Complex Numbers

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.

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

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

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

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

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} $

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

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