Complex Numbers

(P)

$ \begin{aligned} & x^2+x+1=0 \Rightarrow x=\omega, \omega^2 \text { where } \omega=e^{i \frac{2 \pi}{3}} \\ & \Rightarrow(\alpha+1)=1+\omega=-\omega^2 \Rightarrow(\alpha+1)^{2026}=\left(-\omega^2\right)^{2026}=\omega^2 \\ & \Rightarrow \frac{1}{(\alpha+1)^{2026}}=\frac{1}{\omega^2}=\omega \\ & \Rightarrow \beta+1=1+\omega^2=-\omega \Rightarrow(-\omega)^{2026}=(\beta+1)^{2026}=\omega \\ & \Rightarrow \frac{1}{(\beta+1)^{2026}}=\frac{1}{\omega}=\omega^2 \\ & \Rightarrow \frac{1}{(\alpha+1)^{2026}}, \frac{1}{(\beta+1)^{2026}} \text { are roots of } x^2+x+1=0 \\ & P \rightarrow 1 \end{aligned} $

(Q)

$ \begin{aligned} & \frac{1}{(\alpha+1)^{2027}}=\frac{1}{(\alpha+1)^{2026} \cdot(1+\alpha)}=\frac{1}{\omega^2 \cdot\left(-\omega^2\right)}=-\frac{1}{\omega}=-\omega^2 \\ & \frac{1}{(\beta+1)^{2027}}=\frac{1}{(\beta+1)^{2026} \cdot(\beta+1)}=\frac{1}{\omega(-\omega)}=-\omega \\ & \Rightarrow \frac{1}{(\alpha+1)^{2027}}, \frac{1}{(\beta+1)^{2027}} \text { are roots of } x^2-x+1=0 \\ & Q \rightarrow 2 \end{aligned} $

(R)

$ \begin{aligned} & \gamma=-\omega, \delta=-\omega^2, \text { where } \omega=e^{i \frac{2 \pi}{3}} \\ & \Rightarrow \gamma-1=-\omega-1=\omega^2 \Rightarrow(\gamma-1)^{2026}=\left(\omega^2\right)^{2026}=\omega^2 \\ & \Rightarrow \delta-1=-\omega^2-1=\omega \Rightarrow(\delta-1)^{2026}=\omega^{2026}=\omega \\ & \Rightarrow \frac{1}{(\gamma-1)^{2026}}=\omega, \frac{1}{(\delta-1)^{2026}}=\omega^2 \Rightarrow \omega+\omega^2=-1 \\ & \mathrm{R} \rightarrow 4 \end{aligned} $

(S)

$ \begin{aligned} & x^2+x-1=0 \Rightarrow x(x+1)=1 \Rightarrow(x+1)=\frac{1}{x} \\ & \Rightarrow \frac{1}{(p+1)^3}+\frac{1}{(r+1)^3}=p^3+r^3=(p+r)^3-3 p r(p+r)=(-1)^3-3(-1)(-1)=-4 \\ & \mathrm{~S} \rightarrow 5 \end{aligned} $

First, multiply matrices $S$ and $T$:

$ S T = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} $

So, $a = 0$, $b = -1$, $c = 1$, and $d = 1$.

(A)

We calculate:

$ \frac{b + ia}{d + ic} = \frac{-1}{1 + i} \times \frac{1 - i}{1 - i} = \frac{-1 + i}{2} $

The result is a complex number, but this does not satisfy the given condition, so this statement is incorrect.

(B)

Now check the transformation using $\omega$:

$ \frac{a\omega + b}{c\omega + d} = \frac{-1}{\omega + 1} = \frac{-1}{-\omega^2} = \omega $

This equality holds true, so this statement is correct.

(C)

Compute higher powers of $ST$:

$ (ST)^2 = \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ 1 & 0 \end{bmatrix} $

$ (ST)^3 = \begin{bmatrix} -1 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = -I $

Thus, $(ST)^6 = I$. Therefore, if $(ST)^2 = (ST)^M$, then $M = 6\lambda + 2$, where $\lambda$ is an integer. Since the statement does not correctly represent this, it is incorrect.

(D)

For $z = x + iy$, we have:

$ \frac{az + b}{cz + d} = \frac{-1}{z + 1} $

Substitute $z = x + iy$:

$ \frac{-1}{x + iy + 1} = \frac{-1}{(x + 1) + iy} \times \frac{(x + 1) - iy}{(x + 1) - iy} = \frac{-(x + 1) + iy}{(x + 1)^2 + y^2} $

The imaginary part is $\frac{y}{(x + 1)^2 + y^2}$, which is positive since $y > 0$. This means the image of every $z \in H$ stays in $H$. Hence, this statement is correct.

