Complex Numbers

$z = {{{{\left( {1 + i} \right)}^2}} \over {a - i}} \times {{a + i} \over {a + i}}$

$ \Rightarrow z = {{\left( {1 - 1 + 2i} \right)\left( {a + i} \right)} \over {{a^2} + 1}} = {{2ai - 2} \over {{a^2} + 1}}$

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

$ \Rightarrow \sqrt {{{4(1 + {a^2})} \over {{{({a^2} + 1)}^2}}}} = {2 \over {\sqrt {{a^2} + 1} }}$.... (i)

Now given $\left| z \right| = \sqrt {{2 \over 5}} $

so $\sqrt {{2 \over 5}} = {2 \over {\sqrt {1 + {a^2}} }}$ from equation (i)

By squaring both sides

$ \Rightarrow {2 \over 5} = {4 \over {1 + {a^2}}}$

$ \Rightarrow 1 + {a^2} = 10$

a² = 9

$ \Rightarrow $ a = $ \pm $ 3

$ \because $ (a > 0) then a = 3

Now $\overline z = {{{{\left( {1 - i} \right)}^2}} \over {3 + i}}$

= ${{\left( {1 + {i^2} - 2i} \right)} \over {3 + i}}$

= ${{\left( {1 + {i^2} - 2i} \right)\left( {3 - i} \right)} \over {\left( {3 + i} \right)\left( {3 - i} \right)}}$

= ${{\left( { - 2i} \right)\left( {3 - i} \right)} \over {\left( {9 - {i^2}} \right)}}$

= ${{\left( { - 2i} \right)\left( {3 - i} \right)} \over {10}}$

= ${{ - 6i + 2{i^2}} \over {10}}$

= ${{ - 6i - 2} \over {10}}$

= $-{1 \over 5} - {3 \over 5}i$

$\omega = {{5 + 3z} \over {5(1 - z)}}$z

$ \Rightarrow $ $5\omega \left( {1 - z} \right) = 5 + 3z$

$ \Rightarrow $ $5\omega - 5\omega z = 5 + 3z$

$ \Rightarrow $ $5\omega - 5 = 5\omega z + 3z$

$ \Rightarrow $ $z\left( {5\omega + 3} \right) = 5\left( {\omega - 1} \right)$

$ \Rightarrow $ $z = {{5\left( {\omega - 1} \right)} \over {3 + 5\omega }}$

$ \Rightarrow $ $z = {{5\left( {\omega - 1} \right)} \over {5\left( {\omega + {3 \over 5}} \right)}}$

$ \Rightarrow $ $z = {{\left( {\omega - 1} \right)} \over {\left( {\omega + {3 \over 5}} \right)}}$

Given |z| < 1

$ \Rightarrow $ $\left| {{{\omega - 1} \over {\omega + {3 \over 5}}}} \right|$ < 1

$ \Rightarrow $ $\left| {\omega - 1} \right| < \left| {\omega + {3 \over 5}} \right|$

$ \Rightarrow $ $\left| {\omega - 1} \right| < \left| {\omega - \left( { - {3 \over 5}} \right)} \right|$

It says distance of $\omega $ from point 1 is less than distance from point ${\left( { - {3 \over 5}} \right)}$.

x coordinate of perpendicular bisector of points ${\left( { - {3 \over 5},0} \right)}$ and (1, 0),

x = ${{1 - {3 \over 5}} \over 2}$ = ${1 \over 5}$

As $\omega $ is closer to point (1, 0) so $\omega $ should present in the right side of perpendicular bisector.

$ \therefore $ ${\mathop{\rm Re}\nolimits} (\omega ) > {1 \over 5}$

$ \Rightarrow $ 5Re( $\omega$) > 1

Other Method :

$\omega = {{5 + 3z} \over {5(1 - z)}}$z

$ \Rightarrow $ $5\omega \left( {1 - z} \right) = 5 + 3z$

$ \Rightarrow $ $5\omega - 5\omega z = 5 + 3z$

$ \Rightarrow $ $5\omega - 5 = 5\omega z + 3z$

$ \Rightarrow $ $z\left( {5\omega + 3} \right) = 5\left( {\omega - 1} \right)$

$ \Rightarrow $ $z = {{5\left( {\omega - 1} \right)} \over {3 + 5\omega }}$

Given |z| < 1

$ \Rightarrow $ $\left| {{{5\left( {\omega - 1} \right)} \over {5\omega + 3}}} \right|$ < 1

