Complex Numbers

The complex number z satisfying $\left| {z - 2 + i} \right| \ge \sqrt 5 $, which represents the region outside the circle (including the circumference) having centre (2, ${ - 1}$) and radius $\sqrt 5 $ units.



Now, for ${{z_0} \in S{1 \over {\left| {{z_0} - 1} \right|}}}$ is maximum.

When ${\left| {{z_0} - 1} \right|}$ is minimum. And for this it is required that ${{z_0} \in S}$, such that z₀ is collinear with the points (2, $ - $1) and (1, 0) and lies on the circumference of the circle $\left| {z - 2 + i} \right|$ = $\sqrt 5 $.

So let z₀ = x + iy, and from the figure 0 < x < 1 and y >0.

So, ${{4 - {z_0} - {{\overline z }_0}} \over {{z_0} - {{\overline z }_0} + 2i}} = {{4 - x - iy - x + iy} \over {x + iy - x + iy + 2i}} = {{2(2 - x)} \over {2i(y + 1)}} = - i\left( {{{2 - x} \over {y + 1}}} \right)$

$ \because $ ${{{2 - x} \over {y + 1}}}$ is a positive real number, so ${{4 - {z_0} - {{\overline z }_0}} \over {{z_0} - {{\overline z }_0} + 2i}}$ is purely negative imaginary number.

$ \Rightarrow $ $\arg \left( {{{4 - {z_0} - {{\overline z }_0}} \over {{z_0} - {{\overline z }_0} + 2i}}} \right) = - {\pi \over 2}$

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

$ \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}

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.

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

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,

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 }}$

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.

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

$ \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, $\mathrm{Z}_k=\cos \frac{2 k \pi}{10}+i \sin \frac{2 k \pi}{10}, k=1,2,3, \ldots, 9$

$\begin{aligned} & \text { (P) } Z_k \cdot Z_j=1 \\ & \Rightarrow\left(\cos \frac{2 k \pi}{10}+i \sin \frac{2 k \pi}{10}\right) \\ & \left(\cos \frac{2 j \pi}{10}+i \sin \frac{2 j \pi}{10}\right)=1 \end{aligned}$

$\begin{aligned} & \Rightarrow \cos \left(\frac{2 \pi(k+j)}{10}\right)+i \sin \frac{2 \pi(k+j)}{10}=1 \\ & \Rightarrow \cos \frac{2 \pi(k+j)}{10}=1 \text { and } \sin \frac{2 \pi(k+j)}{10}=0 \\ & \Rightarrow \quad \frac{2 \pi(k+j)}{10}=2 \lambda \pi \text { where } \lambda \in \text { Integer } \\ & \Rightarrow \quad k+j=10 \lambda \\ & \Rightarrow \quad j=10 \lambda-k \\ \end{aligned}$

For every value of $k$ there is a possible value of $j$

Hence, $P$ match with 1.

(Q) Given $Z_1 \cdot Z=Z_k$

$\begin{aligned} & \left(\cos \frac{2 \pi}{10}+i \sin \frac{2 \pi}{10}\right) \cdot z=\cos \frac{2 \pi k}{10}+i \sin \frac{2 \pi k}{10} \\ & z=\frac{\cos \frac{2 \pi k}{10}+i \sin \frac{2 \pi k}{10}}{\cos \frac{2 \pi}{10}+i \sin \frac{2 \pi}{10}} \\ & z=\cos \frac{2 \pi(k-1)}{10}+i \sin \frac{2 \pi(k-1)}{10} \end{aligned}$

For every value of $k$, there is a unique value of $z$ is possible.

$\therefore$ The statement $z_1 \cdot z=z_k$ has no solution z in the set of complex numbers is false.

Hence, Q match with 2.

(R) Given, $z_k=\cos \frac{2 \pi k}{10}+i \sin \frac{2 \pi k}{10}, k=1,2,3, \ldots 9$

Here, $z_1, z_2, z_3, \ldots z_9$ are the roots of $z^{10}-1=0$

We know $z^{10}-1=(z-1)$

$\begin{aligned} & \left(z^9+z^8+z^7+z^6+z^5+z^4+z^3+z^2+z+1\right) \\ & =(z-1)\left(z-z_1\right)\left(z-z_2\right) \ldots\left(z-z_9\right) \\ \Rightarrow & 1+z+z^2+\ldots+z^9=\left(z-z_1\right)\left(z-z_2\right) \\ & \left(z-z_3\right) \ldots\left(z-z_9\right) \\ \Rightarrow & \left|1+z+z^2+\cdots+z^9\right| \\ & =\left|\left(z-z_1\right)\left(z-z_2\right)\left(z-z_3\right) \ldots\left(z-z_9\right)\right| \\ \Rightarrow & \left|z-z_1\right| \cdot\left|z-z_2\right|\left|z-z_3\right| \ldots\left|z-z_9\right| \\ & =\left|1+z+z^2+\ldots+z^9\right| \end{aligned}$

