Complex Numbers

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 $\frac{\pi}{6}$.

Let $z = x + iy$

$\therefore$ $\,\,\,\,{z^2} = {x^2} - {y^2} + 2ixy$

Now ${{{z^2}} \over {z - 1}}$ is real

$ \Rightarrow {\mathop{\rm Im}\nolimits} \left( {{{{z^2}} \over {z - 1}}} \right) = 0$

$ \Rightarrow {\mathop{\rm Im}\nolimits} \left( {{{{x^2} - {y^2} + 2ixy} \over {\left( {x - 1} \right) + iy}}} \right) = 0$

$ \Rightarrow {\mathop{\rm Im}\nolimits} \left[ {\left( {{x^2} - {y^2} + 2ixy} \right)\left. {\left( {x - 1} \right) - iy} \right)} \right] = 0$

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

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

$ \Rightarrow y = 0;\,{x^2} + {y^2} - 2x = 0$

$\therefore$ $\,\,\,\,$ $z$ lies either on real axis or on a circle through origin.

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.

${\left( {1 + \omega } \right)^7} = A + B\omega ;\,\,\,\,{\left( { - {\omega ^2}} \right)^7} = A + B\omega $

$ - {\omega ^2} = A + B\omega ;\,\,\,\,\,\,\,\,\,\,1 + \omega = A + B\omega $

$ \Rightarrow A = 1,B = 1.$

As real part of roots is $1$

Let roots are $1 + pi,1 + q$

$\therefore$ sum of roots $ = 1 + pi + 1 + qi = - \alpha $

which is real $ \Rightarrow q = - p\,\,$

or root are $1+pi$ and $1-pi$

product of roots $ = 1 + {p^2} = \beta \in \left( {1,\infty } \right)$

$p \ne 0$ as roots are distinct.

Let $z=x+iy$

$\left| {z - 1} \right| = \left| {z + 1} \right|{\left( {x - 1} \right)^2} + {y^2}$

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

$ \Rightarrow {\mathop{\rm Re}\nolimits} \,z = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow x = 0$

$\left| {z - 1} \right| = \left| {z - i} \right|{\left( {x - 1} \right)^2} + {y^2}$

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

$ \Rightarrow x = y$

$\left| {z + 1} \right| = \left| {z - i} \right|{\left( {x + 1} \right)^2} + {y^2}$

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

Only $(0,0)$ will satisfy all conditions.

$ \Rightarrow $ Number of complex number $z=1$

(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} $

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

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

Given that $\left| {z - {4 \over z}} \right| = 2$

Now $\left| z \right| = \left| {z - {4 \over z} + {4 \over { - z}}} \right| \le \left| {z - {4 \over z}} \right| + {4 \over {\left| z \right|}}$

$ \Rightarrow \left| z \right| \le 2 + {4 \over {\left| z \right|}}$

$ \Rightarrow {\left| z \right|^2} - 2\left| z \right| - 4 \le 0$

$ \Rightarrow \left( {\left| z \right| - {{2 + \sqrt {20} } \over 2}} \right)\left( {\left| z \right| - {{2 - \sqrt {20} } \over 2}} \right) \le 0$

$\left( {\left| z \right| - \left( {1 + \sqrt 5 } \right)} \right)\left( {\left| z \right| - \left( {1 - \sqrt 5 } \right)} \right) \le 0$

$ \Rightarrow \left( { - \sqrt 5 + 1} \right) \le \left| z \right| \le \left( {\sqrt 5 + 1} \right)$

$ \Rightarrow {\left| z \right|_{\max }} = \sqrt 5 + 1$

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 $

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

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

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.

$z$ lies on or inside the circle with center $(-4,0)$ and radius $3$ units.



From the Argand diagram maximum value of $\left| {z + 1} \right|$ is $6$

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)$

${z^2} + z + 1 = 0 \Rightarrow z = \omega \,\,\,$ or $\,\,\,{\omega ^2}$

So, $z + {1 \over z} = \omega + {\omega ^2} = - 1$

${z^2} + {1 \over {{z^2}}} = {\omega ^2} + \omega = - 1,$

${z^3} + {1 \over {{z^3}}} = {\omega ^3} + {\omega ^3} = 2$

${z^4} + {1 \over {{z^4}}} = - 1,$

${z^5} + {1 \over {{z^5}}} = - 1$

and $\,\,\,\,{z^6} + {1 \over {{z^6}}} = 2$

$\therefore$ The given sum $ = 1 + 1 + 4 + 1 + 1 + 4 = 12$

$\sum\limits_{k = 1}^{10} {\left( {\sin {{2k\pi } \over {11}} + i\cos {{2k\pi } \over {11}}} \right)} $

