Complex Numbers — Page 13 | JEE Mathematics

Q121

Let

O

be the origin, the point

A

be

z_1=\sqrt{3}+2 \sqrt{2} i

, the point

B\left(z_2\right)

be such that

\sqrt{3}\left|z_2\right|=\left|z_1\right|

and

\arg \left(z_2\right)=\arg \left(z_1\right)+\dfrac{\pi}{6}

. Then

A area of triangle ABO is

\dfrac{11}{4}

B area of triangle ABO is

\dfrac{11}{\sqrt{3}}

C ABO is a scalene triangle

D ABO is an obtuse angled isosceles triangle

Correct Answer

Option D

Solution

z_1=\sqrt{3}+2 \sqrt{2} i \quad \& \frac{\left|z_2\right|}{\left|z_1\right|}=\frac{1}{\sqrt{3}}

given $\arg \left(\dfrac{z_2}{z_1}\right)=\dfrac{\pi}{6}$

\begin{aligned} & z_2=\frac{\left|z_2\right|}{\left|z_1\right|} \cdot z_1 \mathrm{e}^{i\left(\frac{\pi}{6}\right)} \\ & z_2=\frac{1}{\sqrt{3}} \cdot \frac{(\sqrt{3}+2 \sqrt{2} i)(\sqrt{3}+i)}{2} \\ & z_2=\frac{(3-2 \sqrt{2})+i(2 \sqrt{6}+\sqrt{3})}{2 \sqrt{3}} \end{aligned}

Now,

z_1-z_2=\frac{(3+2 \sqrt{2})+i(2 \sqrt{6}-\sqrt{3})}{2 \sqrt{3}}

$\left|z_1-z_2\right|=\left|z_2\right| \Rightarrow \Delta \mathrm{ABO}$ is isosceles with angles $\dfrac{\pi}{6}, \dfrac{\pi}{6} \& \dfrac{2 \pi}{3}$

Q122

Let

A = \left\{ \theta \in [0, 2\pi] : 1 + 10\operatorname{Re}\left( \dfrac{2\cos\theta + i\sin\theta}{\cos\theta - 3i\sin\theta} \right) = 0 \right\}

. Then

\sum\limits_{\theta \in A} \theta^2

is equal to

A

\dfrac{21}{4} \pi^2

B

6\pi^2

C

\dfrac{27}{4} \pi^2

D

8\pi^2

Correct Answer

Option A

Solution

\begin{aligned} & 1+10 \operatorname{Re}\left(\frac{2 \cos \theta+i \sin \theta}{\cos \theta-3 i \sin \theta}\right)=0 \\ & \therefore \mathrm{z}+\overline{\mathrm{z}}=2 \operatorname{Re}(\mathrm{z}) \\ & \frac{2 \cos \theta+\mathrm{i} \sin \theta}{\cos \theta-3 \mathrm{i} \sin \theta}+\frac{2 \cos \theta-\mathrm{i} \sin \theta}{\cos \theta+3 \mathrm{i} \sin \theta}=2 \times\left(\frac{-1}{10}\right) \\ & \frac{\left(2 \cos ^2 \theta-3 \sin ^2 \theta\right)+\left(2 \cos ^2 \theta\right)-\left(3 \sin ^2 \theta\right)}{\cos ^2 \theta+9 \sin ^2 \theta}=\frac{-2}{10} \\ & \Rightarrow \frac{2 \cos ^2 \theta-3 \sin ^2 \theta}{\cos ^2 \theta+9 \sin ^2 \theta}=\frac{-1}{10} \\ & \Rightarrow 20 \cos ^2 \theta-30 \sin ^2 \theta=-\cos ^2 \theta-9 \sin ^2 \theta \\ & 21 \cos ^2 \theta-21 \sin ^2 \theta=0 \\ & \Rightarrow \cos 2 \theta=0 \\ & 2 \theta=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2} \\ & \Rightarrow \sum \theta^2=\frac{\pi^2}{16}+\frac{9 \pi^2}{16}+\frac{25 \pi^2}{16}+\frac{49 \pi^2}{16}=\frac{84 \pi^2}{16}=\frac{21 \pi^2}{4} \end{aligned}

Q123

Let

z

be a complex number such that

|z|=1

. If

\dfrac{2+\mathrm{k}^2 z}{\mathrm{k}+\bar{z}}=\mathrm{k} z, \mathrm{k} \in \mathbf{R}

, then the maximum distance of

\mathrm{k}+i \mathrm{k}^2

from the circle

|z-(1+2 i)|=1

is :

