Let $$\omega $$ be a complex number such that 2$$\omega $$ + 1 = z where z = $$\sqrt {-3} $$. If $$\left| {\matrix{ 1 & 1 & 1 \cr 1 & { - {\omega ^2} - 1} & {{\omega ^2}} \cr 1 & {

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}} &

Complex Numbers — Page 4 | JEE Mathematics

Q31

Let z

\in

C, the set of complex numbers. Then the equation, 2|z + 3i|

-

|z

-

i| = 0 represents :

A a circle with radius

{8 \over 3}.

B a circle with diameter

{{10} \over 3}.

C an ellipse with length of major axis

{{16} \over 3}.

D an ellipse with length of minor axis

{{16} \over 9}.

Correct Answer

Option A

Solution

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}

Q32

Let

\omega

be a complex number such that 2

\omega

+ 1 = z where z =

\sqrt {-3}

. If

\left| \begin{array}{lll}1 & 1 & 1 \\ 1 & { - {\omega ^2} - 1} & {{\omega ^2}} \\ 1 & {{\omega ^2}} & {{\omega ^7}} \end{array} \right| = 3k

, then k is equal to :

A z

B -1

C 1

D -z

Correct Answer

Option D

Solution

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| \begin{array}{lll}1 & 1 & 1 \\ 1 & { - {\omega ^2} - 1} & {{\omega ^2}} \\ 1 & {{\omega ^2}} & {{\omega ^7}} \end{array} \right| = 3k

$\Rightarrow$

\left| \begin{array}{lll}1 & 1 & 1 \\ 1 & \omega & {{\omega ^2}} \\ 1 & {{\omega ^2}} & \omega \end{array} \right| = 3k

Applying R1 $\to$ R1 + R2 + R3 $\Rightarrow$

\left| \begin{array}{lll}3 & 0 & 0 \\ 1 & \omega & {{\omega ^2}} \\ 1 & {{\omega ^2}} & \omega \end{array} \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

Q33

The set of all

\alpha

\in

R, for which w =

{{1 + \left( {1 - 8\alpha } \right)z} \over {1 - z}}

is purely imaginary number, for all z

\in

C satisfying |z| = 1 and Re z

\ne

1, is :

A an empty set

B {0}

C

\left\{ {0,{1 \over 4}, - {1 \over 4}} \right\}

D equal to R

Correct Answer

Option B

Solution

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}

Q34

If |z

-

3 + 2i|

\le

4 then the difference between the greatest value and the least value of |z| is :

A

2\sqrt {13}

B 8

C 4 +

\sqrt {13}

D

\sqrt {13}

Correct Answer

Option A

Solution

\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}

Q35

If z and w are two complex numbers such that |zw| = 1 and arg(z) – arg(w) =

{\pi \over 2}

, then :

A

z\overline w = {{1 - i} \over {\sqrt 2 }}

B

\overline z w = i

C

z\overline w = {{ - 1 + i} \over {\sqrt 2 }}

D

\overline z w = -i

Correct Answer

Option D

Solution

\left| {zw} \right| = 1

$\Rightarrow$

\left| z \right|\left| w \right| = 1

Let

w = {1 \over r}{e^{i\theta }}

then z =

r{e^{i\left( {\theta + {\pi \over 2}} \right)}}

\overline z w = {e^{ - i\left( {\theta + {\pi \over 2}} \right)}}.{e^{i\theta }} = {e^{ - i(\pi /2)}} = - i

z\overline w = {e^{i\left( {\theta + {\pi \over 2}} \right)}}.{e^{ - i\theta }} = {e^{i\pi /2}} = i

Q36

Let z

\in

C be such that |z| < 1. If

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

z, then :

A 4Im(

\omega

) > 5

B 5Im(

\omega

) < 1

C 5Re(

\omega

) > 4

D 5Re(

\omega

) > 1

Correct Answer

Option D

Solution

\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

Q37

All the points in the set

S = \left\{ {{{\alpha + i} \over {\alpha - i}}:\alpha \in R} \right\}(i = \sqrt { - 1} )

lie on a :

A straight line whose slope is –1

B straight line whose slope is 1.

C circle whose radius is 1.

D circle whose radius is

\sqrt 2

.

Correct Answer

Option C

Solution

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.

Q38

If

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

, then (1 + iz + z5 + iz8)9 is equal to :

A 1

B –1

C 0

D (-1 + 2i)9

Correct Answer

Option B

Solution

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}}}

Q39

Let

\alpha

and

\beta

be two roots of the equation x2 + 2x + 2 = 0 , then

\alpha ^{15}

+

\beta ^{15}

is equal to :

A -256

B 512

C -512

D 256

Correct Answer

Option A

Solution

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

Q40

Let

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

If R(z) and 1(z) respectively denote the real and imaginary parts of z, then :

A R(z) =

-

3

B R(z) < 0 and I(z) > 0

C I(z) = 0

D R(z) > 0 and I(z) > 0

Correct Answer

Option C

Solution

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