Put $z=1$

$\begin{aligned} & \Rightarrow\left|1-z_1\right| \cdot\left|1-z_2\right|\left|1-z_3\right| \ldots\left|1-z_9\right|=10 \\ & \Rightarrow \frac{\left|1-z_1\right|\left|1-z_2\right|\left|1-z_3\right| \ldots\left|1-z_9\right|}{10}=1 \end{aligned}$

Hence, R match with 3.

(S) We know the sum of all roots of $z^{10}-1=0$ is zero.

$\begin{aligned} & \Rightarrow 1+z_1+z_2+z_3+\ldots+z_9=0 \\ & \Rightarrow 1+\left(\cos \frac{2 \pi}{10}+i \sin \frac{2 \pi}{10}\right) \\ & +\left(\cos \left(\frac{4 \pi}{10}\right)+i \sin \left(\frac{4 \pi}{10}\right)\right) \\ & +\left(\cos \frac{6 \pi}{10}+i \sin \frac{6 \pi}{10}\right) \\ & +\ldots+\cos \frac{18 \pi}{10}+i \sin \frac{18 \pi}{10}=0 \\ \end{aligned}$

$\begin{aligned} & \begin{array}{l} \Rightarrow\left(1+\cos \frac{2 \pi}{10}+\cos \frac{4 \pi}{10}+\cdots+\cos \frac{18 \pi}{10}\right) \\ +i\left(\sin \frac{2 \pi}{10}+\sin \frac{4 \pi}{10}+\ldots+\sin \frac{18 \pi}{10}\right) \\ =0+i 0 \end{array} \\ & \Rightarrow 1+\cos \frac{2 \pi}{10}+\cos \frac{4 \pi}{10}+\cos \frac{6 \pi}{10}+\ldots+\cos \frac{18 \pi}{10}=0 \\ & \Rightarrow \sum_{k=1}^9 \cos \frac{2 k \pi}{10}=-1 \\ & \Rightarrow 1-\sum_{k=1}^9 \cos \frac{2 k \pi}{10}=1-(-1)=2 \end{aligned}$

Hint:

(i) Recall $(\cos \theta+i \sin \theta)(\cos \phi+i \sin \phi)=\cos (\theta+\phi)+i \sin (\theta+\phi)$

(ii) $\cos \theta=0$ when $\theta=2 n \pi$, where $n$ is an integer. and $\sin \theta=0$ when $\theta=n \pi$, where $n$ is an integer.

(iii) $1, z_1, z_2, z_3, \ldots z_9$ are the roots of $z^{10}-1=0$.

(iv) $z^{10}-1=(z-1)\left(1+z+z^2+\ldots+z^9\right)$

$=(z-1)\left(z-z_1\right)\left(z-z_2\right) \ldots\left(z-z_9\right)$

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 .$

We know $|1-3 i-z|=|z-(1-3 i)|$ implies distance of $z$ from $(1,-3)$.

Let $P=(1,-3)$

$\min _{z \in \mathrm{S}}|1-3 i-z|=\mathrm{PN}$

$\begin{aligned} & \Rightarrow \min _{z \in S}|1-3 i-z|=\frac{|\sqrt{3} \cdot 1-3|}{\sqrt{(\sqrt{3})^2+1^2}} \\ & \Rightarrow \min _{z \in S}|1-3 i-z|=\frac{3-\sqrt{3}}{2} \end{aligned}$

Hints:

(i) $\left|z_1-z_2\right|$ implies distance between $z_1$ and $z_2$

(ii) The minimum distance of a point $\left(x_1, y_1\right)$ from the line $a x+b y+c=0$ is $\frac{\left|a x_1+b y_1+c\right|}{\sqrt{a^2+b^2}}$.

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

$\begin{aligned} x+i y & =x-i y \\ \Rightarrow \quad x+i y-x+i y & =0 \\ \Rightarrow \quad 2 i y & =0 \\ \Rightarrow \text { Imaginary part } & =0 \end{aligned}$