$ \Rightarrow $ $5\left| {\omega - 1} \right| < \left| {5\omega + 3} \right|$

$ \Rightarrow $ $25\left( {{\omega ^2} - 2\omega + 1} \right) < 25{\omega ^2} + 30\omega + 9$

$ \Rightarrow $ $25\left( {\omega \overline \omega - \omega - \overline \omega + 1} \right)$ <

$25\omega \overline \omega + 15\omega + 15\overline \omega + 9$

(Using ${\left| z \right|^2} = z\overline z $)

$ \Rightarrow $ $16 < 40\omega + 40\overline \omega $

$ \Rightarrow $ $\omega + \overline \omega > {2 \over 5}$

$ \Rightarrow $ $2{\mathop{\rm Re}\nolimits} \left( \omega \right) > {2 \over 5}$

$ \Rightarrow $ 5Re( $\omega$) > 1

Let h + ik = ${{\alpha + i} \over {\alpha - i}}$

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

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

$ \therefore $ h = ${{{\alpha ^2} - 1} \over {{\alpha ^2} + 1}}$ and k = ${{2\alpha } \over {{\alpha ^2} + 1}}$

By squaring and adding we get

h² + k² = 1

$ \therefore $ This is circle whose radius is 1.

$z = {{\sqrt 3 } \over 2} + {i \over 2}$

$ \Rightarrow $ z = $\cos {\pi \over 6}$ + i $\sin {\pi \over 6}$

$ \Rightarrow $ z = ${e^{i{\pi \over 6}}}$

x² – 2x + 2 = 0

$ \therefore $ x = ${{2 \pm \sqrt { - 4} } \over 2} = 1 \pm i$

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

or ${\alpha \over \beta } = {{1 - i} \over {1 + i}} = {{{{\left( {1 - i} \right)}^2}} \over {1 + {i^2}}} = - i$

$ \therefore $ ${\left( { i} \right)^n}$ = 1

and ${\left( { - i} \right)^n}$ = 1

So n must be a multiple of 4

$ \therefore $ Minimum value of n = 4

$\left| {{z_1}} \right| = 9,\,\,\left| {{z_2} - \left( {3 + 4i} \right)} \right| = 4$

${C_1},(0,0)$ radius r₁ = 9

C₂ (3, 4), radius r₂ = 4

C₁C₂ = $\left| {{r_1} - {r_2}} \right| = 5$

$ \therefore $  Circle touches internally

$ \therefore $  ${\left| {{z_1} - {z_2}} \right|_{\min }} = 0$

${{z - \alpha } \over {z + \alpha }} + {{\overline z - \alpha } \over {\overline z + \alpha }} = 0$

$z\overline z + z\alpha - \alpha \overline z - {\alpha ^2} + z\overline z - z\alpha + \overline z \alpha - {\alpha ^2} = 0$

${\left| z \right|^2} = {\alpha ^2},$  $a = \pm 2$

$\left| z \right| + z = 3 + i$

$z = 3 - \left| z \right| + i$

Let  $3 - \left| z \right| = a \Rightarrow \left| z \right| = \left( {3 - a} \right)$

$ \Rightarrow z = a + i \Rightarrow \left| z \right| = \sqrt {{a^2} + 1} $

$ \Rightarrow 9 + {a^2} - 6a = {a^2} + 1 \Rightarrow a = {8 \over 6} = {4 \over 3}$

$ \Rightarrow \left| z \right| = 3 - {4 \over 3} = {5 \over 3}$

${\left( { - 2 - {i \over 3}} \right)^3} = - {{{{\left( {6 + i} \right)}^3}} \over {27}}$

$ = {{ - 198 - 107i} \over {27}} = {{x + iy} \over {27}}$

Hence, $y - x = 198 - 107 = 91$

$z = {\left( {{{\sqrt 3 + i} \over 2}} \right)^5} + {\left( {{{\sqrt 3 - i} \over 2}} \right)^5}$