$ = i\sum\limits_{k = 1}^{10} {\left( {\cos {{2k\pi } \over {11}} - i\,\sin {{2k\pi } \over {11}}} \right)} $

$ = i\sum\limits_{k = 1}^{10} {{e^{ - {{2k\pi } \over {11}}}}} i = i\left\{ {\sum\limits_{k = 0}^{10} {{e^{ - {{2k\pi } \over {11}}}}} - 1} \right\}$

$ = i\left[ {1 + {e^{ - {{2\pi } \over {11}}i}} + e - {{4\pi } \over {11}}i + .....11\,\,terms} \right] - i$

$ = i\left[ {{{1 - {{\left( {{e^{ - {{2\pi } \over {11}}}}} \right)}^{11}}} \over {1 - {e^{ - {{2\pi } \over {11}}i}}}}} \right] - i$

$ = i\left[ {{{1 - {e^{ - 2\pi i}}} \over {1 - {e^{ - {{2\pi } \over {11}}i}}}}} \right] - i$

$ = i \times 0 - i$

[as $\,\,\,\,\,\,$ ${e^{ - 2\pi i}} = 1$ ]

$ = - i$

$w=\alpha+i \beta$

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.

${\left( {x - 1} \right)^3} + 8 = 0$

$ \Rightarrow \left( {x - 1} \right) = \left( { - 2} \right){\left( 1 \right)^{1/3}}$

$ \Rightarrow x - 1 = - 2\,\,\,$ or $\,\,\, - 2\omega \,\,\,\,$ or $\,\,\,\, - 2{\omega ^2}$

or $\,\,\,x = - 1\,\,\,$ or $\,\,\,1 - 2\omega \,\,\,$ or $\,\,\,1 - 2{\omega ^2}.$

Given $\,\omega = {z \over {z - {1 \over 3}i}}\,$ and $\left| \omega \right| = 1$

$\therefore$ ${{\left| z \right|} \over {\left| {z - {1 \over {\sqrt 3 }}i} \right|}} = \left| \omega \right|$

$ \Rightarrow $ ${{\left| z \right|} \over {\left| {z - {1 \over {\sqrt 3 }}i} \right|}} = 1$

$ \Rightarrow $ $\left| z \right| = \left| {z - {1 \over {\sqrt 3 }}i} \right|$ ..........equation (1)

$\left| z \right|$ represent distance of $z$ from point (0, 0) and

$\left| {z - {1 \over {\sqrt 3 }}i} \right|$ represent distance of $z$ from point $\left( {0,{1 \over {\sqrt 3 }}} \right)$.

According to the equation (1) the distance of $z$ from point (0, 0) and $\left( {0,{1 \over {\sqrt 3 }}} \right)$ is equal. Only if z is on a straight line then it will be equal distance from the both the points.

Given that, $\,\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$

$\,\left| {{z_1} + {z_2}} \right|$ is the vector sum of ${z_1}$ and ${z_2}$. So $\,\left| {{z_1} + {z_2}} \right|$ should be $<$ $\left| {{z_1}} \right| + \left| {{z_2}} \right|$ but here they are equal so ${z_1}$ and ${z_2}$ are collinear.

S if ${z_1}$ makes an angle $\theta $ with x axis then ${z_2}$ will also make $\theta $ angle.

$\therefore$ arg ${z_1}$ - arg ${z_2}$ = $\theta $ - $\theta $ = 0

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

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

$\therefore$ 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

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

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

Given $\overline z + i\overline w = 0$

$ \Rightarrow \overline z = - i\overline w $

$ \Rightarrow \overline{\overline z} = - \overline {i\overline w } $

$ \Rightarrow \overline{\overline z} = - \overline i \overline{\overline w} $

$ \Rightarrow z = - \overline i w$

$ \Rightarrow z = - \left( { - i} \right)w$

$ \Rightarrow z = iw$

Now given that Arg(zw) = $\pi $

$ \Rightarrow $ Arg(z$ \times $${z \over i}$) = $\pi $

$ \Rightarrow $ Arg(z²) - Arg(i) = $\pi $

$ \Rightarrow $ 2Arg(z) - ${\pi \over 2}$ = $\pi $

[ $i$ complex number represent (0, 1) point on imaginary axis and Arg($i$) means the angle made by the point (0, 1) with real axis which is ${\pi \over 2}$]

