The complex number $z=\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}$ is equal to :

$\mathrm{Z}=\frac{\mathrm{i}-1}{\cos \frac{\pi}{3}+\mathrm{i} \sin \frac{\pi}{3}}=\frac{\mathrm{i}-1}{\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}}$ $=\frac{i-1}{\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}} \times \frac{\frac{1}{2}-\sqrt{\frac{3}{2}} \mathrm{i}}{\frac{1}{2}-\sqrt{3 / 2} \mathrm{i}}=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{3}+1}{2} \mathrm{i}$ Apply polar form, $r \cos \theta=\frac{\sqrt{3}-1}{2}$ $r \sin \theta=\frac{\sqrt{3}+1}{2}$ Now, $\tan \theta=\frac{\sqrt{3}+1}{\sqrt{3}-1}$ So, $ \the

For all $$z \in C$$ on the curve $$C_{1}:|z|=4$$, let the locus of the point $$z+\frac{1}{z}$$ be the curve $$\mathrm{C}_{2}$$. Then :

Let $\mathrm{w}=\mathrm{z}+\frac{1}{\mathrm{z}}=4 \mathrm{e}^{\mathrm{i} \theta}+\frac{1}{4} \mathrm{e}^{-\mathrm{i} \theta}$ $\Rightarrow \mathrm{w}=\frac{17}{4} \cos \theta+\mathrm{i} \frac{15}{4} \sin \theta$ So locus of $w$ is ellipse $\frac{x^{2}}{\left(\frac{17}{4}\right)^{2}}+\frac{y^{2}}{\left(\frac{15}{4}\right)^{2}}=1$ Locus of $\mathrm{z}$ is circle $\mathrm{x}^{2}+\mathrm{y}^{2}=16$ So intersect at 4 points.

Complex Numbers — Page 9 | JEE Mathematics

Q81

The complex number

z=\dfrac{i-1}{\cos \dfrac{\pi}{3}+i \sin \dfrac{\pi}{3}}

is equal to :

A

\cos \dfrac{\pi}{12}-i \sin \dfrac{\pi}{12}

B

\sqrt{2}\left(\cos \dfrac{\pi}{12}+i \sin \dfrac{\pi}{12}\right)

C

\sqrt{2} i\left(\cos \dfrac{5 \pi}{12}-i \sin \dfrac{5 \pi}{12}\right)

D

\sqrt{2}\left(\cos \dfrac{5 \pi}{12}+i \sin \dfrac{5 \pi}{12}\right)

Correct Answer

Option D

Solution

$\mathrm{Z}=\dfrac{\mathrm{i}-1}{\cos \dfrac{\pi}{3}+\mathrm{i} \sin \dfrac{\pi}{3}}=\dfrac{\mathrm{i}-1}{\dfrac{1}{2}+\dfrac{\sqrt{3}}{2} \mathrm{i}}$ $=\dfrac{i-1}{\dfrac{1}{2}+\dfrac{\sqrt{3}}{2} \mathrm{i}} \times \dfrac{\dfrac{1}{2}-\sqrt{\dfrac{3}{2}} \mathrm{i}}{\dfrac{1}{2}-\sqrt{3 / 2} \mathrm{i}}=\dfrac{\sqrt{3}-1}{2}+\dfrac{\sqrt{3}+1}{2} \mathrm{i}$ Apply polar form, $r \cos \theta=\dfrac{\sqrt{3}-1}{2}$ $r \sin \theta=\dfrac{\sqrt{3}+1}{2}$ Now, $\tan \theta=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}$ So, $\theta=\dfrac{5 \pi}{12}$

Q82

If the center and radius of the circle

\left| {{{z - 2} \over {z - 3}}} \right| = 2

are respectively

(\alpha,\beta)

and

\gamma

, then

3(\alpha+\beta+\gamma)

is equal to :

A 12

B 10

C 11

D 9

Correct Answer

Option A

Solution

\left| {{{z - 2} \over {z - 3}}} \right| = 2

\begin{aligned} & \sqrt{(x-2)^2+y^2}=2 \sqrt{(x-3)^2+y^2} \\\\ & \Rightarrow x^2+y^2-4 x+4=4 x^2+4 y^2-24 x+36 \\\\ & \Rightarrow 3 x^2+3 y^2-20 x+32=0 \\\\ & \Rightarrow x^2+y^2-\frac{20}{3} \mathrm{x}+\frac{32}{3}=0 \\\\ & \Rightarrow (\alpha, \beta)=\left(\frac{10}{3}, 0\right) \\\\ & \gamma=\sqrt{\frac{100}{9}-\frac{32}{3}}=\sqrt{\frac{4}{9}}=\frac{2}{3} \\\\ & 3(\alpha + \beta + \gamma)=3\left(\frac{10}{3}+\frac{2}{3}\right) \\\\ & =12 \end{aligned}

