,\,\,$ 2 $\,\left| \, \right.$ x + iy + 3i $\,\left| \, \right.$ = $\,\left| \, \right.$ x + iy $-$ i $\,\left| \, \right.$

$ \Rightarrow $$\,\,\,$ 2 $\,\left| \, \right.$ x + i (y + 3)$\,\left| \, \right.$ = $\,\left| \, \right.$ x + i (y $-$ 1)$\,\left| \, \right.$

$ \Rightarrow $$\,\,\,$ 2 $\sqrt {{x^2} + {{\left( {y + 3} \right)}^2}} $ = $\sqrt {{x^2} + {{\left( {y - 1} \right)}^2}} $

$ \Rightarrow $$\,\,\,$ 4 (x² + y² + 6y + 9) = x² + y² $-$ 2y + 1

$ \Rightarrow $$\,\,\,$ 3x² + 3y² + 26y + 35 = 0

$ \Rightarrow $$\,\,\,$ x² + y² + ${{26} \over 3}$ y + ${{35} \over 3}$ = 0

This is a equation of circle with center ($-$ ${{13} \over 3}$, 0)

$\therefore\,\,\,$ Radius = $\sqrt {0 + {{169} \over 9} - {{35} \over 3}} $

= $\sqrt {{{64} \over 9}} $

= ${8 \over 3}$

Given 2$\omega $ + 1 = z;

z = $\sqrt 3 i$

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

$ \Rightarrow $ As $\omega $ is complex cube root of unity.

${\omega ^3} = 1$

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

$\left| {\matrix{ 1 & 1 & 1 \cr 1 & { - {\omega ^2} - 1} & {{\omega ^2}} \cr 1 & {{\omega ^2}} & {{\omega ^7}} \cr } } \right| = 3k$

$ \Rightarrow $ $\left| {\matrix{ 1 & 1 & 1 \cr 1 & \omega & {{\omega ^2}} \cr 1 & {{\omega ^2}} & \omega \cr } } \right| = 3k$

Applying R₁ $ \to $ R₁ + R₂ + R₃

$ \Rightarrow $ $\left| {\matrix{ 3 & 0 & 0 \cr 1 & \omega & {{\omega ^2}} \cr 1 & {{\omega ^2}} & \omega \cr } } \right| = 3k$

$ \Rightarrow $ $3\left( {{\omega ^2} - {\omega ^4}} \right) = 3k$

$ \Rightarrow $ $\left( {{\omega ^2} - \omega } \right) = k$

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

$ \Rightarrow $ k = ${ - \sqrt 3 i}$ = -z

It is given that $z = x + iy$ satisfies ${\mathop{\rm Im}\nolimits} \left( {{{az + b} \over {z + 1}}} \right) = y$.

Therefore,

${\mathop{\rm Im}\nolimits} \left( {{{a(x + iy) + b} \over {(x + iy) + 1}}} \right) = y$

$ \Rightarrow Im\left( {{{ax + iay + b} \over {x + iy + 1}}} \right) = y$

Rationalizing the above equation and multiplying and dividing LHS by (x + 1 $-$ iy), we get

${\mathop{\rm Im}\nolimits} \left( {{{(ax + iay + b)} \over {(x + iy + 1)}} \times {{(x + 1 - iy)} \over {(x + 1 - iy)}}} \right) = y$

Using ${a^2} - {b^2} = (a + b)(a - b)$, we get

${\mathop{\rm Im}\nolimits} \left( {{{(ax + iay + b)} \over {{{(x + 1)}^2} - {{(iy)}^2}}}} \right) = y$

${\mathop{\rm Im}\nolimits} \left( {{{a{x^2} + ax - iayx + iaxy - {i^2}a{y^2} + bx + b - iby} \over {{{(x + 1)}^2} + {y^2}}}} \right) = y$ (as ${i^2} = - 1$)

${\mathop{\rm Im}\nolimits} \left( {{{(a{x^2} + ax + a{y^2} + bx + b) + i(axy - ayx + ay - by)} \over {{{(x + 1)}^2} + {y^2}}}} \right) = y$

Rearranging LHS, we get

${\mathop{\rm Im}\nolimits} \left( {{{[(a{x^2} + bx) + (ax + b) + a{y^2}] + i(ay - by)} \over {{{(x + 1)}^2} + {y^2}}}} \right) = y$

$ \Rightarrow {{ay - by} \over {{{(x + 1)}^2} + {y^2}}} = y$ (as imaginary value in bracket is coefficient of i)