$ \Rightarrow $ 2Arg(z) = ${{3\pi } \over 2}$

$ \Rightarrow $ Arg(z) = ${{3\pi } \over 4}$

Given $\,\left| {{z^2} - 1} \right| = {\left| z \right|^2} + 1$,

By squaring both sides we get,

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

$ \Rightarrow $ $\left( {{z^2} - 1} \right)$$\overline {\left( {{z^2} - 1} \right)} $ = ${\left( {{{\left| z \right|}^2} + 1} \right)^2}$ [ as ${{{\left| z \right|}^2}}$ = $z\overline z $ ]

$ \Rightarrow $ $\left( {{z^2} - 1} \right)$$\left( {{{\left( {\overline z } \right)}^2} - 1} \right)$ = ${\left( {{{\left| z \right|}^2} + 1} \right)^2}$

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

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

$ \Rightarrow $ ${{z^2}}$ $+$${{{\left( {\overline z } \right)}^2}}$ $+$ 2${z\overline z }$ = 0

$ \Rightarrow $ ${\left( {z + \overline z } \right)^2}$ = 0

$ \Rightarrow $ ${z + \overline z }$ = 0

$ \Rightarrow $ $z$ = $-$ ${\overline z }$

If $z$ = x + iy

then ${\overline z }$ = x - iy

$\therefore$ x + iy = - (x - iy)

$ \Rightarrow $ x + iy = - x + iy

$ \Rightarrow $ x = 0

$\therefore$ z is purely imaginary.

So, it is lie on the imaginary axis.

Given ${z^{{1 \over 3}}} = p + iq$

$ \Rightarrow $ z = (p + iq)³

= p³ + (iq)³ +3p(iq)(p + iq)

= p³ - iq³ +3ip²q - 3pq²

= p(p² - 3q²) - iq(q² - 3p²)

Given that $z = x - iy$

$\therefore$ $x - iy$ = p(p² - 3q²) - iq(q² - 3p²)

By comparing both sides we get,

${x \over p} = {p^2} - 3{q^2}$ and ${y \over q} = {q^2} - 3{p^2}$

$\therefore$ ${{\left( {{x \over p} + {y \over q}} \right)} \over {\left( {{p^2} + {q^2}} \right)}}$

= ${{{p^2} - 3{q^2} + {q^2} - 3{p^2}} \over {{p^2} + {q^2}}}$

= ${{ - 2{q^2} - 2{p^2}} \over {{p^2} + {q^2}}}$

= ${{ - 2\left( {{q^2} + {p^2}} \right)} \over {{p^2} + {q^2}}}$

= $-2$

Given that,
$\left| {z\omega } \right| = 1$
$ \Rightarrow $ $\left| z \right|\left| \omega \right|$ = 1
$ \Rightarrow $ $\left| z \right|$ = ${1 \over {\left| \omega \right|}}$

and $Arg(z) - Arg(\omega ) = {\pi \over 2}$
$ \Rightarrow $ $Arg\left( {{z \over \omega }} \right)$ $= {\pi \over 2}$

When argument of a complex number is ${\pi \over 2}$, it means it is making an angle of ${\pi \over 2}$ with the real axis in the counterclockwise, so it is along the imaginary axis and positive side of imaginary axis.

So, ${{z \over \omega }}$ is a purely imaginary number that means there is no real part in this complex number.

So we can assume,
${{z \over \omega }}$ = $ki$

$ \Rightarrow $ ${\left| {{z \over \omega }} \right|}$ = $\left| {ki} \right|$

$ \Rightarrow $ ${\left| {{z \over \omega }} \right|}$ = $\left| k \right|\left| i \right|$

$ \Rightarrow $ ${\left| {{z \over \omega }} \right|}$ = $k$       [ as $\left| i \right|$ = 1 ]

$ \Rightarrow $ $\left| z \right|$$ \times $${1 \over {\left| \omega \right|}}$ = $k$

$ \Rightarrow $ $\left| z \right|$$ \times $ $\left| z \right|$ = $k$       [ as ${1 \over {\left| \omega \right|}}$ = $\left| z \right|$ ]

$ \Rightarrow $ ${\left| z \right|^2}$ = $k$

$ \Rightarrow $ $\left| z \right|$ = $\sqrt k $

$\therefore$ $\left| \omega \right|$ = ${1 \over {\sqrt k }}$

As ${{z \over \omega }}$ is imaginary so we can write,

${{z \over \omega }}$ = $ - {{\overline z } \over {\overline \omega }}$
[ When $z$ is imaginary then $z$ = $-\overline z $ ]

