Complex Numbers

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.