$z = {\left( {{e^{i\pi /6}}} \right)^5} + {\left( {{e^{ - i\pi /6}}} \right)^5}$

$ = {e^{i5\pi /6}} + {e^{ - i5\pi /6}}$

$ = \cos {{5\pi } \over 6} + i{{\sin 5\pi } \over 6} + \cos \left( {{{ - 5\pi } \over 6}} \right) + i\sin \left( {{{ - 5\pi } \over 6}} \right)$

$ = 2\cos {{5\pi } \over 6} < 0$

${\rm I}(z) = 0$ and ${\mathop{\rm Re}\nolimits} (z) < 0$

Given, $3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|$

$ \Rightarrow $ ${{\left| {{z_1}} \right|} \over {\left| {{z_2}} \right|}} = {4 \over 3}$

$ \Rightarrow $ ${{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}} = {4 \over 3} \times {3 \over 2} = 2$

As we know, for any compled number

${{3{z_1}} \over {2{z_2}}} = {{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}}$(cos$\theta $ + i sin$\theta $)

= 2(cos$\theta $ + i sin$\theta $)

$ \therefore $ ${{2{z_2}} \over {3{z_1}}}$ = ${1 \over {2\left( {\cos \theta + i\sin \theta } \right)}}$

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

= ${\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}$
Now, given

$z = {{3{z_1}} \over {2{z_2}}} + {{2{z_2}} \over {3{z_1}}}$

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

= ${{5 \over 2}\cos \theta + {3 \over 2}i\sin \theta }$

So, |z| = $\sqrt {{{25} \over 4}{{\cos }^2}\theta + {9 \over 4}{{\sin }^2}\theta } $

= ${1 \over 2}\sqrt {9 + 16{{\cos }^2}\theta } $

z is neither purely real nor purely imaginary and |z| depends on $\theta $.

1 + x + x² = 0

x = ${{ - 1 \pm \sqrt {1 - 4} } \over 2} = {{ - 1 \pm i\sqrt 3 } \over 2}$

z₀ = w, w²

Now  

z = 3 + 6iz$_0^{81}$ $-$ 3iz$_0^{93}$

z = 3 + 6iw⁸¹ $-$ 3iw⁹³      (w⁹³ = w⁸¹ = 1)

$ \Rightarrow $  z = 3 + 3i

then arg(z) = tan^$-$1$\left( {{3 \over 3}} \right)$ = tan^$-$1 (1) = ${\pi \over 4}$

Given complex number,

${{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}$

$ = {{\left( {3 + 2i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$

$ = {{3 + 6i\sin \theta + 2i\sin \theta - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}$

$ = {{\left( {3 - 4{{\sin }^2}\theta } \right) + i\left( {8\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$

As complex number is purely imaginary, So real part of this complex number is zero.

$ \therefore $   ${{3 - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}$ = 0

$ \Rightarrow $   $3 - 4{\sin ^2}\theta = 0$

$ \Rightarrow $   $\sin \theta = \pm {{\sqrt 3 } \over 2}$

as   $\theta $ $ \in $ $\left( { - {\pi \over 2},\pi } \right)$

$ \therefore $   $\theta $ $=$ $-$ ${\pi \over 3},{\pi \over 3},{{2\pi } \over 3}$

$ \therefore $   Sum of those values of A is

$ = - {\pi \over 3} + {\pi \over 3} + {{2\pi } \over 3}$

$ = {{2\pi } \over 3}$

Given equation,

x² + 2x + 2 = 0

$ \therefore $  x = ${{ - 2 \pm \sqrt {4 - 4.1.2} } \over {2.1}}$

x = $-$ 1 $ \pm $ i

$ \therefore $  $\alpha $ = $-$ 1 + i

and $\beta $ = $-$ 1 $-$ i

Note :

x + iy = r (cos$\theta $ + isin$\theta $)

$ \therefore $  (x + iy)ⁿ = rⁿ (cosn$\theta $ + isinn$\theta $)

$ \therefore $  $-$ 1 + i = $\sqrt 2 $ [cos${{3\pi } \over 4}$ + isin${{3\pi } \over 4}$ ]

$ \Rightarrow $  ($-$ 1 + i)¹⁵ = ${\left( {\sqrt 2 } \right)^{15}}$ [cos$\left( {{{15.3\pi } \over 4}} \right) + i\sin \left( {{{15.3\pi } \over 4}} \right)$]