A

\sqrt{5}+1

B 3

C

\sqrt{3}+1

D 2

Correct Answer

Option A

Solution

\begin{aligned} &\begin{aligned} & \frac{2+k^2 z}{k+\bar{z}}=k z \\ & \Rightarrow 2+k^2 z=k^2 z+k \bar{z} \\ & \Rightarrow 2+k|z|^2 \quad\left(z \bar{z}=|z|^2,|z|=1\right) \\ & \Rightarrow 2=k \\ & \therefore k+k^2 i=2+4 i \end{aligned}\\ &\therefore \quad \text{ The maximum distance is }\\ &\begin{aligned} & =\sqrt{(4-2)+(2-1)^2}+\text{ radius } \\ & =\sqrt{(2)^2+(1)^2}+1 \\ & =\sqrt{5}+1 \end{aligned} \end{aligned}

Q124

If the locus of z ∈ ℂ, such that Re

\left( \dfrac{z - 1}{2z + i} \right) + \text{Re} \left( \dfrac{\overline{z} - 1}{2\overline{z} - i} \right) = 2

, is a circle of radius r and center

(a, b)

, then

\dfrac{15ab}{r^2}

is equal to :

A 16

B 24

C 12

D 18

Correct Answer

Option D

Solution

$\operatorname{Re}\left(\dfrac{\mathrm{z}-1}{2 \mathrm{z}+\mathrm{i}}\right)+\operatorname{Re}\left(\dfrac{\overline{\mathrm{z}}-1}{2 \bar{z}-\mathrm{i}}\right)=2$ Here, $\dfrac{\mathrm{z}-1}{2 \mathrm{z}+\mathrm{i}}=\left(\dfrac{\overline{\bar{z}-1}}{2 \overline{\mathrm{z}}-\mathrm{i}}\right)=2$ $=\operatorname{Re}\left(\dfrac{z-1}{2 z+i}\right)+\operatorname{Re}\left(\overline{\dfrac{z-1}{2 z+i}}\right)=2$ $=2 \operatorname{Re}\left(\dfrac{z-1}{2 z+1}\right)=2 \Rightarrow \operatorname{Re}\left(\dfrac{z-1}{2 z+i}\right)=1$

\begin{aligned} & \text{ Let } z=x+i y \\ & \operatorname{Re}\left(\frac{(x-1)+i y}{2 x+i(2 y+1)}\right)=1 \Rightarrow \operatorname{Re}\left[\frac{((x-1)+i y)(2 x-i(y+1)}{(2 x+i(2 y+1)(2 x-i(2 y+1))}\right]=1 \\ & \Rightarrow \frac{2 x(x-1)+y(2 y+1)}{4 x^2+(2 y+1)^2}=1 \\ & \Rightarrow 2 x^2-2 x+2 y^2+y=4 x^2+4 y^2+1+4 y \\ & \Rightarrow 2 x^2+2 y^2+3 y+2 x+1=0 \\ & \Rightarrow x^2+y^2+x+\frac{3}{2} y+\frac{1}{2}=0 \\ & \text{ centre }=\left(\frac{-1}{2}, \frac{-3}{4}\right), r=\sqrt{\frac{1}{4}+\frac{9}{16}-\frac{1}{2}}=\frac{\sqrt{5}}{4} \\ & a=\frac{-1}{2}, b=\frac{-3}{4}, r^2=\frac{5}{16} \end{aligned}

$15 \dfrac{\mathrm{ab}}{\mathrm{r}^2}=15 \times\left(\dfrac{-1}{2}\right) \times\left(\dfrac{-3}{4}\right) \times \dfrac{16}{5}=18$

Q125

Among the statements (S1) : The set

\left\{z \in \mathbb{C}-\{-i\}:|z|=1\right.

and

\dfrac{z-i}{z+i}

is purely real

\}

contains exactly two elements, and (S2) : The set

\left\{z \in \mathbb{C}-\{-1\}:|z|=1\right.

and

\dfrac{z-1}{z+1}

is purely imaginary

\}

contains infinitely many elements.