Q83

For all

z \in C

on the curve

C_{1}:|z|=4

, let the locus of the point

z+\dfrac{1}{z}

be the curve

\mathrm{C}_{2}

. Then :

A the curves

C_{1}

and

C_{2}

intersect at 4 points

B the curve

C_{2}

lies inside

C_{1}

C the curve

C_{1}

lies inside

C_{2}

D the curves

C_{1}

and

C_{2}

intersect at 2 points

Correct Answer

Option A

Solution

Let $\mathrm{w}=\mathrm{z}+\dfrac{1}{\mathrm{z}}=4 \mathrm{e}^{\mathrm{i} \theta}+\dfrac{1}{4} \mathrm{e}^{-\mathrm{i} \theta}$ $\Rightarrow \mathrm{w}=\dfrac{17}{4} \cos \theta+\mathrm{i} \dfrac{15}{4} \sin \theta$ So locus of $w$ is ellipse $\dfrac{x^{2}}{\left(\dfrac{17}{4}\right)^{2}}+\dfrac{y^{2}}{\left(\dfrac{15}{4}\right)^{2}}=1$ Locus of $\mathrm{z}$ is circle $\mathrm{x}^{2}+\mathrm{y}^{2}=16$ So intersect at 4 points.

Q84

For two non-zero complex numbers

z_{1}

and

z_{2}

, if

\operatorname{Re}\left(z_{1} z_{2}\right)=0

and

\operatorname{Re}\left(z_{1}+z_{2}\right)=0

, then which of the following are possible? A.

\operatorname{Im}\left(z_{1}\right)>0

and

\operatorname{Im}\left(z_{2}\right) > 0

B.

\operatorname{Im}\left(z_{1}\right) 0

C.

\operatorname{Im}\left(z_{1}\right) > 0

and

\operatorname{Im}\left(z_{2}\right) D.

\operatorname{Im}\left(z_{1}\right) Choose the correct answer from the options given below :

A A and C

B A and B

C B and D

D B and C

Correct Answer

Option D

Solution

Let,

{z_1} = {x_1} + i{y_1}

and

{z_2} = {x_2} + i{y_2}

$\therefore$

{z_1}{z_2} = {x_1}{x_2} - {y_1}{y_2} + i({x_1}{y_2} + {x_2}{y_1})

Given,

{\mathop{\rm Re}\nolimits} ({z_1} + {z_2}) = 0

\Rightarrow {x_1} + {x_2} = 0

...... (1) also given,

{\mathop{\rm Re}\nolimits} ({z_1}{z_2}) = 0

\Rightarrow {x_1}{x_2} - {y_1}{y_2} = 0

\Rightarrow {x_1}{x_2} = {y_1}{y_2}

\Rightarrow {y_1}{y_2} = - x_1^2

[ $\because$

{x_2} = - {x_1}

] So, multiplication of imaginary part's of z1 and z2 is negative.

It means sign of y1 and y2 are opposite of each other.

$\therefore$ B and C are correct.

Q85

If

z=x+i y, x y \neq 0

, satisfies the equation

z^2+i \bar{z}=0

, then

\left|z^2\right|

is equal to :

A 9

B

\dfrac{1}{4}

C 4

D 1

Correct Answer

Option D

Solution

\begin{aligned} & z^2=-i \bar{z} \\ & \left|z^2\right|=|i \bar{z}| \\ & \left|z^2\right|=|z| \\ & |z|^2-|z|=0 \\ & |z|(|z|-1)=0 \\ & |z|=0 \text{ (not acceptable) } \\ & \therefore|z|=1 \\ & \therefore|z|^2=1 \end{aligned}

Q86

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( {1 - i\sqrt 3 } \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( {\sqrt 3 + i} \right)