And $-$1 $-$ i = $\sqrt 2 $ $\left[ {\cos \left( { - {{3\pi } \over 4}} \right) + i\sin \left( { - {{3\pi } \over 4}} \right)} \right]$

= $\sqrt 2 \left[ {\cos {{3\pi } \over 4} - \sin {{3\pi } \over 4}} \right]$

$ \therefore $  ($-$1 $-$ i)¹⁵ = ${\left( {\sqrt 2 } \right)^{15}}\left[ {\cos \left( {{{15.3\pi } \over 4}} \right) - i\sin \left( {{{15.3\pi } \over 4}} \right)} \right]$

Now

$\alpha $¹⁵ + $\beta $¹⁵

= ($-$1 + i)¹⁵ + ($-$ 1 $-$ i)¹⁵

=  ${\left( {\sqrt 2 } \right)^{15}}$ $\left[ {2\cos \left( {{{15.3\pi } \over 4}} \right)} \right]$

= ${\left( {\sqrt 2 } \right)^{15}}\left[ {2\cos \left( {11\pi + {\pi \over 4}} \right)} \right]$

= ${\left( {\sqrt 2 } \right)^{15}}\left[ {2\left( { - \cos {\pi \over 4}} \right)} \right]$

= ${\left( {\sqrt 2 } \right)^{15}} \times 2 \times - {1 \over {\sqrt 2 }}$

= $ - {\left( {\sqrt 2 } \right)^{14}}.2$

= $-$ 2⁷ $ \times $ 2

= $-$ 2⁸

= $-$ 256

${\left( {{{1 + i\sqrt 3 } \over {1 - i\sqrt 3 }}} \right)^n} = 1$

$ \Rightarrow \,\,\,\,\,{\left( {{{{{1 + i\sqrt 3 } \over 2}} \over {{{1 - i\sqrt 3 } \over 2}}}} \right)^n} = 1$

We know, $\omega $ = $-$ ${{1 - i\sqrt 3 } \over 2}$

and $\omega $² = $-$ ${{\left( {1 + i\sqrt 3 } \right)} \over 2}$

$ \Rightarrow $ $\,\,\,\,$ ${\left( {{{ - {\omega ^2}} \over { - \omega }}} \right)^n} = 1$

$ \Rightarrow $ ($\omega $)ⁿ = 1 = $\omega $³

$ \Rightarrow $ $\,\,\,\,$ n = 3

Given equation,

x² $-$ x + 1 = 0

Roots of this equation

x = ${{1 \pm \sqrt 3 i} \over 2}$

$\therefore\,\,\,$ $ \propto \, = \,{{1 + \sqrt 3 \,i} \over 2}$

and $\beta = \,{{1 - \sqrt 3 \,i} \over 2}$

We know;

$\omega = {{ - 1 + \sqrt 3 \,i} \over 2} = - \left( {{{1 - \sqrt 3 \,i} \over 2}} \right) = - \beta $

and ${\omega ^2} = {{ - 1 - \sqrt 3 \,i} \over 2} = - \left( {{{1 + \sqrt 3 \,i} \over 2}} \right) = - \propto $

$\therefore\,\,\,$ $ \propto \, = - {\omega ^2}$ and $\beta \, = \, - \omega $

$\therefore\,\,\,$ ${ \propto ^{101}} + {\beta ^{107}}$

$ = {\left( { - {\omega ^2}} \right)^{101}} + {\left( { - \omega } \right)^{107}}$

$ = {\left( { - 1} \right)^{101}}.{\left( {{\omega ^2}} \right)^{101}} + {\left( { - 1} \right)^{107}}.{\left( \omega \right)^{107}}$

$ = - 1.{\left( {{\omega ^2}} \right)^{101}} - {\omega ^{107}}$

$ = - \left( {{\omega ^{202}} + {\omega ^{107}}} \right)$

$ = - \left( {{\omega ^{3.67}}.\omega + {\omega ^{3.35}}.{\omega ^2}} \right)$

$ = - \left( {\omega + {\omega ^2}} \right)\,\,\,$ [ as $\,\,\,$ ${\omega ^{3n}} = 1$]

$ = - \left( { - 1} \right)$ $\,\,\,\,\,\,$ [as $\,\,\,$ $1 + \omega + {\omega ^2} = 0$ ]

$ = 1$

$\left| {z - \left( {3 - 2i} \right)} \right| \le 4$ represents a circle whose center is (3, $-$2) and radius = 4.