$\begin{aligned} \text{(A)} \quad & (-1+\sqrt{2})^{\mathrm{n}}=\mathrm{m}+\sqrt{2} \mathrm{n}, \mathrm{m}, \mathrm{n} \in \mathbb{Z} \\ & (1+\sqrt{2})^{\mathrm{n}}=\mathrm{m}_1+\sqrt{2} \mathrm{n}_1, \mathrm{~m}_1, \mathrm{n}_1 \in \mathbb{Z} \\ & \Rightarrow \mathbb{Z} \cup \mathrm{T}_1 \cup \mathrm{T}_2 \subseteq \mathrm{S} \end{aligned}$

but $b \sqrt{2} \in S$ for negative $b \in \mathbb{Z}$.

So $\quad \mathbb{Z} \cup T_1 \cup T_2 \subset \mathrm{S}$

$\begin{aligned} \text{(B)}\quad& (\sqrt{2}-1)^{\mathrm{n}}=\frac{1}{(\sqrt{2}+1)^{\mathrm{n}}}<\frac{1}{2024} \\ & \Rightarrow 2024<(\sqrt{2}+1)^{\mathrm{n}}, \exists \mathrm{n} \in \mathbb{N} \\ & \Rightarrow \mathrm{T}_1 \cap\left(0, \frac{1}{2024}\right) \neq \phi \end{aligned}$

$\begin{aligned} \text{(C)}\quad& (1+\sqrt{2})^{\mathrm{n}}>2024, \exists \mathrm{n} \in \mathbb{N} \\ & \Rightarrow \mathrm{T}_2 \cap(2024, \infty) \neq \phi \end{aligned}$

$\mathrm{(D)}$ $\quad \sin (\pi(a+b \sqrt{2})=0) \Rightarrow b=0, a \in \mathbb{Z}$.

$\Rightarrow$ Options (A), (C), (D) are Correct.

$ \because|z|^3+2 z^2+4 \bar{z}-8=0 $ .........(1)

Take conjugate both sides

$ \Rightarrow|\mathrm{z}|^3+2 \overline{\mathrm{z}}^2+4 \mathrm{z}-8=0 $ ........(2)

$ \begin{aligned} & \text { By }(1)-(2) \\\\ & \Rightarrow 2\left(\mathrm{z}^2-\bar{z}^2\right)+4(\bar{z}-\mathrm{z})=0 \\\\ & \Rightarrow \mathrm{z}+\overline{\mathrm{z}}=2 \\\\ & \Rightarrow|\mathrm{z}+\overline{\mathrm{z}}|=2 \end{aligned} $

Let $z=x+i y$

$ \therefore \mathrm{x}=1 \quad \therefore \mathrm{z}=1+\mathrm{iy} $

Put in (1)

$ \begin{aligned} & \Rightarrow\left(1+y^2\right)^{3 / 2}+2\left(1-y^2+2 i y\right)+4(1-i y)-8=0 \\\\ & \Rightarrow\left(1+y^2\right)^{3 / 2}=2\left(1+y^2\right) \\\\ & \Rightarrow \sqrt{1+y^2}=2=|z| \end{aligned} $

Also $\mathrm{y}= \pm \sqrt{3}$

$ \begin{aligned} & \therefore \mathrm{z}=1 \pm \mathrm{i} \sqrt{3} \\\\ & \Rightarrow \mathrm{z}-\overline{\mathrm{z}}= \pm 2 \mathrm{i} \sqrt{3} \\\\ & \Rightarrow|\mathrm{z}-\overline{\mathrm{z}}|=2 \sqrt{3} \\\\ & \Rightarrow|\mathrm{z}-\overline{\mathrm{z}}|^2=12 \\\\ & \text { Now } \mathrm{z}+1=2+\mathrm{i} \sqrt{3} \\\\ & |\mathrm{z}+1|^2=4+3=7 \\\\ & \therefore \mathrm{P} \rightarrow 2 ; \mathrm{Q} \rightarrow 1 ; \mathrm{R} \rightarrow 3 ; \mathrm{S} \rightarrow 5 \\\\ & \therefore \text { Option }[\mathrm{B}] \text { is correct. } \end{aligned} $

Let, complex number ${\left( {\overline z } \right)^2} + {1 \over {{z^2}}}$ is a new complex number $w$.