A both are incorrect

B both are correct

C only (S2) is correct

D only (S1) is correct

Correct Answer

Option C

Solution

\begin{aligned} & \frac{z-i}{z+i}=\frac{\bar{z}+i}{\bar{z}-i} \\ & =z \bar{z}-i \bar{z}-i z-1=z \bar{z}+z i+i \bar{z}-1 \\ & =z+\bar{z}=0 \\ & =2 x=0 \\ & =x=0 \quad \text{ (y-axis) } \end{aligned}

\begin{aligned} & |z|=1 \\ & \therefore \quad z=i \quad(z \neq-i \text{ is given }) \end{aligned}

Statement 1 is incorrect

\begin{aligned} & \frac{z-i}{z+i}+\frac{\bar{z}-1}{\bar{z}+1}=0 \\ & =z \bar{z}-\bar{z}+z-1+z \bar{z}-z+\bar{z}-1=0 \\ & =z \bar{z}=1 \\ & =|z|=1 \end{aligned}

Statement 2 is correct

Q126

Let the product of

\omega_1=(8+i) \sin \theta+(7+4 i) \cos \theta

and

\omega_2=(1+8 i) \sin \theta+(4+7 i) \cos \theta

be

\alpha+i \beta

,

i=\sqrt{-1}

. Let p and q be the maximum and the minimum values of

\alpha+\beta

respectively. Then

\mathrm{p}+\mathrm{q}

is equal to :

A 130

B 150

C 160

D 140

Correct Answer

Option A

Solution

\begin{aligned} & \omega_1=(8 \sin \theta+7 \cos \theta)+i(\sin \theta+4 \cos \theta) \\ & \omega_2=(\sin \theta+4 \cos \theta)+i(8 \sin \theta+7 \cos \theta) \\ & \alpha=(8 \sin \theta+7 \cos \theta)+(\sin \theta+4 \cos \theta) \\ & \quad-(\sin \theta+4 \cos \theta)+(8 \sin \theta+7 \cos \theta)=0 \\ & \beta=(8 \sin \theta+7 \cos \theta)^2+(\sin \theta+4 \cos \theta)^2 \end{aligned}

\begin{aligned} & =65 \sin ^2 \theta+65 \cos ^2 \theta+56 \sin 2 \theta+4 \sin 2 \theta \\ & =65+60 \sin 2 \theta \\ & (\alpha+\beta)_{\max }=125=p \\ & (\alpha+\beta)_{\min }=5=q \\ & p+q=130 \end{aligned}

Q127

If\,\,{z_1},{z_2},{z_3} \in \,\,are\,\,the\,\,vertices\,\,of\,\,an\,\,equilateral\,\,triangle,\,\,whose\,\,centroid\,\,is\,\,{z_0},\,\,then\,\,\sum\limits_{k = 1}^3 {{{\left( {{z_k} - {z_0}} \right)}^2}\,is\,\,equal\,\,to}

A 0

B 1

C i

D -i

Correct Answer

Option A

Solution

\begin{aligned} & z_0=z_1+z_2+z_3 \\ & \sum_{k=1}^3\left(z_k-z_0\right)^2=\left(z_1-z_0\right)^2+\left(z_2-z_0\right)^2+\left(z_3-z_0\right)^2 \end{aligned}

Let $z_0$ is origin $\Rightarrow z_1, z_2, z_3$ lies on a circle having $\left|z_0-z_i\right|=R$

\begin{aligned} & \therefore z_1=R e^{i 2 \pi / 3} z_2=R e^{i 4 \pi / 3} z_3=R e^{i 6 \pi / 3} \\ & \Rightarrow z_1^2+z_2^2+z_3^2=R^2\left[e^{i 4 \pi / 3}+e^{i 8 \pi / 3}+e^{i 12 \pi / 3}\right] \\ & =0 \\ & \therefore \sum_{k=1}^3\left(z_k-z_0\right)^2=0 \end{aligned}

Q128

Let A =

\left\{ {\theta \in \left( { - {\pi \over 2},\pi } \right):{{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}is\,purely\,imaginary} \right\}

. Then the sum of the elements in A is :

A

{5\pi \over 6}

B

\pi

C

{3\pi \over 4}

D

{{2\pi } \over 3}

Correct Answer

Option D

Solution

Given complex number,

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

= {{\left( {3 + 2i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}

= {{3 + 6i\sin \theta + 2i\sin \theta - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}

= {{\left( {3 - 4{{\sin }^2}\theta } \right) + i\left( {8\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}

As complex number is purely imaginary, So real part of this complex number is zero. $\therefore$

{{3 - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}

= 0 $\Rightarrow$

3 - 4{\sin ^2}\theta = 0

$\Rightarrow$

\sin \theta = \pm {{\sqrt 3 } \over 2}

as $\theta$

\in

\left( { - {\pi \over 2},\pi } \right)

$\therefore$ $\theta$ $=$ $-$

{\pi \over 3},{\pi \over 3},{{2\pi } \over 3}

$\therefore$ Sum of those values of A is

= - {\pi \over 3} + {\pi \over 3} + {{2\pi } \over 3}

= {{2\pi } \over 3}

Q129

Let

S_{1}=\left\{z_{1} \in \mathbf{C}:\left|z_{1}-3\right|=\frac{1}{2}\right\}

and

S_{2}=\left\{z_{2} \in \mathbf{C}:\left|z_{2}-\right| z_{2}+1||=\left|z_{2}+\right| z_{2}-1||\right\}