Correct Answer

Option B

Solution

$z=\left(\dfrac{1+\sin \dfrac{2 \pi}{9}+i \cos \dfrac{2 \pi}{9}}{1+\sin \dfrac{2 \pi}{9}-i \cos \dfrac{2 \pi}{9}}\right)^{3}$

\begin{aligned} & 1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}=1+\cos \frac{5 \pi}{18}+i \sin \frac{5 \pi}{18} \\\\ & =1+2 \cos ^{2} \frac{5 \pi}{36}-1+2 i \sin \frac{5 \pi}{36} \cos \frac{5 \pi}{36} \\\\ & =2 \cos \frac{5 \pi}{36}\left(\cos \frac{5 \pi}{36}+i \sin \frac{5 \pi}{36}\right)=2 \cos \frac{5 \pi}{36} e^{i \frac{5 \pi}{36}} \\\\ & z=-\frac{\sqrt{3}}{2}+\frac{1}{2} i=\frac{1}{2}(i-\sqrt{3})=-\frac{1}{2}(\sqrt{3}-i) \end{aligned}

Q87

Let

\mathrm{p,q\in\mathbb{R}}

and

{\left( {1 - \sqrt 3 i} \right)^{200}} = {2^{199}}(p + iq),i = \sqrt { - 1}

then

\mathrm{p+q+q^2}

and

\mathrm{p-q+q^2}

are roots of the equation.

A

{x^2} + 4x - 1 = 0

B

{x^2} - 4x + 1 = 0

C

{x^2} + 4x + 1 = 0

D

{x^2} - 4x - 1 = 0

Correct Answer

Option B

Solution

{\left( {1 - \sqrt 3 i} \right)^{200}}

= {\left[ {2\left( {{1 \over 2} - {{\sqrt 3 } \over 2}i} \right)} \right]^{200}}

= {2^{200}}{\left( {\cos {\pi \over 3} - i\sin {\pi \over 3}} \right)^{200}}

= {2^{200}}\left( {\cos {{200\pi } \over 3} - i\sin {{200\pi } \over 3}} \right)

= {2^{200}}\left( {\cos \left( {66\pi + {{2\pi } \over 3}} \right) - i\sin \left( {66\theta + {{2\pi } \over 3}} \right)} \right)

= {2^{200}}\left( {\cos {{2\pi } \over 3} - i\sin {{2\pi } \over 3}} \right)

= {2^{200}}\left( { - {1 \over 2} - {{\sqrt 3 } \over 2}i} \right)

= {2^{199}}\left( { - 1 - \sqrt 3 i} \right)

= {2^{199}}\left( {p + iq} \right)

$\therefore$ p = $-$ 1 and q =

- \sqrt3

Now,

p - q + {q^2} = - 1 + \sqrt 3 + 3 = 2 + \sqrt 3 = \alpha

and

p + q + {q^2} = - 1 - \sqrt 3 + 3 = 2 - \sqrt 3 = \beta

$\therefore$

\alpha + \beta = 4

\alpha \beta = \left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right) = 4 - 3 = 1

$\therefore$ Quadratic equation is

{x^2} - (\alpha + \beta )x + \alpha \beta = 0

\Rightarrow {x^2} - 4x + 1 = 0

Q88

If the set

\left\{\operatorname{Re}\left(\dfrac{z-\bar{z}+z \bar{z}}{2-3 z+5 \bar{z}}\right): z \in \mathbb{C}, \operatorname{Re}(z)=3\right\}

is equal to the interval

(\alpha, \beta]

, then

24(\beta-\alpha)

is equal to :

A 36

B 27

C 42

D 30

Correct Answer

Option D

Solution

Let $z_1=\left(\dfrac{z-\bar{z}+z \bar{z}}{2-3 z+5 \bar{z}}\right)$ Let $\mathrm{z}=3+\mathrm{iy}$ $\bar{z}=3-i y$

\begin{aligned} & z_1=\frac{2 i y+\left(9+y^2\right)}{2-3(3+i y)+5(3-i y)} \\\\ & =\frac{9+y^2+i(2 y)}{8-8 i y} \\\\ & =\frac{\left(9+y^2\right)+i(2 y)}{8(1-i y)} \\\\ & \operatorname{Re}\left(z_1\right)=\frac{\left(9+y^2\right)-2 y^2}{8\left(1+y^2\right)} \end{aligned}