$\therefore$ $w = {\left( {\overline z } \right)^2} + {1 \over {{z^2}}}$

Now, let $z = r{e^{i\theta }}$ where $r = |z|$ and $\theta$ = argument

$\therefore$ $w = {\left( {r{e^{ - i\theta }}} \right)^2} + {1 \over {{r^2}}}({e^{ - 2i\theta }})$

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

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

$\therefore$ Real part of $w = {\mathop{\rm Re}\nolimits} (w) = \left( {{r^2} + {1 \over {{r^2}}}} \right)\cos 2\theta $

Imaginary part of $w = {\mathop{\rm Im}\nolimits} (w) = \left( {{r^2} + {1 \over {{r^2}}}} \right)( - \sin 2\theta )$

Given both ${\mathop{\rm Re}\nolimits} (w)$ and ${\mathop{\rm Im}\nolimits} (w)$ are integer.

$\therefore$ Let $Re(w) = {I_1}$

and ${\mathop{\rm Im}\nolimits} (w) = {I_2}$

$\therefore$ $I_1^2 + I_2^2 = {\left( {{r^2} + {1 \over {{r^2}}}} \right)^2} = \alpha $ (where $\alpha$ is an integer)

[Note : As I₁ and I₂ are integer so $I_1^2 + I_2^2$ is also an integer.]

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

$ \Rightarrow {r^4} + 2 + {1 \over {{r^4}}} = \alpha $

$ \Rightarrow {r^8} + 2{r^4} + 1 = {r^4}\alpha $

$ \Rightarrow {r^8} - (\alpha - 2){r^4} + 1 = 0$

$\therefore$ ${r^4} = {{(\alpha - 2)\, \pm \,\sqrt {{{(\alpha - 2)}^2} - 4\,.\,1\,.\,1} } \over {2\,.\,1}}$

$ \Rightarrow r = {\left( {{{(\alpha - 2)\, \pm \,\sqrt {{\alpha ^2} - 4\alpha } } \over 2}} \right)^{{1 \over 4}}}$

$ \Rightarrow |z| = r = {\left( {{{(\alpha - 2)\, \pm \,\sqrt {{\alpha ^2} - 4\alpha } } \over 2}} \right)^{{1 \over 4}}}$

In option only positive sign is given so ignoring negative sign we get,

$|z| = r = {\left( {{{(\alpha - 2) + \sqrt {{\alpha ^2} - 4\alpha } } \over 2}} \right)^{{1 \over 4}}}$ ....... (1)

From option (A),

$|z| = {\left( {{{43 + 3\sqrt {205} } \over 2}} \right)^{{1 \over 4}}}$

Comparing with (1), we get

$\alpha - 2 = 43$

$ \Rightarrow \alpha = 45$

Putting $\alpha = 45$ in (1), we get

$|z| = {\left( {{{43 + \sqrt {45(41)} } \over 2}} \right)^{{1 \over 4}}}$

$ = {\left( {{{43 + \sqrt {9 \times 5 \times 41} } \over 2}} \right)^{{1 \over 4}}}$

$ = {\left( {{{43 + 3\sqrt {205} } \over 2}} \right)^{{1 \over 4}}}$

$\therefore$ Option (A) is correct.

We can re-write

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

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

Comparing with option (B) we get,

$2(\alpha - 2) = 7$

$ \Rightarrow 2\alpha = 11$

$ \Rightarrow \alpha = {{11} \over 2}$ $\ne$ Integer

$\therefore$ Option (B) is incorrect.

Similarly option (C) and (D) also incorrect.

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

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

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

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

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.

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

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

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.

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

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

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 $\frac{\left|a x_1+b y_1+c\right|}{\sqrt{a^2+b^2}}$.

Given, $S_1=\{z \in C:|z|<4\}$ i.e. $S_1$ lies inside the circle of centre $(0,0)$ and radius 4.

Given, $S_3=\{z \in C: \operatorname{Re}(z)>0\}$ i.e. $S_3$ lies right side of the line $x=0$.

Also given $S_2=\left\{z \in C: \operatorname{Im}\left(\frac{z-1+\sqrt{3} i}{1-\sqrt{3} i}\right)>0\right\}$

Put $z=x+i y$ in $\operatorname{Im}\left(\frac{z-1+\sqrt{3} i}{1-\sqrt{3} i}\right)>0$

$\begin{aligned} & \Rightarrow \operatorname{Im}\left(\frac{x-1+i(y+\sqrt{3})}{1-\sqrt{3} i}\right)>0 \\ & \Rightarrow \operatorname{Im}\left(\frac{(x-1)+i(y+\sqrt{3})}{1-\sqrt{3} i} \times \frac{1+\sqrt{3} i}{1+\sqrt{3} i}\right)>0 \\ & \Rightarrow \operatorname{Im}\left[\begin{array}{l} \frac{(x-1)-\sqrt{3}(y+\sqrt{3})}{4}+ \\ i\left(\frac{\sqrt{3}(x-1)+(y+\sqrt{3})}{4}\right) \end{array}\right]>0 \\ \end{aligned}$