$\left| z \right|$ = $\left| z -0\right|$  represents the distance of point 'z' from origin (0, 0)



Suppose RS is the normal of the circle passing through origin 'O' and G is its center (3, $-$2).

Here, OR is the least distance

and OS is in the greatest distance

OR = RG $-$ OG and OS = OG + GS . . . . .(1)

As, RG = GS = 4

OG = $\sqrt {{3^2} + \left( { - 2{)^2}} \right)} $ = $\sqrt {9 + 4} $ = $\sqrt {13} $

From (1), OR = 4 $-$ $\sqrt {13} $ and OS = 4 + $\sqrt {13} $

So, required difference = $\left( {4 + \sqrt {13} } \right)$ $-$ $\left( {4 - \sqrt {13} } \right)$

= $\sqrt {13} + \sqrt {13} $ = $2\sqrt {13} $

As w = ${{1 + \left( {1 - 8\alpha } \right)z} \over {1 - z}}$, w is purely imaginary

$ \therefore w$ + $\bar w$ = 0

$ \Rightarrow $ ${{1 + \left( {1 - 8\alpha } \right)z} \over {1 - z}}$ + ${{1 + \left( {1 - 8\alpha } \right)\bar z} \over {1 - \bar z}}$ = 0

$ \Rightarrow $ [1 + (1 - 8$\alpha $)][1 - $\bar z$] + [1 + ( 1 - 8$\alpha $)][1 - z] = 0

$ \Rightarrow $ 2 - (z + $\bar z$) + (1 - 8$\alpha $)(z + $\bar z$) - 2(1 - 8$\alpha $) = 0

$ \Rightarrow $ 2 - (z + $\bar z$) + (z + $\bar z$) - 8$\alpha $(z + $\bar z$) - 2 + 16$\alpha $ = 0

$ \Rightarrow $ 16$\alpha $ = 8$\alpha $(z + $\bar z)$

z + $\bar z$ = 2 or $\alpha $ = 0

but z + $\bar z$ = 2 is not possible as Re(Z) $ \ne $ 1

$ \therefore $ $\alpha $ = 0

$ \therefore $ $\alpha $ $ \in $ {0}

Let z = x + iy

Then,

Im $\left( {{{iz - 2} \over {z - i}}} \right)$ + 1 = 0

$ \Rightarrow $ ${\mathop{\rm Im}\nolimits} \left[ {\left( {{{i\left( {x + iy} \right) - 2} \over {x + iy - i}}} \right)} \right] + 1 = 0$

$ \Rightarrow $${\mathop{\rm Im}\nolimits} \left[ {\left( {{{ix - y - 2} \over {x + i\left( {y - 1} \right)}}} \right)} \right] + 1 = 0$

$ \Rightarrow $${\mathop{\rm Im}\nolimits} \left[ {\left( {{{ix - y - 2} \over {x + i\left( {y - 1} \right)}}} \right)\left( {{{x - i\left( {y - 1} \right)} \over {x - i\left( {y - 1} \right)}}} \right)} \right] + 1 = 0$

$ \Rightarrow $${\mathop{\rm Im}\nolimits} \left[ {\left( {{\matrix{ i{x^2} - {i^2}x\left( {y - 1} \right) - xy + \hfill \cr \,\,\,\,iy\left( {y - 1} \right) - 2x + i2\left( {y - 1} \right) \hfill \cr} \over {{x^2} - {i^2}{{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0$

$ \Rightarrow $ ${\mathop{\rm Im}\nolimits} \left[ {\left( {{\matrix{ i{x^2} + x\left( {y - 1} \right) - xy \hfill \cr \,\,\, - 2x + i\left( {y - 1} \right)\left( {y + 2} \right) \hfill \cr} \over {{x^2} + {{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0$

$ \Rightarrow $${\mathop{\rm Im}\nolimits} \left[ {\left( {{\matrix{ x\left( {y - 1} \right) - xy\, - 2x \hfill \cr \,\, + i\left[ {\left( {y - 1} \right)\left( {y + 2} \right) + {x^2}} \right] \hfill \cr} \over {{x^2} + {{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0$