$ \Rightarrow y(a - b) = y({(x + 1)^2} + {y^2}) \Rightarrow (a - b) = {(x + 1)^2} + {y^2}$

It is given that a $-$ b = 1 and y $\ne$ 0

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

$ \Rightarrow 1 - {y^2} = {(x + 1)^2} \Rightarrow (x + 1) = \pm \sqrt {1 - {y^2}} $ (as ${x^2} = b \Rightarrow x = \pm \sqrt 6 $)

$ \Rightarrow 1 = - 1 \pm \sqrt {1 - {y^2}} $

or, $x = - 1 + \sqrt {1 - {y^2}} $ and $x = - 1 - \sqrt {1 - {y^2}} $

Here,

z $-$ (3 + 3i) = $2\sqrt 2 $ (cos($-$135^o) + i sin ($-$ 135^o))

= $2\sqrt 2 $ ($-$ ${1 \over {\sqrt 2 }}$ $-$${i \over {\sqrt 2 }}$)

= $-$ 2 $-$ 2i

$ \Rightarrow $   z = 3 + 3 i $-$ 2 $-$ 2 i = 1 + i

Note :

Polar form of a complex number :

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

Here r = modulus of z and $\theta $ argument of z.

Rationalizing the given expression

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

For the given expression to be purely imaginary, real part of the above expression should be equal to zero.

$ \Rightarrow {{2 - 6{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }} = 0$

$ \Rightarrow {\sin ^2}\theta = {1 \over 3}$

$ \Rightarrow \sin \theta = \pm {1 \over {\sqrt 3 }}$

Here, $x + iy = {1 \over {a + ibt}} \times {{a - ibt} \over {a - ibt}}$

$\therefore$ $x + iy = {{a - ibt} \over {{a^2} + {b^2}{t^2}}}$

Let a $\ne$ 0, b $\ne$ 0

$\therefore$ $x = {a \over {{a^2} + {b^2}{t^2}}}$ and $y = {{ - bt} \over {{a^2} + {b^2}{t^2}}}$

$ \Rightarrow {y \over x} = {{ - bt} \over a} \Rightarrow t = {{ay} \over {bx}}$

On putting $x = {a \over {{a^2} + {b^2}{t^2}}}$, we get

$x\left( {{a^2} + {b^2}\,.\,{{{a^2}{y^2}} \over {{b^2}{x^2}}}} \right) = a$

$ \Rightarrow {a^2}({x^2} + {y^2}) = ax$

or, ${x^2} + {y^2} - {x \over a} = 0$ ...... (i)

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

$\therefore$ Option (a) is correct.

For a $\ne$ 0 and b = 0,

$x + iy = {1 \over a} \Rightarrow x = {1 \over a},\,y = 0$

$\Rightarrow$ z lies on X-axis.

$\therefore$ Option (c) is correct.

For a = 0 and b $\ne$ 0,

$x + iy = {1 \over {ibt}}$

$ \Rightarrow x = 0,\,y = - {1 \over {bt}}$

$\Rightarrow$ z lies on Y-axis.

$\therefore$ Option (d) is correct.

$\left| {{{{z_1} - 2{z_2}} \over {2 - {z_1}{{\overline z }_2}}}} \right| = 1 \Rightarrow {\left| {{z_1} - 2{z_2}} \right|^2} = {\left| {2 - {z_1}{{\overline z }_2}} \right|^2}$

$ \Rightarrow \left( {{z_1} - 2{z_2}} \right)\left( {\overline {{z_1} - 2{z_2}} } \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ = \left( {2 - {z_1}{{\overline z }_2}} \right)\left( {\overline {2 - {z_1}{{\overline z }_2}} } \right)$

$ \Rightarrow \left( {{z_1} - 2{z_1}} \right)\left( {{{\overline z }_1} - 2{{\overline z }_2}} \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ = \left( {2 - {z_1}\overline {{z_2}} } \right)\left( {2 - {{\overline z }_1}{z_2}} \right)$

$ \Rightarrow \left( {{z_1}{{\overline z }_1}} \right) - 2{z_1}{\overline z _2} - 2{\overline z _1}{z_2} + 4{z_2}{\overline z _2}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ = 4 - 2{\overline z _1}{z_2} - 2{z_1}{\overline z _2} + {z_1}{\overline z _1}{z_2}{\overline z _2}$

$ \Rightarrow {\left| {{z_1}} \right|^2} + 4{\left| {{z_2}} \right|^2}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ = 4 + {\left| {{z_1}} \right|^2}{\left| {{z_2}} \right|^2}$