$\begin{aligned} & \Rightarrow \quad \frac{\sqrt{3} x+y}{4}>0 \\ & \Rightarrow \sqrt{3} x+y>0 \end{aligned}$

$\text { So, } S_2 \text { lies above the line } \sqrt{3} x+y=0$

$\text { Given, } S=S_1 \cap S_2 \cap S_3$

Here, $\angle \mathrm{AOC}=\frac{\pi}{2}+\frac{\pi}{3}=\frac{5 \pi}{6}$

Area of $S=\frac{1}{2}\left(\frac{5 \pi}{6}\right) \cdot 4^2=\frac{20 \pi}{3}$ Sq. Units

Hints:

(i) $|z| < a$ implies $z$ lies inside the circle of centre $(0,0)$ and radius $a$

(ii) $\operatorname{Re}(z)>0$ implies $z$ lies on the right side of the line $x=0$ i.e. $y$-axis.

(iii) Put $z=x+i y$ in the expression $\operatorname{Im}\left(\frac{z-1+\sqrt{3} i}{1-\sqrt{3} i}\right)>0$

(iv) If a circular arc $A B$ form $\theta$ angle at the centre of circle of radius R, then the area of this circular section is $\frac{1}{2} \theta \cdot r^2$

If $z=x+i y$, then $\left(x-x_0\right)^2+\left(y-y_0\right)^2=r^2$ and $\left(x-x_0\right)^2+\left(y-y_0\right)^2=4 r^2$ can be written as $\left|Z-Z_0\right|^2=r^2$ and $\left|Z-Z_0\right|^2=4 r^2$ respectively where, $\mathrm{Z}_0=x_0+i y_0$ (given)

$\Rightarrow\left|Z-Z_0\right|^2=r^2$ and $\left|Z-Z_0\right|^2=4 r^2$

$\Rightarrow\left(Z-Z_0\right)\left(\bar{Z}-\bar{Z}_0\right)=r^2$ and $\left(Z-Z_0\right)\left(\bar{Z}-\bar{Z}_0\right)=4 r^2\left(|Z|^2=Z \bar{Z}\right)$

$\begin{aligned} & \text { Given, } \alpha \text { and } \frac{1}{\bar{\alpha}} \text { line on circles }\left(x-x_0\right)^2+\left(y-y_0\right)^2 \\ & =r^2 \text { and }\left(x-x_0\right)^2+\left(y-y_0\right)^2=4 r^2 \text { respectively } \\ & \begin{array}{l} \therefore \quad\left(\alpha-Z_0\right)\left(\bar{\alpha}-\bar{Z}_0\right)=r^2 \text { and } \\ \quad\left(\frac{1}{\bar{\alpha}}-Z_0\right)\left(\frac{1}{\alpha}-\bar{Z}_0\right)=4 r^2 \end{array} \end{aligned}$

$\begin{aligned} \Rightarrow & |\alpha|^2-\alpha \bar{Z}_0-\bar{\alpha} Z_0+\left|Z_0\right|^2=r^2 \text { and } \\ & \frac{1}{|\alpha|^2}-\frac{1}{\bar{\alpha}} \bar{Z}_0-\frac{1}{\alpha} Z_0+\left|\bar{Z}_0\right|^2=4 r^2 \\ \Rightarrow & \alpha Z_0+\bar{\alpha} Z_0=|\alpha|^2+\left|Z_0\right|^2-r^2 \text { and } \\ & \frac{\alpha \bar{Z}_0}{|\alpha|^2}+\frac{\bar{\alpha} Z_0}{|\alpha|^2}=\frac{1}{|\alpha|^2}+\left|Z_0\right|^2-4 r^2 \\ \Rightarrow & \alpha \bar{Z}_0+\bar{\alpha} Z_0=|\alpha|^2+\left|Z_0\right|^2-r^2 =|\alpha|^2\left(\frac{1}{\left|\alpha^2\right|}+\left|Z_0^2\right|-4 r^2\right) \end{aligned}$

$\Rightarrow|\alpha|^2+\left|Z_0\right|^2-r^2=1+|\alpha|^2\left|Z_0\right|^2-4|\alpha|^2 r^2 \quad\text{... (i)}$

Given, $2\left|Z_0\right|^2=r^2+2$

Put $\left|Z_0\right|^2=\frac{r^2}{2}+1$ in the equation (i)

