Complex Numbers

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.