Complex Numbers — Page 3 | JEE Mathematics

Q21

If z is a complex number such that

\,\left| z \right| \ge 2\,

, then the minimum value of

\,\,\left| {z + {1 \over 2}} \right|

:

A is strictly greater that

{{5 \over 2}}

B is strictly greater that

{{3 \over 2}}

but less than

{{5 \over 2}}

C is equal to

{{5 \over 2}}

D lie in the interval (1, 2)

Correct Answer

Option B

Solution

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}

Q22

Let

{Z_1}

and

{Z_2}

be two roots of the equation

{Z^2} + aZ + b = 0

, Z being complex. Further , assume that the origin,

{Z_1}

and

{Z_2}

form an equilateral triangle. Then :

A

{a^2} = 4b

B

{a^2} = b

C

{a^2} = 2b

D

{a^2} = 3b

Correct Answer

Option D

Solution

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}

" ]

Q23

Let z and w be complex numbers such that

\overline z + i\overline w = 0

and arg zw =

\pi

. Then arg z equals :

A

{{5\pi } \over 4}

B

{{\pi } \over 2}

C

{{3\pi } \over 4}

D

{{\pi } \over 4}

Correct Answer

Option C

Solution

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}

Q24

If

\alpha

and

\beta

be the roots of the equation x2 – 2x + 2 = 0, then the least value of n for which

{\left( {{\alpha \over \beta }} \right)^n} = 1

is :

A 2

B 5

C 4

D 3

Correct Answer

Option C

Solution

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

Q25

If

\,\omega = {z \over {z - {1 \over 3}i}}\,

and

\left| \omega \right| = 1

, then

z

lies on :

A an ellipse

B a circle

C a straight line

D a parabola

Correct Answer

Option C

Solution

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.

Q26

If

\,\left| {{z^2} - 1} \right| = {\left| z \right|^2} + 1

, then z lies on :

A an ellipse

B the imaginary axis

C a circle

D the real axis

Correct Answer

Option B

Solution

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.

Q27

If

\,\left| {z - {4 \over z}} \right| = 2,

then the maximum value of

\,\left| z \right|

is equal to :

A

\sqrt 5 + 1

B 2

C

2 + \sqrt 2

D

\sqrt 3 + 1

Correct Answer

Option A

Solution

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

Q28

A complex number z is said to be unimodular if

\,\left| z \right| = 1

. Suppose

{z_1}

and

{z_2}

are complex numbers such that

{{{z_1} - 2{z_2}} \over {2 - {z_1}\overline {{z_2}} }}

is unimodular and

{z_2}

is not unimodular. Then the point

{z_1}

lies on a :

A circle of radius 2.

B circle of radius

{\sqrt 2 }

.

C straight line parallel to x-axis

D straight line parallel to y-axis.

Correct Answer

Option A

Solution

\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.

Q29

A value of

\theta \,

for which

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

is purely imaginary, is :

A

{\sin ^{ - 1}}\left( {{{\sqrt 3 } \over 4}} \right)

B

{\sin ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)\,

C

{\pi \over 3}

D

{\pi \over 6}

Correct Answer

Option B

Solution

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 }}

Q30

The equation Im

\left( {{{iz - 2} \over {z - i}}} \right)

+ 1 = 0, z

\in

C, z

\ne

i represents a part of a circle having radius equal to :

A 2

B 1

C

{3 \over 4}

D

{1 \over 2}

Correct Answer

Option C

Solution

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( {\begin{array}{ll}i{x^2} - {i^2}x\left( {y - 1} \right) - xy + \\ \,\,\,\,iy\left( {y - 1} \right) - 2x + i2\left( {y - 1} \right) \end{array} \over {{x^2} - {i^2}{{\left( {y - 1} \right)}^2}}} \right)} \right] + 1 = 0

$\Rightarrow$

{\mathop{\rm Im}\nolimits} \left[ {\left( {\begin{array}{ll}i{x^2} + x\left( {y - 1} \right) - xy \\ \,\,\, - 2x + i\left( {y - 1} \right)\left( {y + 2} \right) \end{array} \over {{x^2} + {{\left( {y - 1} \right)}^2}}} \right)} \right] + 1 = 0

$\Rightarrow$

{\mathop{\rm Im}\nolimits} \left[ {\left( {\begin{array}{ll}x\left( {y - 1} \right) - xy\, - 2x \\ \,\, + i\left[ {\left( {y - 1} \right)\left( {y + 2} \right) + {x^2}} \right] \end{array} \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}}