$\begin{aligned} & \Rightarrow|\alpha|^2+\frac{r^2}{2}+1-r^2=1+|\alpha|^2\left(\frac{r^2}{2}+1\right)-4|\alpha|^2 r^2 \\ & \Rightarrow|\alpha|^2-\frac{r^2}{2}=\frac{1}{2}|\alpha|^2 r^2+|\alpha|^2-4|\alpha|^2 r^2 \end{aligned}$

$\begin{array}{ll} \Rightarrow & -\frac{r^2}{2}=\frac{1}{2}|\alpha|^2 r^2-4|\alpha|^2 r^2 \\ \Rightarrow & -\frac{1}{2}=\frac{1}{2}(\alpha)^2-4|\alpha|^2 \\ \Rightarrow & \frac{-1}{2}=-\frac{7}{2}|\alpha|^2 \\ \Rightarrow & |\alpha|^2=\frac{1}{7} \\ \Rightarrow & |\alpha|=\frac{1}{\sqrt{7}} \end{array}$

Hints :

(i) Apply the property $Z \cdot \bar{Z}=|\bar{Z}|^2$

(ii) If $Z=(x+i y)$ and $Z_0=\left(x_0+i y_0\right)$ then $\left(x-x_0\right)^2+\left(y-y_0\right)^2=r^2$ and $\left(x-x_0\right)^2+ \left(y-y_0\right)^2=4 r^2$ can be written as $\left|Z-Z_0\right|^2=r^2$ and $\left|Z-Z_0\right|^2=4 r^2$ respectively.

Given, $w=\frac{\sqrt{3}}{2}+\frac{i}{2}=e^{\frac{i \pi}{6}}$

And $\mathrm{P}=\left\{w^n: n=1,2,3, \ldots\right\}=e^{\text {in } \frac{\pi}{6}}, n=1,2,3, \ldots$

Also given

$\begin{aligned} & H_1=\left\{Z \in C: \operatorname{Re}(Z)>\frac{1}{2}\right\}, \\ & H_2=\left\{Z \in C: \operatorname{Re}(Z)<\frac{-1}{2}\right\} \end{aligned}$

And $\mathrm{Z}_1 \in \mathrm{P} \cap \mathrm{H}_1, \mathrm{Z}_2 \in \mathrm{P} \cap \mathrm{H}_2$

Here, Possible values of P are $e^{\frac{i \pi}{6}}, e^{\frac{i \pi}{3}}, e^{\frac{i \pi}{2}}, e^{\frac{i 2 \pi}{3}}, e^{\frac{i 5 \pi}{6}}, e^{i \pi}, e^{\frac{i 7 \pi}{6}}, e^{\frac{i 4 \pi}{6}}, e^{\frac{i 3 \pi}{2}}, e^{\frac{i 5 \pi}{3}}, e^{\frac{i 11 \pi}{6}}, e^{i 2 \pi}$ And $\mathrm{H}_1$ and $\mathrm{H}_2$ are the set of all points at lies right side of $x=\frac{1}{2}$ and left side of $x=-\frac{1}{2}$ respectively

Here, $\mathrm{Z}_1=e^{\frac{i \pi}{6}}$ or $e^{\frac{i 11 \pi}{6}}$ or $e^{i 2 \pi}$ and $\mathrm{Z}_2=e^{\frac{i 5 \pi}{6}} \text { or } e^{i \pi} \text { or } e^{\frac{i 7 \pi}{6}}$

$\text { Now, } \angle \mathrm{Z}_1 \mathrm{OZ}_2=\frac{2 \pi}{3} \text { or } \frac{5 \pi}{6} \text { or } \pi$

Hints:

(i) $\mathrm{H}_1$ are the set of all points that lies right side of line $x=\frac{1}{2}$ and $\mathrm{H}_2$ are the set of all points that lies left of the line $x=-\frac{1}{2}$.

(ii) $\mathrm{P}=\mathrm{W}^n$ has 12 different roots lies on a unit circle and angle between two successive roots is $\frac{\pi}{6}$.

Given, $\quad a=z^2+z+1$

$\Rightarrow z^2+z+1-a=0$

Using Sridharacharya Rule

$\begin{aligned} & \qquad z=\frac{-1 \pm \sqrt{1^2-4(1-a)}}{2} \\ & \Rightarrow \quad z=\frac{-1 \pm \sqrt{1-4+4 a}}{2} \\ & \Rightarrow \quad z=\frac{-1 \pm \sqrt{4 a-3}}{2} \end{aligned}$

Now, we can observe that if $\mathrm{a}=\frac{3}{4}$, then $z= \frac{-1}{2}$ which will become purely real.