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

$ \Rightarrow $2x² + 2y² - y - 1 = 0

$ \Rightarrow $x² + y² - $\left( {{1 \over 2}} \right)$y - $\left( {{1 \over 2}} \right)$ = 0

$ \therefore $ Center of the circle is $\left( {0,{1 \over 4}} \right)$

$ \therefore $ Radius = $\sqrt {{0^2} + {{\left( {{1 \over 4}} \right)}^2} + {1 \over 2}} $

= $\sqrt {{1 \over {16}} + {1 \over 2}} $

= $\sqrt {{9 \over {16}}} $

= ${{3 \over 4}}$

Given,

2 $\,\left| \, \right.$z + 3i$\,\left| \, \right.$ = $\,\left| \, \right.$z $-$i$\,\left| \, \right.$

Let z = x + iy

$ \Rightarrow $$\,\,\,$ 2 $\,\left| \, \right.$ x + iy + 3i $\,\left| \, \right.$ = $\,\left| \, \right.$ x + iy $-$ i $\,\left| \, \right.$

$ \Rightarrow $$\,\,\,$ 2 $\,\left| \, \right.$ x + i (y + 3)$\,\left| \, \right.$ = $\,\left| \, \right.$ x + i (y $-$ 1)$\,\left| \, \right.$

$ \Rightarrow $$\,\,\,$ 2 $\sqrt {{x^2} + {{\left( {y + 3} \right)}^2}} $ = $\sqrt {{x^2} + {{\left( {y - 1} \right)}^2}} $

$ \Rightarrow $$\,\,\,$ 4 (x² + y² + 6y + 9) = x² + y² $-$ 2y + 1

$ \Rightarrow $$\,\,\,$ 3x² + 3y² + 26y + 35 = 0

$ \Rightarrow $$\,\,\,$ x² + y² + ${{26} \over 3}$ y + ${{35} \over 3}$ = 0

This is a equation of circle with center ($-$ ${{13} \over 3}$, 0)

$\therefore\,\,\,$ Radius = $\sqrt {0 + {{169} \over 9} - {{35} \over 3}} $

= $\sqrt {{{64} \over 9}} $

= ${8 \over 3}$

Given 2$\omega $ + 1 = z;

z = $\sqrt 3 i$

$ \Rightarrow $ $\omega = {{\sqrt 3 i - 1} \over 2}$

$ \Rightarrow $ As $\omega $ is complex cube root of unity.

${\omega ^3} = 1$

$1 + \omega + {\omega ^2} = 0$

$\left| {\matrix{ 1 & 1 & 1 \cr 1 & { - {\omega ^2} - 1} & {{\omega ^2}} \cr 1 & {{\omega ^2}} & {{\omega ^7}} \cr } } \right| = 3k$

$ \Rightarrow $ $\left| {\matrix{ 1 & 1 & 1 \cr 1 & \omega & {{\omega ^2}} \cr 1 & {{\omega ^2}} & \omega \cr } } \right| = 3k$

Applying R₁ $ \to $ R₁ + R₂ + R₃

$ \Rightarrow $ $\left| {\matrix{ 3 & 0 & 0 \cr 1 & \omega & {{\omega ^2}} \cr 1 & {{\omega ^2}} & \omega \cr } } \right| = 3k$

$ \Rightarrow $ $3\left( {{\omega ^2} - {\omega ^4}} \right) = 3k$

$ \Rightarrow $ $\left( {{\omega ^2} - \omega } \right) = k$

$ \therefore $ $k = \left( {{{ - 1 - \sqrt 3 i} \over 2}} \right) - \left( {{{ - 1 + \sqrt 3 i} \over 2}} \right)$

$ \Rightarrow $ k = ${ - \sqrt 3 i}$ = -z

Here,

z $-$ (3 + 3i) = $2\sqrt 2 $ (cos($-$135^o) + i sin ($-$ 135^o))

= $2\sqrt 2 $ ($-$ ${1 \over {\sqrt 2 }}$ $-$${i \over {\sqrt 2 }}$)

= $-$ 2 $-$ 2i

$ \Rightarrow $   z = 3 + 3 i $-$ 2 $-$ 2 i = 1 + i

Note :

Polar form of a complex number :

z = r (cos$\theta $ + i sin$\theta $)

Here r = modulus of z and $\theta $ argument of z.