$ \Rightarrow $ $\overline z \omega $ = $ - z\overline \omega $

$ \Rightarrow $ $\overline z \omega $ = $-{{z \over \omega }}$.$\overline \omega $.$\omega $

$ \Rightarrow $ $\overline z \omega $ = $-{{z \over \omega }}$.${\left| \omega \right|^2}$

$ \Rightarrow $ $\overline z \omega $ = $-ki$.${\left( {{1 \over {\sqrt k }}} \right)^2}$

$ \Rightarrow $ $\overline z \omega $ = $-ki$.${1 \over k}$

$ \Rightarrow $ $\overline z \omega $ = $-i$

Method 2 :

Given that,
$\left| {z\omega } \right| = 1$
$ \Rightarrow $ $\left| z \right|\left| \omega \right|$ = 1
$ \Rightarrow $ $\left| z \right|$ = ${1 \over {\left| \omega \right|}}$

and $Arg(z) - Arg(\omega ) = {\pi \over 2}$
$ \Rightarrow $ $Arg\left( {{z \over \omega }} \right)$ $= {\pi \over 2}$

When argument of a complex number is ${\pi \over 2}$, it means it is making an angle of ${\pi \over 2}$ with the real axis in the counterclockwise, so it is along the imaginary axis and positive side of imaginary axis.

So, ${{z \over \omega }}$ is a purely imaginary number that means there is no real part in this complex number.

So we can assume,
${{z \over \omega }}$ = $ki$

$ \Rightarrow $ ${\left| {{z \over \omega }} \right|}$ = $\left| {ki} \right|$

$ \Rightarrow $ ${\left| {{z \over \omega }} \right|}$ = $\left| k \right|\left| i \right|$

$ \Rightarrow $ ${\left| {{z \over \omega }} \right|}$ = $k$       [ as $\left| i \right|$ = 1 ]

$ \Rightarrow $ $\left| z \right|$$ \times $${1 \over {\left| \omega \right|}}$ = $k$

$ \Rightarrow $ $\left| z \right|$$ \times $ $\left| z \right|$ = $k$       [ as ${1 \over {\left| \omega \right|}}$ = $\left| z \right|$ ]

$ \Rightarrow $ ${\left| z \right|^2}$ = $k$

$ \Rightarrow $ $\left| z \right|$ = $\sqrt k $

$\therefore$ $\left| \omega \right|$ = ${1 \over {\sqrt k }}$

(1) Magnitude of $\overline z \omega $

= $\left| {\overline z } \right|\left| \omega \right|$

= $\left| z \right|\left| \omega \right|$ [ as $\left| z \right|$ = $\left| {\overline z } \right|$ ]

= $\sqrt k $.${{1 \over {\sqrt k }}}$

= 1

$\therefore$ The distance from the origin of ${\overline z \omega }$ is 1.

(2) Argument of ${\overline z \omega }$ = $Arg\left( {\overline z \omega } \right)$

= $Arg\left( {\overline z } \right) + Arg\left( \omega \right)$

= $-Arg\left( z \right) + Arg\left( \omega \right)$

= $ - \left( {Arg\left( z \right) - Arg\left( \omega \right)} \right)$

= $ - {\pi \over 2}$

$\therefore$ ${\overline z \omega }$ is at (0, -1) on the negative side of imaginary axis and making an angle of ${\pi \over 2}$ clockwise.

$\therefore$ ${\overline z \omega }$ = 0 + (-1)$ \times $$i$ = $-i$

Given quadratic equation,
${Z^2} + aZ + b = 0$
and two roots are ${Z_1}$ and ${Z_2}$.

$\therefore$ ${Z_1}$ + ${Z_2}$ = $-a$ and ${Z_1}$${Z_2}$ = $b$

Question says,
There are three complex numbers:
1. Origin (0)
2. ${Z_1}$
3. ${Z_2}$
and they form an equilateral triangle. So They are the vertices of the triangle.

