Complex Numbers — Page 6 | JEE Mathematics

Q51

If

{\mathop{\rm Re}\nolimits} \left( {{{z - 1} \over {2z + i}}} \right) = 1

, where z = x + iy, then the point (x, y) lies on a :

A straight line whose slope is

{3 \over 2}

B straight line whose slope is

-{2 \over 3}

C circle whose diameter is

{{\sqrt 5 } \over 2}

D circle whose centre is at

\left( { - {1 \over 2}, - {3 \over 2}} \right)

Correct Answer

Option C

Solution

{\mathop{\rm Re}\nolimits} \left( {{{z - 1} \over {2z + i}}} \right) = 1

Put z = x + iy $\therefore$

{\mathop{\rm Re}\nolimits} \left( {{{\left( {x + iy} \right) - 1} \over {2\left( {x + iy} \right) + i}}} \right) = 1

$\Rightarrow$

{\mathop{\rm Re}\nolimits} \left( {\left( {{{\left( {x - 1} \right) + iy} \over {2x + i\left( {2y + 1} \right)}}} \right)\left( {{{2x - i\left( {2y + 1} \right)} \over {2x - i\left( {2y + 1} \right)}}} \right)} \right) = 1

$\Rightarrow$

{\mathop{\rm Re}\nolimits} \left( {{{\left\{ {\left( {x - 1} \right) + iy} \right\}\left\{ {2x - i\left( {2y + 1} \right)} \right\}} \over {4{x^2} + {{\left( {2y + 1} \right)}^2}}}} \right) = 1

Real part of this equation is = 1 $\therefore$

{{2x\left( {x - 1} \right) + y\left( {2y + 1} \right)} \over {4{x^2} + {{\left( {2y + 1} \right)}^2}}}

= 1 $\Rightarrow$ 2x2 + 2y2 +2x + 3y + 1 = 0 $\Rightarrow$ x2 + y2 +x +

{3 \over 2}

y +

{1 \over 2}

= 0 This is an equation of circle. $\therefore$ Locus is a circle whose center is

\left( { - {1 \over 2}, - {3 \over 4}} \right)

and radius

{{\sqrt 5 } \over 4}

$\therefore$ Diameter = 2 $\times$

{{\sqrt 5 } \over 4}

=

{{\sqrt 5 } \over 2}

Q52

If

z

is a complex number, then the number of common roots of the equations

z^{1985}+z^{100}+1=0

and

z^3+2 z^2+2 z+1=0

, is equal to

A 0

B 2

C 1

D 3

Correct Answer

Option B

Solution

\begin{array}{ll} \text{ } & z^{1985}+z^{100}+1=0 \& z^3+2 z^2+2 z+1=0 \\ & (z+1)\left(z^2-z+1\right)+2 z(z+1)=0 \\ & (z+1)\left(z^2+z+1\right)=0 \\ \Rightarrow \quad & z=-1, \quad z=w, w^2 \\ & \text{ Now putting } z=-1 \text{ not satisfy } \\ & \text{ Now put } z=w \\ \Rightarrow \quad & w^{1985}+w^{100}+1 \\ \Rightarrow \quad & w^2+w+1=0 \\ \Rightarrow \quad & \text{ Also, } z=w^2 \\ \Rightarrow \quad & w^{3970}+w^{200}+1 \\ \Rightarrow & w+w^2+1=0 \end{array}

Two common root

Q53

Let the lines (2

-

i)z = (2 + i)

\overline z

and (2

+

i)z + (i

-

2)

\overline z

-

4i = 0, (here i2 =

-

1) be normal to a circle C. If the line iz +

\overline z

+ 1 + i = 0 is tangent to this circle C, then its radius is :

A

{3 \over {2\sqrt 2 }}

B

3\sqrt 2

C

{1 \over {2\sqrt 2 }}

D

{3 \over {\sqrt 2 }}

Correct Answer

Option A

Solution

(2 - i)z = (2 + i)\overline z

\Rightarrow (2 - i)(x + iy) = (2 + i)(x - iy)

