MCQ Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 8

The locus of

$z$ such that

$\left| {\dfrac{{z + i}}{{z - 1}}} \right| = 2$

0%
straight line
0%
circle with radius $2$
0%
circle with radius $\frac{{2\sqrt 2 }}{3}$
0%
none of these

Explanation

$\left|\dfrac{z+i}{z-1}\right|=2$

Put,

$z = x + iy$

$\therefore\,\left|\dfrac{x + iy+i}{x + iy-1}\right|=2$

$\Rightarrow\,\left|\dfrac{x + i\left(y+1\right)}{\left(x-1\right)+iy}\right|=2$

$\Rightarrow\,\left|x + i\left(y+1\right)\right|=2\left|\left(x-1\right)+iy\right|$

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

Squaring both sides, we get

$\Rightarrow\,{x}^{2}+{\left(y+1\right)}^{2}=4{\left(x-1\right)}^{2}+4{y}^{2}$

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

$\Rightarrow\,\left(x-2x+2\right)\left(x+2x-2\right)=\left(2y-y-1\right)\left(2y+y+1\right)$

$\Rightarrow\,\left(-x+2\right)\left(3x-2\right)=\left(y-1\right)\left(3y+1\right)$

$\Rightarrow\,-3{x}^{2}+2x+6x-4=3{y}^{2}+y-3y-1$

$\Rightarrow\,-3{x}^{2}+8x-4=3{y}^{2}-2y-1$

$\Rightarrow\,-3{x}^{2}+8x-4-3{y}^{2}+2y+1=0$

$\Rightarrow\,3{x}^{2}+3{y}^{2}-8x-2y+3=0$

$\Rightarrow\,{x}^{2}+{y}^{2}-\dfrac{8}{3}x-\dfrac{2}{3}y+1=0$ is of the form

${x}^{2}+{y}^{2}+2gx+2fy+c=0$

where

$g=\dfrac{4}{3},\,f=\dfrac{1}{3}$ and

$c=1$

Radius

$=r=\sqrt{{\left(\dfrac{4}{3}\right)}^{2}+{\left(\dfrac{1}{3}\right)}^{2}+1}=\sqrt{\dfrac{16}{9}+\dfrac{1}{9}-1}=\sqrt{\dfrac{17-9}{9}}=\dfrac{\sqrt{8}}{3}=\dfrac{2\sqrt{2}}{3}$

The locus of

$z$ is a circle with radius

$\dfrac{2\sqrt{2}}{3}$

$\left(\dfrac{1 + i}{1 - i}\right)^4 + \left(\dfrac{1 - i}{1 + i}\right)^4 =$

0%
$0$
0%
$1$
0%
$2$
0%
$4$

Explanation

$\left( \dfrac{1+i}{1-i} \right)^{4}+ \left( \dfrac{1-i}{1+i} \right)^{4}$

$= \left( \dfrac{(1+i)^{2}}{1-i^{2}} \right)^{4}+ \left( \dfrac{(1-i)^{2}}{1-i^{2}} \right)^{4}$

$=\dfrac{1}{2^{4}} (1+i^{2} +2 i)^{4}+ \dfrac{1}{2^{4}} (1+i^{2}-2i)^{4}$

$=\dfrac{1}{2^{4}} (1-1+ 2i)^{4}+ \dfrac{1}{2^{4}} (1-1-2i)^{4}$

$=\dfrac{2^{4}}{2^{4}} (i^{4})+ \dfrac{2^{4}}{2^{4}} (i)^{4}$

$= 1+1 =2$

$|z_1+z_2|=|z_1|+|z_2|$ where

$z_1$ and

$z_2$ are different non - zero complex number, then ?