$ \Rightarrow {\left| {{z_1}} \right|^2} + 4{\left| {{z_2}} \right|^2} - 4 - {\left| {{z_1}} \right|^2}{\left| {{z_2}} \right|^2} = 0$

$\left( {{{\left| {{z_1}} \right|}^2} - 4} \right)\left( {1 - {{\left| {{z_2}} \right|}^2}} \right) = 0$

As $\,\,\,\left| {{z_2}} \right| \ne 1\,\,\,$ $\therefore$ $\,\,\,{\left| {{z_1}} \right|^2} = 4 \Rightarrow \left| {{z_1}} \right| = 2$

$ \Rightarrow \,$ Point $\,{z_1}\,$ lies on circle of radius $2.$

We know minimum value of

$\,\,\,\left| {{Z_1} + {Z_2}} \right|\,\,\,$ is $\,\,\,\left| {\left| {{Z_1}} \right| - \left| {{Z_2}} \right|} \right|$

Thus minimum value of

$\,\,\,\left| {Z + {1 \over 2}} \right|\,\,\,$ is $\,\,\,\left| {\left| Z \right| - {1 \over 2}} \right|$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \le \left| {Z + {1 \over 2}} \right| \le \left| Z \right| + {1 \over 2}$

Since, $\,\,\,\left| Z \right| \ge 2$

$\therefore$ $\,\,\,2 - {1 \over 2} < \left| {Z + {1 \over 2}} \right| < 2 + {1 \over 2}$

$ \Rightarrow {3 \over 2} < \left| {Z + {1 \over 2}} \right| < {5 \over 2}$

Given, $\mathrm{Z}_k=\cos \frac{2 k \pi}{10}+i \sin \frac{2 k \pi}{10}, k=1,2,3, \ldots, 9$

$\begin{aligned} & \text { (P) } Z_k \cdot Z_j=1 \\ & \Rightarrow\left(\cos \frac{2 k \pi}{10}+i \sin \frac{2 k \pi}{10}\right) \\ & \left(\cos \frac{2 j \pi}{10}+i \sin \frac{2 j \pi}{10}\right)=1 \end{aligned}$

$\begin{aligned} & \Rightarrow \cos \left(\frac{2 \pi(k+j)}{10}\right)+i \sin \frac{2 \pi(k+j)}{10}=1 \\ & \Rightarrow \cos \frac{2 \pi(k+j)}{10}=1 \text { and } \sin \frac{2 \pi(k+j)}{10}=0 \\ & \Rightarrow \quad \frac{2 \pi(k+j)}{10}=2 \lambda \pi \text { where } \lambda \in \text { Integer } \\ & \Rightarrow \quad k+j=10 \lambda \\ & \Rightarrow \quad j=10 \lambda-k \\ \end{aligned}$

For every value of $k$ there is a possible value of $j$

Hence, $P$ match with 1.

(Q) Given $Z_1 \cdot Z=Z_k$

$\begin{aligned} & \left(\cos \frac{2 \pi}{10}+i \sin \frac{2 \pi}{10}\right) \cdot z=\cos \frac{2 \pi k}{10}+i \sin \frac{2 \pi k}{10} \\ & z=\frac{\cos \frac{2 \pi k}{10}+i \sin \frac{2 \pi k}{10}}{\cos \frac{2 \pi}{10}+i \sin \frac{2 \pi}{10}} \\ & z=\cos \frac{2 \pi(k-1)}{10}+i \sin \frac{2 \pi(k-1)}{10} \end{aligned}$

For every value of $k$, there is a unique value of $z$ is possible.

$\therefore$ The statement $z_1 \cdot z=z_k$ has no solution z in the set of complex numbers is false.

Hence, Q match with 2.

(R) Given, $z_k=\cos \frac{2 \pi k}{10}+i \sin \frac{2 \pi k}{10}, k=1,2,3, \ldots 9$

Here, $z_1, z_2, z_3, \ldots z_9$ are the roots of $z^{10}-1=0$

We know $z^{10}-1=(z-1)$

$\begin{aligned} & \left(z^9+z^8+z^7+z^6+z^5+z^4+z^3+z^2+z+1\right) \\ & =(z-1)\left(z-z_1\right)\left(z-z_2\right) \ldots\left(z-z_9\right) \\ \Rightarrow & 1+z+z^2+\ldots+z^9=\left(z-z_1\right)\left(z-z_2\right) \\ & \left(z-z_3\right) \ldots\left(z-z_9\right) \\ \Rightarrow & \left|1+z+z^2+\cdots+z^9\right| \\ & =\left|\left(z-z_1\right)\left(z-z_2\right)\left(z-z_3\right) \ldots\left(z-z_9\right)\right| \\ \Rightarrow & \left|z-z_1\right| \cdot\left|z-z_2\right|\left|z-z_3\right| \ldots\left|z-z_9\right| \\ & =\left|1+z+z^2+\ldots+z^9\right| \end{aligned}$