Hence, if $a \neq \frac{3}{4}$, otherwise $z$ will be purely real.

(i) Make quadratic equation in $z$ and apply Sridharacharya rule.

(ii) Analyse the value of '$a$' for which '$z$' becomes purely real.

(A) z is equidistant from the points $i|z|$ and $ - i|z|$, whose perpendicular bisector is ${\mathop{\rm Im}\nolimits} (z) = 0$.

(B) Sum of distance of z from (4, 0) and ($-$4, 0) is a constant 10, hence locus of z is ellipse with semi-major axis 5 and focus at ($\pm$ 4, 0), ae = 4.

$\therefore$ $e = {4 \over 5}$

(C) $|z| \le |w| + \left| {{1 \over w}} \right| = {5 \over 2} < 3$

(D) $|z| \le |w| + \left| {{1 \over w}} \right| = 2$

$\therefore$ ${\mathop{\rm Re}\nolimits} (z) \le |z| \le 2$

Given, $z = {{(1 - t){z_1} + t\,{z_2}} \over {(1 - t) + t}}$

Clearly, z divides z₁ and z₂ in the ratio of t : (1 $-$ t), 0 < t < 1

$\Rightarrow$ AP + BP = AB

i.e., $|z - {z_1}| + |z - {z_2}| = |{z_1} - {z_2}|$

$\Rightarrow$ Option (a) is true.

and $\arg (z - {z_1}) = \arg ({z_2} - z) = \arg ({z_2} - {z_1})$

$\Rightarrow$ (b) is false and (d) is true.

Also, $\arg (z - {z_1}) = \arg ({z_2} - {z_1})$

$ \Rightarrow \arg \left( {{{z - {z_1}} \over {{z_2} - {z_1}}}} \right) = 0$

$\therefore$ ${{z - {z_1}} \over {{z_2} - {z_1}}}$ is purely real.

$ \Rightarrow {{z - {z_1}} \over {{z_2} - {z_1}}} = {{\overline z - {{\overline z }_1}} \over {{{\overline z }_2} - {{\overline z }_1}}}$

or, $\left| {\matrix{ {z - {z_1}} & {\overline z - {{\overline z }_1}} \cr {{z_2} - {z_1}} & {{{\overline z }_2} - {{\overline z }_1}} \cr } } \right| = 0$

$\therefore$ Option (c) is correct.

Hence, (a, c, d) is the correct option.

$ \begin{aligned} & \text { Given, } z=(1-t) z_1+t z_2 \\\\ & \Rightarrow \frac{z-z_1}{z_2-z_1}=t \\\\ & \Rightarrow \arg \left(\frac{z-z_1}{z_2-z_1}\right) = 0 .......(i) \end{aligned} $

$\begin{aligned} \Rightarrow \arg \left(z-z_1\right) & =\arg \left(z_2-z_1\right) \\\\ \frac{z-z_1}{z_2-z_1} & =\frac{\bar{z}-\bar{z}_1}{\bar{z}_2-\bar{z}_1} \\\\ \left|\begin{array}{cc}z-z_1 & \bar{z}-\bar{z}_1 \\ z_2-z_1 & \bar{z}_2-\bar{z}_1\end{array}\right| & =0\end{aligned}$

$\begin{gathered}\mathrm{AP}+\mathrm{PB}=\mathrm{AB} \\\\ \Rightarrow\left|z-z_1\right|+\left|z-z_2\right|=\left|z_1-z_2\right|\end{gathered}$

We have

$\overline z {z^3} + z{\overline z ^3} = 350$

Substituting $z = x + iy$, we get

$({x^2} + {y^2})({x^2} - {y^2}) = 175$

$({x^2} + {y^2})({x^2} - {y^2}) = 5 \times 5 \times 7$

${x^2} + {y^2} = 25$

${x^2} - {y^2} = 7$

whose solutions are $x = \pm 4$ and $y = \pm 3;x,y \in I$.

Therefore, that is, area is found as $8 \times 6 = 48$ sq. unit.

