Complex Numbers — Page 7 | JEE Mathematics

Q61

Let n denote the number of solutions of the equation z2 + 3

\overline z

= 0, where z is a complex number. Then the value of

\sum\limits_{k = 0}^\infty {{1 \over {{n^k}}}}

is equal to :

A 1

B

{4 \over 3}

C

{3 \over 2}

D 2

Correct Answer

Option B

Solution

z2 + 3

\overline z

= 0 Put z = x + iy $\Rightarrow$ x2 $-$ y2 + 2ixy + 3(x $-$ iy) = 0 $\Rightarrow$ (x2 $-$ y2 + 3x) + i(2xy $-$ 3y) = 0 + i0 $\therefore$ x2 $-$ y2 + 3x = 0 ..... (1) 2xy $-$ 3y = 0 ..... (2) x =

{3 \over 2}

, y = 0 Put x =

{3 \over 2}

in equation (1)

{9 \over 4} - {y^2} + {9 \over 2} = 0

{y^2} = {{27} \over 4} \Rightarrow y = \pm {{3\sqrt 3 } \over 2}

$\therefore$

(x,y) = \left( {{3 \over 2},{{3\sqrt 3 } \over 2}} \right),\left( {{3 \over 2},{{ - 3\sqrt 3 } \over 2}} \right)

Put y = 0 $\Rightarrow$ x2 $-$ 0 + 3x = 0 x = 0, $-$ 3 $\therefore$ (x, y) = (0, 0), ( $-$ 3, 0) $\therefore$ No of solutions = n = 4

\sum\limits_{K = 0}^\infty {\left( {{1 \over {{n^k}}}} \right)} = \sum\limits_{K = 0}^\infty {\left( {{1 \over {4{n^k}}}} \right)}

= {1 \over 1} + {1 \over 4} + {1 \over {16}} + {1 \over {64}} + ......

= {1 \over {1 - {1 \over 4}}} = {4 \over 3}

Q62

Let C be the set of all complex numbers. Let S1 = {z

\in

C : |z

-

2|

\le

1} and S2 = {z

\in

C : z(1 + i) +

\overline z

(1

-

i)

\ge

4}. Then, the maximum value of

{\left| {z - {5 \over 2}} \right|^2}

for z

\in

S1

\cap

S2 is equal to :

A

{{3 + 2\sqrt 2 } \over 4}

B

{{5 + 2\sqrt 2 } \over 2}

C

{{3 + 2\sqrt 2 } \over 2}

D

{{5 + 2\sqrt 2 } \over 4}

Correct Answer

Option D

Solution

|t $-$ 2| $\le$ 1 Put t = x + iy (x $-$ 2)2 + y2 $\le$ 1 Also, t(1 + i) +

\overline t

(1 $-$ i) $\ge$ 4 x $-$ y $\ge$ 2 Let point on circle be A(2 + cos $\theta$ , sin $\theta$ )

\theta \in \left[ { - {{3\pi } \over 4},{\pi \over 4}} \right]

{(AP)^2} = {\left( {2 + \cos \theta - {5 \over 2}} \right)^2} + \sin \theta

= {\cos ^2}\theta - \cos \theta + {1 \over 4} + {\sin ^2}\theta

= {5 \over 4} - \cos \theta

For (AP)2 maximum

\theta = - {{3\pi } \over 4}

{(AP)^2} = {5 \over 4} + {1 \over {\sqrt 2 }} = {{5\sqrt 2 + 4} \over {4\sqrt 2 }}

Q63

If

S = \left\{ {z \in C:{{z - i} \over {z + 2i}} \in R} \right\}

, then :

A S contains exactly two elements

B S contains only one element

C S is a circle in the complex plane

D S is a straight line in the complex plane

Correct Answer

Option D

Solution

Given

{{z - i} \over {z + 2i}} \in R

Then

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

is 0 or $\pi$ $\Rightarrow$ S is straight line in complex

Q64

If z is a complex number such that

{{z - i} \over {z - 1}}

is purely imaginary, then the minimum value of | z

-

(3 + 3i) | is :

A

2\sqrt 2 - 1

B

3\sqrt 2

C

6\sqrt 2

D

2\sqrt 2

Correct Answer

Option D

Solution

{{z - i} \over {z - 1}}

is purely imaginary number Let

z = x + iy

$\therefore$

{{x + i(y - 1)} \over {(x - 1) + i(y)}} \times {{(x - 1) - iy} \over {(x - 1) - iy}}