Put $z=1$

$\begin{aligned} & \Rightarrow\left|1-z_1\right| \cdot\left|1-z_2\right|\left|1-z_3\right| \ldots\left|1-z_9\right|=10 \\ & \Rightarrow \frac{\left|1-z_1\right|\left|1-z_2\right|\left|1-z_3\right| \ldots\left|1-z_9\right|}{10}=1 \end{aligned}$

Hence, R match with 3.

(S) We know the sum of all roots of $z^{10}-1=0$ is zero.

$\begin{aligned} & \Rightarrow 1+z_1+z_2+z_3+\ldots+z_9=0 \\ & \Rightarrow 1+\left(\cos \frac{2 \pi}{10}+i \sin \frac{2 \pi}{10}\right) \\ & +\left(\cos \left(\frac{4 \pi}{10}\right)+i \sin \left(\frac{4 \pi}{10}\right)\right) \\ & +\left(\cos \frac{6 \pi}{10}+i \sin \frac{6 \pi}{10}\right) \\ & +\ldots+\cos \frac{18 \pi}{10}+i \sin \frac{18 \pi}{10}=0 \\ \end{aligned}$

$\begin{aligned} & \begin{array}{l} \Rightarrow\left(1+\cos \frac{2 \pi}{10}+\cos \frac{4 \pi}{10}+\cdots+\cos \frac{18 \pi}{10}\right) \\ +i\left(\sin \frac{2 \pi}{10}+\sin \frac{4 \pi}{10}+\ldots+\sin \frac{18 \pi}{10}\right) \\ =0+i 0 \end{array} \\ & \Rightarrow 1+\cos \frac{2 \pi}{10}+\cos \frac{4 \pi}{10}+\cos \frac{6 \pi}{10}+\ldots+\cos \frac{18 \pi}{10}=0 \\ & \Rightarrow \sum_{k=1}^9 \cos \frac{2 k \pi}{10}=-1 \\ & \Rightarrow 1-\sum_{k=1}^9 \cos \frac{2 k \pi}{10}=1-(-1)=2 \end{aligned}$

Hint:

(i) Recall $(\cos \theta+i \sin \theta)(\cos \phi+i \sin \phi)=\cos (\theta+\phi)+i \sin (\theta+\phi)$

(ii) $\cos \theta=0$ when $\theta=2 n \pi$, where $n$ is an integer. and $\sin \theta=0$ when $\theta=n \pi$, where $n$ is an integer.

(iii) $1, z_1, z_2, z_3, \ldots z_9$ are the roots of $z^{10}-1=0$.

(iv) $z^{10}-1=(z-1)\left(1+z+z^2+\ldots+z^9\right)$

$=(z-1)\left(z-z_1\right)\left(z-z_2\right) \ldots\left(z-z_9\right)$

Given $\,\,\,\,\left| z \right| = 1,\,\,\arg \,z = \theta $

As we know, $\,\,\,\,\overrightarrow z = {1 \over z}$

$\therefore$ $\,\,\,\,\arg \left( {{{1 + z} \over {1 + \overrightarrow z }}} \right) = \arg \left( {{{1 + z} \over {1 + {1 \over z}}}} \right)$

$ = \arg \left( z \right) = \theta .$

We know $|1-3 i-z|=|z-(1-3 i)|$ implies distance of $z$ from $(1,-3)$.

Let $P=(1,-3)$

$\min _{z \in \mathrm{S}}|1-3 i-z|=\mathrm{PN}$

$\begin{aligned} & \Rightarrow \min _{z \in S}|1-3 i-z|=\frac{|\sqrt{3} \cdot 1-3|}{\sqrt{(\sqrt{3})^2+1^2}} \\ & \Rightarrow \min _{z \in S}|1-3 i-z|=\frac{3-\sqrt{3}}{2} \end{aligned}$

Hints:

(i) $\left|z_1-z_2\right|$ implies distance between $z_1$ and $z_2$

(ii) The minimum distance of a point $\left(x_1, y_1\right)$ from the line $a x+b y+c=0$ is $\frac{\left|a x_1+b y_1+c\right|}{\sqrt{a^2+b^2}}$.