[ Important Point : If ${Z_1}$, ${Z_2}$ and ${Z_3}$ are the vertices of an equilateral triangle then -
$Z_1^2$ + $Z_2^2$ + $Z_3^2$ = ${Z_1}{Z_2}$ + ${Z_2}{Z_3}$ + ${Z_3}{Z_1}$ ]

In this question,
${Z_1}$ = 0, ${Z_2}$ = ${Z_1}$ and ${Z_3}$ = ${Z_2}$

By putting those values in the equation we get,

${0^2}$ + $Z_1^2$ + $Z_2^2$ = $0$ + ${Z_1}{Z_2}$ + 0

$ \Rightarrow $ $Z_1^2$ + $Z_2^2$ = ${Z_1}{Z_2}$

$ \Rightarrow $ $Z_1^2$ + $Z_2^2$ = $b$ [ as ${Z_1}$${Z_2}$ = $b$ ]

$ \Rightarrow $ ${\left( {{Z_1} + {Z_2}} \right)^2}$ - $2{Z_1}{Z_2}$ = $b$

$ \Rightarrow $ ${\left( {{Z_1} + {Z_2}} \right)^2}$ - $2b$ = $b$

$ \Rightarrow $ ${\left( {{Z_1} + {Z_2}} \right)^2}$ = $3b$

$ \Rightarrow $ ${\left( { - a} \right)^2}$ = $3b$

$ \Rightarrow $ ${a^2}$ = $3b$

So Option (D) is correct.

[ Note : This question is asked to check if you know the following formula -

"If ${Z_1}$, ${Z_2}$ and ${Z_3}$ are the vertices of an equilateral triangle then -
$Z_1^2$ + $Z_2^2$ + $Z_3^2$ = ${Z_1}{Z_2}$ + ${Z_2}{Z_3}$ + ${Z_3}{Z_1}$" ]

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

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

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

$ \Rightarrow $ ${\left[ {{{1 + 2i + {i^2}} \over {1 + 1}}} \right]^x} = 1$

$ \Rightarrow $ ${\left[ {{{1 + 2i - 1} \over 2}} \right]^x} = 1$

$ \Rightarrow {\left( i \right)^x} = 1$

We know $i = \sqrt { - 1} $

$\therefore$ ${i^2} = - 1$

$ \Rightarrow $ ${i^3} = - 1 \times i = - i$

$ \Rightarrow $ ${i^4} = - i \times i = - {i^2} = - \left( { - 1} \right) = 1$

So when power of $i$ is 4 or multiple of 4 then it's value is = 1

$\therefore$ ${\left( i \right)^x} = 1$ $ = {\left( i \right)^{4n}}$ where n is a positive integer.

Let $\left| z \right| = \left| \omega \right| = r$

$\therefore$ $z = r{e^{i\theta }},\omega = r{e^{i\phi }}$

where $\,\,\theta + \phi = \pi .$

$\therefore$ $z = r{e^{i\left( {\pi - \phi } \right)}} = r{e^{i\pi }}.$ ${e^{ - i\phi }} = - r{e^{ - i\phi }} = - \overline \omega .$

[as $\,\,\,\,\overline \omega = r{e^{ - i\phi }}$ ]

Given $\left| {z - 4} \right| < \left| {z - 2} \right|$

Let $\,\,\,z = x + iy$

$ \Rightarrow \left| {\left. {\left( {x - 4} \right) + iy} \right)} \right| < \left| {\left( {x - 2} \right) + iy} \right|$

$ \Rightarrow {\left( {x - 4} \right)^2} + {y^2} < {\left( {x - 2} \right)^2} + {y^2}$

$ \Rightarrow {x^2} - 8x + 16 < {x^2} - 4x + 4$

$ \Rightarrow 12 < 4x$

$ \Rightarrow x > 3$

$ \Rightarrow {\mathop{\rm Re}\nolimits} \left( z \right) > 3$

Let the circle be $\left| {z - {z_3}} \right| = r.$

Then according to given conditions

$\left| {{z_3} - {z_1}} \right| = r + a$ (Shown in the image)

and $\left| {{z_3} - {z_2}} \right| = r + b.$ (Shown in the image)

Eliminating $r,$ we get

$\left| {{z_3} - {z_1}} \right| - \left| {{z_3} - {z_2}} \right| = a - b.$

$\therefore$ Locus of center ${z_3}$ is

$\left| {z - {z_1}} \right| - \left| {z - {z_2}} \right| = a - b$ = constant.

Definition of hyperbola says, when difference of distance between two points is constant from a particular point then that particular point will lie on a hyperbola.

Here distance of z₁ from z₃ is = $r + a$ and distance of z₂ from z₃ is = $r + b$

Now their difference = ($r + a$) - ($r + b$) = $a - b$ = a constant

$\therefore$ Locus of z₃ is a hyperbola.