We have

$X = \sin \theta + \sin 3\theta \, + \,...\, + \,\sin 29\theta $

$2(\sin \theta )X = 1 - \cos 2\theta + \cos 2\theta - \cos 4\theta \, + \,...\, + \,\cos 28\theta - \cos 30\theta $

$X = {{1 - \cos 30\theta } \over {2\sin \theta }} = {1 \over {4\sin 2^\circ }}$

given $-$ $\alpha$ particle

In the direction of, or, Now rotation about origin through angle of means multiply by

${z_0} = 1 + 2i = (1,2) = ({x_0},{y_0})$

${z_1} = ({x_0} + 5,{y_0} + 3)$

$ = (6,5) = 6 + 5i$

$\Rightarrow 2$ in the direction of $i = j$

${x_1} = 2\cos 45^\circ $

${y_1} = 2\sin 45^\circ $

${z_2}(7 + 6i)$

Now, rotation about origin through an angle of 2$\pi$ means multiply z$_2$ by $ \to {e^{i\pi /2}}$

$\cos 90^\circ + i\sin 90^\circ = i$

${z_3} = i(7 + 6i)$

$ = - 6 + 7i$

$|z + 1 - i{|^2} + |z - 5 - i{|^2}$

The points ($-1,1$) and (5, 1) are the extremities of the diameter of the given circle of radius 3

Hence, PA$^2$ + PB$^2$ = AB$^2$ = 36

Since, $|w - (2 + i)| < 3$

$|w| - |2 + i| < 3$

$ - 3 + \sqrt 5 < |w| < 3 + \sqrt 5 $

$ - 3 - \sqrt 5 < |w| < 3 - \sqrt 5 $

Also, $|z - (2 + i)| = 3$

$ - 3 + \sqrt 5 \le |z| \le 3 + \sqrt 5 $

$\therefore$ $ - 3 < |z| - |w| + 3 < 9$

In the Cartesian coordinates sets A, B and C defined the regions given by

$A:y\ge1,B:(x-2)+(y-1)^2=9$

B and C being a circle and a straight line intersect in two points, out of which only one satisfies $y > -1$.

Thus, the no. of elements in the set A $\cap$ B $\cap$ C is 1.

Given, $|z|=1$ and $z \neq \pm 1$

To find : values of $\frac{z}{1-z^{2}}$ lies on ?

Any complex number $z$ can be written as :

Since, $|z|=1$

$z=r(\cos \theta+i \sin \theta), \text { where } r=|z| $

Then, $z=(\cos \theta+i \sin \theta)$

Consider, $\frac{z}{1-z^{2}}$

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

$\begin{aligned}& =\frac{(\cos \theta+i \sin \theta)(\sin \theta+i \cos \theta)}{2 \sin \theta} \\ & {\left[\frac{1}{(\sin \theta-i \cos \theta)}=\sin \theta+i \cos \theta\right]} \\& =\frac{\sin \theta \cos \theta+i \cos ^{2} \theta+i \sin ^{2} \theta+i^{2} \sin \theta \cos \theta}{2 \sin \theta} \\ & =\frac{i\left(\cos ^{2} \theta+\sin ^{2} \theta\right)}{2 \sin \theta}=\frac{i}{2 \sin \theta}\end{aligned}$

Thus, $\frac{z}{1-z^{2}}$ lies on the imaginary axis that is the $\mathrm{Y}$-axis in argand plane.

$\overrightarrow {OP} = \overrightarrow {OA} + \overrightarrow {OP} $

$\overrightarrow {OP} = \overrightarrow {OA} + \overrightarrow {OB} $

$\overrightarrow {OP} = 3{e^{{{i\pi } \over 4}}} + 4{e^{i\left( {{\pi \over 2} + {\pi \over 4}} \right)}}$

$ = 3{e^{{{i\pi } \over 4}}} + 4{e^{{{i\pi } \over 2}}}.\,{e^{{{i\pi } \over 4}}}$

$ = 3{e^{{{i\pi } \over 4}}} + 4i{e^{{{i\pi } \over 4}}}$

$ = {e^{{{i\pi } \over 4}}}(3 + 4i)$

$w=\alpha+i \beta$

since, $\left(\frac{w-\bar{w} z}{1-z}\right)$ is purely real

then, $\quad \frac{w-\bar{w} z}{1-z}=\frac{\overline{w-\bar{w} z}}{\overline{1-z}}=\frac{\bar{w}-w \bar{z}}{1-\bar{z}}$

$\Rightarrow(w-\bar{w} z)(1-\bar{z})=(\bar{w}-w \bar{z})(1-z)$

$\Rightarrow(w-\bar{w} z)-w \bar{z}+\bar{w}|z|^{2}$

$=\bar{w}-w \bar{z}-\bar{w} z+w|z|^{2}$

$\Rightarrow \quad w+\bar{w}|z|^{2}=\bar{w}+w|z|^{2}$

$\Rightarrow \quad|z|^{2}(w-\bar{w})=w-\bar{w}$

$\Rightarrow \quad|z|^{2}=\frac{w-\bar{w}}{w-\bar{w}}$

$\Rightarrow \quad|z|^{2}=1$

Since, $\quad w-\bar{w}=\alpha+i \beta-\alpha+i \beta$

$=2 i \beta$ and $\beta \neq 1$

$\therefore \quad|z|=1$ but $|z| \neq 1$

Shortcut Method:

Since $z=\bar{z}$ then

$\begin{aligned} x+i y & =x-i y \\ \Rightarrow \quad x+i y-x+i y & =0 \\ \Rightarrow \quad 2 i y & =0 \\ \Rightarrow \text { Imaginary part } & =0 \end{aligned}$

Here, equation of $C_2:(x-1)^2+(y-1)^2=(\sqrt{2})^2$ and $C_1:(x-1)^2+(y-1)^2=(1)^2$

$\begin{aligned} & \therefore P(1+\cos \theta, 1+\sin \theta) \text { and } Q(1+\sqrt{2} \sin \theta, 1+\sqrt{2} \sin \theta) \\\\ & \therefore P A^2+P B^2+P C^2+P D^2\end{aligned}$

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

Similarly,

$ \begin{array}{ll} Q A^2+Q B^2+Q C^2+Q D^2 & =12 \\ \therefore \frac{\Sigma Q A^2}{\Sigma P A^2}=\frac{12}{16}=0.75 \end{array} $

Hence (a) is the correct answer.

$|z-1|=\sqrt{2}$

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

$\therefore$ Centre $=(1,0)$

i.e., $1+i \cdot 0\left(\right.$ say $\left.z_{1}\right)$

$\begin{aligned} z_{2} & =2+\sqrt{3} i \\ \text { radius } & =\sqrt{2} \end{aligned}$

To obtain B, OA is rotated with $90^{\circ}$

$\begin{aligned} \frac{z_{3}-z_{1}}{z_{2}-z_{1}} & =\left|\frac{z_{3}-z_{1}}{z_{2}-z_{1}}\right| e^{i \pi / 2} \\ \Rightarrow \quad \frac{z_{3}-z_{1}}{z_{2}-z_{1}} & =\left|\frac{z_{3}-z_{1}}{z_{2}-z_{1}}\right|\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right) \\ \Rightarrow \quad \frac{z_{3}-z_{1}}{z_{2}-z_{1}} & =\frac{\mathrm{OB}}{\mathrm{OA}}(0+i) \\ \Rightarrow \quad \frac{z_{3}-z_{1}}{z_{2}-z_{1}} & =1 . i\quad \because (\mathrm{OA=OB (radius))} \\ \Rightarrow \quad\left(z_{3}-z_{1}\right) & =i\left(z_{2}-z_{1}\right) \\ \Rightarrow \quad z_{3}-1 & =i(2+\sqrt{3} i-1) \\ \Rightarrow \quad z_{3} & =1+i+\sqrt{3} i^{2} \\ & =1-\sqrt{3}+i \end{aligned}$

Similarly, we can calculate Co-ordinate D

$\begin{aligned} \frac{z_{5}-z_{1}}{z_{2}-z_{1}} & =\left|\frac{z_{5}-z_{1}}{z_{2}-z_{1}}\right| e^{i\left(\frac{-\pi}{2}\right)} \\ & =\frac{\mathrm{OD}}{\mathrm{OA}} \cos \left(\frac{-\pi}{2}\right)+i\left(-\sin \frac{\pi}{2}\right) \\ & =0-i \\ \Rightarrow \quad z_{5}-z_{1} & =\left(z_{2}-z_{1}\right)(-i) \\ \Rightarrow \quad z_{5} & =z_{1}+(1+i \sqrt{3})(-i) \\ & =1+(-i)+\sqrt{3} \\ & =(1+\sqrt{3})-i \end{aligned}$

Similarly, we can calculate co-ordinate C

$\begin{aligned} \frac{z_{4}-z_{1}}{z_{2}-z_{1}} & =\left|\frac{z_{4}-z_{1}}{z_{2}-z_{1}}\right| e^{i \pi} \\ \Rightarrow \quad \frac{z_{4}-z_{1}}{z_{2}-z_{1}} & =\frac{\mathrm{OC}}{\mathrm{OA}}(-1+0)\end{aligned}$

$\begin{aligned} \Rightarrow \quad z_{4}-z_{1} & =\left(z_{2}-z_{1}\right)(-1) \\ & =(-1)(1+\sqrt{3} i)=-1-\sqrt{3} i \\ \Rightarrow \quad z_{4} & =z_{1}-1-\sqrt{3} i \\ & =1-1-\sqrt{3} i=-\sqrt{3} i \end{aligned}$