\Rightarrow 2x - ix + 2iy + y = 2x + ix - 2 - iy + y

\Rightarrow 2ix - 4iy = 0

{L_1}:x - 2y = 0

\Rightarrow (2 + i)z + (i - 2)\overline z - 4i = 0

\Rightarrow (2 + i)(x + iy) + (i - 2)(x - iy) - 4i = 0

\Rightarrow 2x + ix + 2iy - y + ix - 2x + y + 2iy - 4i = 0

\Rightarrow 2ix + 4iy - 4i = 0

{L_2}:x + 2y - 2 = 0

Solve L1 and L2: x = 1,

4y = 2,y = {1 \over 2}

$\therefore$

x = 1

Centre

\left( {1,{1 \over 2}} \right)

{L_3}:iz + \overline z + 1 + i = 0

\Rightarrow i(x + iy) + x - iy + 1 + i = 0

\Rightarrow ix - y + x - iy + 1 + i = 0

\Rightarrow (x - y + 1) + i(x - y + 1) = 0

Radius = distance from

\left( {1,{1 \over 2}} \right)

to

x - y + 1 = 0

$\Rightarrow$

r = {{1 - {1 \over 2} + 1} \over {\sqrt 2 }}

$\Rightarrow$

r = {3 \over {2\sqrt 2 }}

Q54

Let a complex number z, |z|

\ne

1, satisfy

{\log _{{1 \over {\sqrt 2 }}}}\left( {{{|z| + 11} \over {{{(|z| - 1)}^2}}}} \right) \le 2

. Then, the largest value of |z| is equal to ____________.

A 5

B 8

C 6

D 7

Correct Answer

Option D

Solution

{{|z| + 11} \over {{{(|z| - 1)}^2}}} \ge {1 \over 2}

2|z| + 22 \ge {(|z| - 1)^2}

2|z| + 22 \ge \,|z{|^2} - 2|z| + 1

|z{|^2} - 4|z| - 21 \le 0

(|z| - 7)(|z| + 3) \le 0

\Rightarrow \,|z| \le 7

$\therefore$

|z{|_{\max }} = 7

Q55

The least value of |z| where z is complex number which satisfies the inequality

\exp \left( {{{(|z| + 3)(|z| - 1)} \over {||z| + 1|}}{{\log }_e}2} \right) \ge {\log _{\sqrt 2 }}|5\sqrt 7 + 9i|,i = \sqrt { - 1}

, is equal to :

A 8

B 3

C 2

D

\sqrt 5

Correct Answer

Option B

Solution

Let | z | = t, t $\ge$ 0

{e^{{{(t + 3)(t - 1)} \over {t + 1}}{{\log }_e}2}} \ge {\log _{\sqrt 2 }}16 = 8

( $\because$ t + 1 > 0)

{2^{{{(t + 3)(t - 1)} \over {t + 1}}}} \ge {2^3}

{{(t + 3)(t - 1)} \over {t + 1}} \ge 3

{t^2} + 2t - 3 \ge 3t + 3

{t^2} - t - 6 \ge 0

t \in ( - \infty , - 2) \cup [3,\infty )

But t $\ge$ 0 $\therefore$

t \in [3,\infty )

Q56

The area of the triangle with vertices A(z), B(iz) and C(z + iz) is :

A 1

B

{1 \over 2}

| z |2

C

{1 \over 2}

| z + iz |2

D

{1 \over 2}

Correct Answer

Option B

Solution

Each side length = |z| Area of

\Delta

=

{1 \over 2}

(area of square) =

{1 \over 2}

|z|2

Q57

If the equation

a|z{|^2} + \overline {\overline \alpha z + \alpha \overline z } + d = 0

represents a circle where a, d are real constants then which of the following condition is correct?