Rationalizing the given expression

${{\left( {2 + 3i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$

For the given expression to be purely imaginary, real part of the above expression should be equal to zero.

$ \Rightarrow {{2 - 6{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }} = 0$

$ \Rightarrow {\sin ^2}\theta = {1 \over 3}$

$ \Rightarrow \sin \theta = \pm {1 \over {\sqrt 3 }}$

$\left| {{{{z_1} - 2{z_2}} \over {2 - {z_1}{{\overline z }_2}}}} \right| = 1 \Rightarrow {\left| {{z_1} - 2{z_2}} \right|^2} = {\left| {2 - {z_1}{{\overline z }_2}} \right|^2}$

$ \Rightarrow \left( {{z_1} - 2{z_2}} \right)\left( {\overline {{z_1} - 2{z_2}} } \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ = \left( {2 - {z_1}{{\overline z }_2}} \right)\left( {\overline {2 - {z_1}{{\overline z }_2}} } \right)$

$ \Rightarrow \left( {{z_1} - 2{z_1}} \right)\left( {{{\overline z }_1} - 2{{\overline z }_2}} \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ = \left( {2 - {z_1}\overline {{z_2}} } \right)\left( {2 - {{\overline z }_1}{z_2}} \right)$

$ \Rightarrow \left( {{z_1}{{\overline z }_1}} \right) - 2{z_1}{\overline z _2} - 2{\overline z _1}{z_2} + 4{z_2}{\overline z _2}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ = 4 - 2{\overline z _1}{z_2} - 2{z_1}{\overline z _2} + {z_1}{\overline z _1}{z_2}{\overline z _2}$

$ \Rightarrow {\left| {{z_1}} \right|^2} + 4{\left| {{z_2}} \right|^2}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ = 4 + {\left| {{z_1}} \right|^2}{\left| {{z_2}} \right|^2}$

$ \Rightarrow {\left| {{z_1}} \right|^2} + 4{\left| {{z_2}} \right|^2} - 4 - {\left| {{z_1}} \right|^2}{\left| {{z_2}} \right|^2} = 0$

$\left( {{{\left| {{z_1}} \right|}^2} - 4} \right)\left( {1 - {{\left| {{z_2}} \right|}^2}} \right) = 0$

As $\,\,\,\left| {{z_2}} \right| \ne 1\,\,\,$ $\therefore$ $\,\,\,{\left| {{z_1}} \right|^2} = 4 \Rightarrow \left| {{z_1}} \right| = 2$

$ \Rightarrow \,$ Point $\,{z_1}\,$ lies on circle of radius $2.$

We know minimum value of

$\,\,\,\left| {{Z_1} + {Z_2}} \right|\,\,\,$ is $\,\,\,\left| {\left| {{Z_1}} \right| - \left| {{Z_2}} \right|} \right|$

Thus minimum value of

$\,\,\,\left| {Z + {1 \over 2}} \right|\,\,\,$ is $\,\,\,\left| {\left| Z \right| - {1 \over 2}} \right|$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \le \left| {Z + {1 \over 2}} \right| \le \left| Z \right| + {1 \over 2}$

Since, $\,\,\,\left| Z \right| \ge 2$

$\therefore$ $\,\,\,2 - {1 \over 2} < \left| {Z + {1 \over 2}} \right| < 2 + {1 \over 2}$

$ \Rightarrow {3 \over 2} < \left| {Z + {1 \over 2}} \right| < {5 \over 2}$

Given $\,\,\,\,\left| z \right| = 1,\,\,\arg \,z = \theta $

As we know, $\,\,\,\,\overrightarrow z = {1 \over z}$

$\therefore$ $\,\,\,\,\arg \left( {{{1 + z} \over {1 + \overrightarrow z }}} \right) = \arg \left( {{{1 + z} \over {1 + {1 \over z}}}} \right)$

$ = \arg \left( z \right) = \theta .$

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.

${\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 $z=1$

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$

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

$z$ 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 $6$

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

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

$ \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

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.