\Rightarrow {{x(x - 1) + y(y - 1) + i( - y - x + 1)} \over {{{(x - 1)}^2} + {y^2}}}

is purely imaginary number

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

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

$\therefore$

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

= {5 \over {\sqrt 2 }} - {1 \over {\sqrt 2 }} = 2\sqrt 2

Q65

Let

\alpha

and

\beta

be the roots of the equation x2 + (2i

-

1) = 0. Then, the value of |

\alpha

8 +

\beta

8| is equal to :

A 50

B 250

C 1250

D 1500

Correct Answer

Option A

Solution

Given equation,

{x^2} + (2i - 1) = 0

\Rightarrow {x^2} = 1 - 2i

Let $\alpha$ and $\beta$ are the two roots of the equation. As, we know roots of a equation satisfy the equation so

{\alpha ^2} = 1 - 2i

and

{\beta ^2} = 1 - 2i

$\therefore$

{\alpha ^2} = {\beta ^2} = 1 - 2i

$\therefore$

|{\alpha ^2}| = \sqrt {{1^2} + {{( - 2)}^2}} = \sqrt {15}

Now,

|{\alpha ^8} + {\beta ^8}|

|{\alpha ^8} + {\alpha ^8}|

= 2|{\alpha ^8}|

= 2|{\alpha ^2}{|^4}

= 2{\left( {\sqrt 5 } \right)^4}

= 2 \times 25

= 50

Q66

Let

z

be a complex number such that

\left| {{{z - 2i} \over {z + i}}} \right| = 2,z \ne - i

. Then

z

lies on the circle of radius 2 and centre :

A (0,

-

2)

B (0, 0)

C (0, 2)

D (2, 0)

Correct Answer

Option A

Solution

$\left|\dfrac{z-2 i}{z+i}\right|=2$ $\Rightarrow (z-2 i)(\bar{z}+2 i)=4(z+i)(\bar{z}-i)$ $\Rightarrow z \bar{z}+2 i z-2 i \bar{z}+4=4(z \bar{z}-z i+\overline{z i}+1)$ $\Rightarrow 3 z \bar{z}-6 i z+6 i \bar{z}=0$ $\Rightarrow z \bar{z}-2 i z+2 i \bar{z}=0$ $\therefore$ Centre $(-2 i)$ or $(0,-2)$ Other Method : $\left|\dfrac{z-2 i}{z+i}\right|=2, z \neq-i$ Put $z=x+i y$

\begin{aligned} & \left|\frac{x+i y-2 i}{x+i y+i}\right|=2 \\\\ & \Rightarrow \left|\frac{x+i(y-2)}{x+i(y+1)}\right|^2=4 \\\\ & \Rightarrow x^2+(y-2)^2=4\left[x^2+(y+1)^2\right] \\\\ & \Rightarrow x^2+y^2+4-4 y=4\left[x^2+y^2+1+2 y\right] \end{aligned}

$\Rightarrow$ $x^2+y^2+4 y=0$ or $x^2+(y+2)^2=2^2$ $\therefore$ Centre is $(0,-2)$ .

Q67

Let arg(z) represent the principal argument of the complex number z. Then, |z| = 3 and arg(z

-

1)

-

arg(z + 1) =

{\pi \over 4}

intersect :

A exactly at one point.

B exactly at two points.

C nowhere.

D at infinitely many points.

Correct Answer

Option C

Solution

Let

z = x + iy

$\therefore$

|z| = \sqrt {{x^2} + {y^2}}

Given,

|z| = 3

$\therefore$

\sqrt {{x^2} + {y^2}} = 3

\Rightarrow {x^2} + {y^2} = 9 = {3^2}

This represent a circle with center at (0, 0) and radius = 3 Now, given

\arg (z - 1) - \arg (z + 1) = {\pi \over 4}

\Rightarrow \arg (x + iy - 1) - \arg (x + iy + 1) = {\pi \over 4}

\Rightarrow \arg (x - 1 + iy) - \arg (x + 1 + iy) = {\pi \over 4}

\Rightarrow {\tan ^{ - 1}}\left( {{y \over {x - 1}}} \right) - {\tan ^{ - 1}}\left( {{y \over {x + 1}}} \right) = {\pi \over 4}