A |

\alpha

|2

-

ad

\ne

0

B |

\alpha

|2

-

ad > 0 and a

\in

R

-

{0}

C |

\alpha

|2

-

ad

\ge

0 and a

\in

R

D

\alpha

= 0, a, d

\in

R+

Correct Answer

Option B

Solution

a|z{|^2} + \alpha \overline z + \overline \alpha z + d = 0

$\Rightarrow$

z\overline z + \left( {{\alpha \over a}} \right)\overline z + \left( {{{\overline \alpha } \over a}} \right)z + {d \over a} = 0

$\therefore$ Centre

= - {\alpha \over a}

r = \sqrt {{{\left| {{\alpha \over a}} \right|}^2} - {d \over a}}

\Rightarrow {\left| {{\alpha \over a}} \right|^2} \ge {d \over a}

\Rightarrow {\left| \alpha \right|^2} \ge ad

Q58

Let a complex number be w = 1

-

{\sqrt 3 }

i. Let another complex number z be such that |zw| = 1 and arg(z)

-

arg(w) =

{\pi \over 2}

. Then the area of the triangle with vertices origin, z and w is equal to :

A 4

B

{1 \over 4}

C 2

D

{1 \over 2}

Correct Answer

Option D

Solution

Given, w = 1 $-$

\sqrt 3

i

\Rightarrow |w| = \sqrt {{{(1)}^2} + {{( - \sqrt 3 )}^2}} = 2

Also, given | zw | = 1 $\Rightarrow$ | z | | w | = 1 (using property) $\Rightarrow$ | z | =

{1 \over 2}

Also, arg(z) $-$ arg(w) =

{{\pi {} } \over 2}

$\therefore$ Angle between two complex number z and w is

{{\pi {} } \over 2}

. $\therefore$

\angle zow = {{\pi {} } \over 2}

$\therefore$

\Delta

zow is a right angle triangle with base

ow = 2

and height

oz = {1 \over 2}

$\therefore$ Area =

{1 \over 2} \times 2 \times {1 \over 2} = {1 \over 2}

Q59

If z and

\omega

are two complex numbers such that

\left| {z\omega } \right| = 1

and

\arg (z) - \arg (\omega ) = {{3\pi } \over 2}

, then

\arg \left( {{{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }}} \right)

is : (Here arg(z) denotes the principal argument of complex number z)

A

{\pi \over 4}

B

- {{3\pi } \over 4}

C

- {\pi \over 4}

D

{{3\pi } \over 4}

Correct Answer

Option B

Solution

As

\left| {z\omega } \right| = 1

$\Rightarrow$ If

\left| z \right| = r

, then

\left| \omega \right| = {1 \over r}

Let

\arg (z) = \theta

$\therefore$

\arg (\omega ) = \left( {\theta - {{3\pi } \over 2}} \right)

So,

z = r{e^{i\theta }}

\Rightarrow \overline z = r{e^{i\theta }}

\omega = {1 \over r}{e^{i\left( {\theta - {{3\pi } \over 2}} \right)}}

Now, consider

{{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }} = {{1 - 2{e^{i\left( { - {{3\pi } \over 2}} \right)}}} \over {1 + 3{e^{i\left( { - {{3\pi } \over 2}} \right)}}}} = \left( {{{1 - 2i} \over {1 + 3i}}} \right)

= {{(1 - 2i)(1 - 3i)} \over {(1 + 3i)(1 - 3i)}} = - {1 \over 2}(1 + i)

$\therefore$

prin\arg \left( {{{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }}} \right)

= prin\arg \left( {{{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }}} \right)

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

= - \left( {\pi - {\pi \over 4}} \right) = {{ - 3\pi } \over 4}

So, option (2) is correct.

Q60

Let S1, S2 and S3 be three sets defined as S1 = {z

\in

C : |z

-

1|

\le

\sqrt 2

} S2 = {z

\in

C : Re((1

-

i)z)

\ge

1} S3 = {z

\in

C : Im(z)

\le

1} Then the set S1

\cap

S2

\cap

S3 :

A has exactly three elements

B is a singleton

C has infinitely many elements

D has exactly two elements

Correct Answer

Option C

Solution

Let, z = x + iy S1 $\equiv$ (x $-$ 1)2 + y2 $\le$ 2 .....

(1) S2 $\equiv$ x + y $\ge$ 1 .....

(2) S3 $\equiv$ y $\le$ 1 ....

(3) $\Rightarrow$ S1 $\cap$ S2 $\cap$ S3 has infinitely many elements.