\begin{aligned} & =\frac{9-y^2}{8\left(1+y^2\right)} \\\\ & =\frac{1}{8}\left[\frac{10-\left(1+y^2\right)}{\left(1+y^2\right)}\right] \\\\ & =\frac{1}{8}\left[\frac{10}{1+y^2}-1\right] \\\\ & 1+y^2 \in[1, \infty] \\\\ & \frac{1}{1+y^2} \in(0,1] \\\\ & \frac{10}{1+y^2} \in(0,10] \\\\ & \frac{10}{1+y^2}-1 \in(-1,9] \\\\ & \operatorname{Re}\left(\mathrm{z}_1\right) \in\left(\frac{-1}{8}, \frac{9}{8}\right] \\\\ & \alpha=\frac{-1}{8}, \beta=\frac{9}{8} \\\\ & 24(\beta-\alpha)=24\left(\frac{9}{8}+\frac{1}{8}\right)=30 \end{aligned}

Q89

Let

S=\left\{z \in \mathbb{C}: \bar{z}=i\left(z^{2}+\operatorname{Re}(\bar{z})\right)\right\}

. Then

\sum\limits_{z \in \mathrm{S}}|z|^{2}

is equal to :

A

\dfrac{7}{2}

B 4

C 3

D

\dfrac{5}{2}

Correct Answer

Option B

Solution

Let $z=x+i y$

\begin{aligned} &\bar{z}=i\left(z^2+\operatorname{Re}(\bar{z})\right) \\\\ &\Rightarrow x-i y=i\left(x^2-y^2+2 i x y+x\right) \\\\ & x-i y=i\left(x^2-y^2+x\right)-2 x y \\\\ &x=-2 x y \Rightarrow x(2 y+1)=0 \\\\ &\Rightarrow x=0, y=\frac{-1}{2} ........(i)\\\\ &-y=x^2-y^2+x ...........(ii) \end{aligned}

Case (I) $x=0$ $\Rightarrow-y=-y^2 \Rightarrow y^2-y=0 \Rightarrow y=0,1$ $\therefore$ $z=0, i$ Case (II) $y=\dfrac{-1}{2}$ $\Rightarrow \dfrac{1}{2}=x^2-\dfrac{1}{4}+x \Rightarrow x^2+x-\dfrac{3}{4}=0$ $\Rightarrow$ $4 x+4 x-3=0 \Rightarrow(2 x-1)(2 x+3)=0$ $\Rightarrow$ $x=\dfrac{1}{2}, \dfrac{-3}{2}$

\begin{aligned} & z=\frac{1}{2}-\frac{1}{2} i, \frac{-3}{2}-\frac{1}{2} i \\\\ & \sum|z|^2=0+1+\frac{1}{2}+\frac{5}{2}=4 \end{aligned}

Q90

Let

\mathrm{C}

be the circle in the complex plane with centre

\mathrm{z}_{0}=\dfrac{1}{2}(1+3 i)

and radius

r=1

. Let

\mathrm{z}_{1}=1+\mathrm{i}

and the complex number

z_{2}

be outside the circle

C

such that

\left|z_{1}-z_{0}\right|\left|z_{2}-z_{0}\right|=1

. If

z_{0}, z_{1}

and

z_{2}

are collinear, then the smaller value of

\left|z_{2}\right|^{2}

is equal to :

A

\dfrac{3}{2}

B

\dfrac{5}{2}

C

\dfrac{13}{2}

D

\dfrac{7}{2}

Correct Answer

Option B

Solution

Given,

z_0=\frac{1+3 i}{2}, z_1=(1+i)

\left|z_1-z_0\right|=\left|\frac{1-i}{2}\right|=\frac{1}{\sqrt{2}}

\begin{aligned} & \left|z_1-z_0\right|\left|z_2-z_0\right|=1 \\\\ & \Rightarrow \frac{1}{\sqrt{2}}\left|z_2-z_0\right|=1 \\\\ & \Rightarrow\left|z_2-z_0\right|=\sqrt{2} \end{aligned}

\begin{aligned} & \frac{z_2-z_0}{z_1-z_0}=\frac{\left|z_2-z_0\right|}{\left|z_1-z_0\right|}( \pm 1)= \pm 2 \\\\ & z_2=z_0 \pm 2\left(z_1-z_0\right) \end{aligned}

z_2=2 z_1-z_0=\frac{3}{2}+\frac{1}{2} i \Rightarrow\left|z_2\right|^2=\frac{5}{2}

OR

z_2=3 z_0-2 z_1=\frac{-1}{2}+\frac{5}{2} i \Rightarrow\left|z_2\right|^2=\frac{13}{2}