\Rightarrow {\tan ^{ - 1}}\left( {{{{y \over {x - 1}} - {y \over {x + 1}}} \over {1 + {y \over {x - 1}} \times {y \over {x + 1}}}}} \right) = {\pi \over 4}

\Rightarrow {\tan ^{ - 1}}\left( {{{{{xy + y - xy + y} \over {{x^2} - 1}}} \over {{{{x^2} - 1 + {y^2}} \over {{x^2} - 1}}}}} \right) = {\pi \over 4}

\Rightarrow {\tan ^{ - 1}}\left( {{{xy + y - xy + y} \over {{x^2} - 1 + {y^2}}}} \right) = {\pi \over 4}

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

\Rightarrow 2y = {x^2} + {y^2} - 1

\Rightarrow {x^2} + {y^2} - 2y - 1 = 0

\Rightarrow {x^2} + {(y - 1)^2} = {(\sqrt 2 )^2}

This represent a circle with center at (0, 1) and radius

\sqrt 2

. From diagram you can see both the circles do not cut anywhere.

Q68

The number of points of intersection of

|z - (4 + 3i)| = 2

and

|z| + |z - 4| = 6

, z

\in

C, is :

A 0

B 1

C 2

D 3

Correct Answer

Option C

Solution

{C_1}:|z - (4 + 3i)| = 2

and

{C_2}:|z| + |z - 4| = 6

,

z \in C

C1 represents a circle with centre (4, 3) and radius 2 and C2 represents a ellipse with focii at (0, 0) and (4, 0) and length of major axis = 6, and length of semi-major axis 2

\sqrt5

and (4, 2) lies inside the both C1 and C2 and (4, 3) lies outside the C2 $\therefore$ number of intersection points = 2

Q69

The area of the polygon, whose vertices are the non-real roots of the equation

\overline z = i{z^2}

is :

A

{{3\sqrt 3 } \over 4}

B

{{3\sqrt 3 } \over 2}

C

{3 \over 2}

D

{3 \over 4}

Correct Answer

Option A

Solution

\overline z = i{z^2}

Let

z = x + iy

x - iy = i({x^2} - {y^2} + 2xiy)

x - iy = i({x^2} - {y^2}) - 2xy

$\therefore$

x = - 2yx

or

{x^2} - {y^2} = - y

x = 0

or

y = - {1 \over 2}

Case - I

x = 0

- {y^2} = - y

y = 0,\,\,1

Case - II

y = - {1 \over 2}

\Rightarrow {x^2} - {1 \over 4} = {1 \over 2} \Rightarrow x = \pm {{\sqrt 3 } \over 2}

x = \left\{ {0,i,{{\sqrt 3 } \over 2} - {i \over 2},{{ - \sqrt 3 } \over 2} - {i \over 2}} \right\}

Area of polygon

= {1 \over 2}\left| \begin{array}{lll}0 & 1 & 1 \\ {{{\sqrt 3 } \over 2}} & {{{ - 1} \over 2}} & 1 \\ {{{ - \sqrt 3 } \over 2}} & {{{ - 1} \over 2}} & 1 \end{array} \right|

= {1 \over 2}\left| { - \sqrt 3 \,\,\, - {{\sqrt 3 } \over 2}} \right| = {{3\sqrt 3 } \over 4}

Q70

Let $$A = \left\{ {z \in C:\left| {{{z + 1} \over {z - 1}}} \right|

A a portion of a circle centred at

\left( {0, - {1 \over {\sqrt 3 }}} \right)

that lies in the second and third quadrants only

B a portion of a circle centred at

\left( {0, - {1 \over {\sqrt 3 }}} \right)

that lies in the second quadrant only

C an empty

D a portion of a circle of radius

{2 \over {\sqrt 3 }}

that lies in the third quadrant only

Correct Answer

Option B

Solution

\left|\frac{z+1}{z-1}\right|<1 \Rightarrow|z+1|<|z-1| \Rightarrow \operatorname{Re}(z)<0

and $\arg \left(\dfrac{z-1}{z+1}\right)=\dfrac{2 \pi}{3}$ is a part of circle as shown.