. Then, for

z_{1} \in S_{1}

and

z_{2} \in S_{2}

, the least value of

\left|z_{2}-z_{1}\right|

is :

A 0

B

\dfrac{1}{2}

C

\dfrac{3}{2}

D

\dfrac{5}{2}

Correct Answer

Option C

Solution

$\because$

{\left| {{Z_2} + |{Z_2} - 1|} \right|^2} = {\left| {{Z_2} - |{Z_2} + 1|} \right|^2}

\Rightarrow \left( {{Z_2} + |{Z_2} - 1|} \right)\left( {{{\overline Z }_2} + |{Z_2} - 1|} \right) = \left( {{Z_2} - |{Z_2} + 1|} \right)\left( {{{\overline Z }_2} - |{Z_2} + 1|} \right)

\Rightarrow {Z_2}\left( {|{Z_2} - 1| + |{Z_2} + 1|} \right) + {\overline Z _2}\left( {|{Z_2} - 1| + |{Z_2} + 1|} \right) = |{Z_2} + 1{|^2} - |{Z_2} - 1{|^2}

\Rightarrow \left( {{Z_2} + {{\overline Z }_2}} \right)\left( {|{Z_2} + 1| + |{Z_2} - 1|} \right) = 2\left( {{Z_2} + {{\overline Z }_2}} \right)

$\Rightarrow$ Either

{Z_2} + {\overline Z _2} = 0

or

|{Z_2} + 1| + |{Z_2} - 1| = 2

So, Z2 lies on imaginary axis or on real axis within

[ - 1,1]

Also

|{Z_1} - 3| = {1 \over 2} \Rightarrow {Z_1}

lies on the circle having center 3 and radius

{1 \over 2}

. Clearly

|{Z_1} - {Z_2}{|_{\min }} = {3 \over 2}

Q130

The value of

{\left( {{{1 + \sin {{2\pi } \over 9} + i\cos {{2\pi } \over 9}} \over {1 + \sin {{2\pi } \over 9} - i\cos {{2\pi } \over 9}}}} \right)^3}

is :

A

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

B -

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

C

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

D

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

Correct Answer

Option B

Solution

{\left( {{{1 + \sin {{2\pi } \over 9} + i\cos {{2\pi } \over 9}} \over {1 + \sin {{2\pi } \over 9} - i\cos {{2\pi } \over 9}}}} \right)^3}

=

{\left( {{{1 + \sin \left( {{\pi \over 2} - {{5\pi } \over {18}}} \right) + i\cos \left( {{\pi \over 2} - {{5\pi } \over {18}}} \right)} \over {1 + \sin \left( {{\pi \over 2} - {{5\pi } \over {18}}} \right) - i\cos \left( {{\pi \over 2} - {{5\pi } \over {18}}} \right)}}} \right)^3}

=

{\left( {{{1 + \cos \left( {{{5\pi } \over {18}}} \right) + i\sin \left( {{{5\pi } \over {18}}} \right)} \over {1 + \cos \left( {{{5\pi } \over {18}}} \right) - i\sin \left( {{{5\pi } \over {18}}} \right)}}} \right)^3}

=

{\left( {{{2{{\cos }^2}\left( {{{5\pi } \over {36}}} \right) + 2i\sin \left( {{{5\pi } \over {36}}} \right)\cos \left( {{{5\pi } \over {36}}} \right)} \over {2{{\cos }^2}\left( {{{5\pi } \over {36}}} \right) - 2i\sin \left( {{{5\pi } \over {36}}} \right)\cos \left( {{{5\pi } \over {36}}} \right)}}} \right)^3}

=

{\left( {{{\cos \left( {{{5\pi } \over {36}}} \right) + i\sin \left( {{{5\pi } \over {36}}} \right)} \over {\cos \left( {{{5\pi } \over {36}}} \right) - i\sin \left( {{{5\pi } \over {36}}} \right)}}} \right)^3}

=

{\left( {{{{e^{i{{5\pi } \over {36}}}}} \over {{e^{ - i{{5\pi } \over {36}}}}}}} \right)^3}

=

{\left( {{e^{i{{5\pi } \over {18}}}}} \right)^3}

=

{{e^{i{{5\pi } \over {18}} \times 3}}}

=

{{e^{i{{5\pi } \over 6}}}}

=

\cos {{5\pi } \over 6} + i\sin {{5\pi } \over 6}

=

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