0%
$Re\left(\dfrac{z_1}{z_2}\right)=0$
0%
$Im\left(\dfrac{z_1}{z_2}\right)=0$
0%
$z_1+z_2=0$
0%
None

Explanation

$let\quad { z }_{ 1 }=a+ib\quad \quad and\quad \quad { z }_{ 2 }=c+id\\ \Rightarrow \quad \left| \quad { z }_{ 1 }+\quad { z }_{ 2 } \right| \quad \quad \quad \quad =\quad \quad \quad \left| \quad { z }_{ 1 } \right| +\left| \quad { z }_{ 2 } \right| \\ \Rightarrow \quad \quad \left| a+ib+c+id \right| \quad =\quad \quad \left| a+ib \right| +\left| c+id \right| \\ \Rightarrow \left| a+c+i\left( b+d \right) \right| \quad \quad \quad =\quad \quad \quad \left| a+ib \right| +\left| c+id \right| \\ \Rightarrow \sqrt { \left( a+c \right) ^{ 2 }+\left( b+d \right) ^{ 2 } } =\quad \quad \sqrt { { a }^{ 2 }+{ b }^{ 2 } } +\sqrt { { c }^{ 2 }+d^{ 2 } } \\ \quad \quad \quad \quad squaring\quad both\quad sides\\ \Rightarrow \left( a+c \right) ^{ 2 }+\left( b+d \right) ^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+d^{ 2 }+2\sqrt { { a }^{ 2 }+{ b }^{ 2 } } .\sqrt { { c }^{ 2 }+d^{ 2 } } \\ \Rightarrow 2ac+2bd\quad \quad =\quad \quad 2\sqrt { { a }^{ 2 }+{ b }^{ 2 } } .\sqrt { { c }^{ 2 }+d^{ 2 } } \\ \Rightarrow ac+bd\quad \quad =\quad \quad \sqrt { { a }^{ 2 }+{ b }^{ 2 } } .\sqrt { { c }^{ 2 }+d^{ 2 } } \\ \quad \quad \quad again\quad squaring\quad both\quad sides\\ \Rightarrow { \left( ac \right) }^{ 2 }+{ \left( bd \right) }^{ 2 }+2abcd\quad =\quad \left( { a }^{ 2 }+{ b }^{ 2 } \right) \left( { c }^{ 2 }+d^{ 2 } \right) \\ \Rightarrow { \left( ac \right) }^{ 2 }+{ \left( bd \right) }^{ 2 }+2abcd\quad =\quad { \left( ac \right) }^{ 2 }+{ \left( ad \right) }^{ 2 }+{ \left( bc \right) }^{ 2 }+{ \left( bd \right) }^{ 2 }\\ \Rightarrow { \left( ad \right) }^{ 2 }+{ \left( bc \right) }^{ 2 }-2abcd\quad =\quad 0\\ \Rightarrow \left( ad-bc \right) ^{ 2 }\quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 0\\ \Rightarrow ad-bc\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 0\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ad\quad \quad =\quad \quad \quad bc\quad \quad \quad \quad \longrightarrow \left( 1 \right) \\ Now\quad \quad \dfrac { { z }_{ 1 } }{ { z }_{ 2 } } \quad =\quad \dfrac { a+ib }{ c+id } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \dfrac { \left( a+ib \right) \left( c-id \right) }{ \left( c+id \right) \left( c-id \right) } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \dfrac { ac-iad+ibc-{ i }^{ 2 }bd }{ { c }^{ 2 }+{ d }^{ 2 } } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \dfrac { ac+bd+i\left( bc-ad \right) }{ { c }^{ 2 }+{ d }^{ 2 } } \quad \quad \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \dfrac { ac+bd }{ { c }^{ 2 }+{ d }^{ 2 } } \quad +\quad 0\quad \quad \quad using\quad equation\quad \left( 1 \right) \\ Imaginary\quad part\quad is\quad zero\quad as\quad there\quad is\quad only\quad real\quad part\\ \therefore \quad Im\left( \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right) =\quad 0$

Modulus of

$\dfrac{\cos \theta - i\sin \theta}{\sin \theta - i \cos \theta}$ is

0%
$0$
0%
$2\theta$
0%
$\pi - 2\theta$
0%
None of these

Explanation

$\cfrac { \cos { \theta } -i\sin { \theta } }{ \sin { \theta } -i\cos { \theta } }$

$\cfrac { \cos { \theta } -i\sin { \theta } }{ \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } } \left( \sin { \theta } +i\cos { \theta } \right) \Rightarrow \left( \cos { \theta } -i\sin { \theta } \right) \left( \sin { \theta } +i\cos { \theta } \right)$

$\Rightarrow \cos { \theta } .\sin { \theta } +\sin { \theta } .\cos { \theta } +i\cos ^{ 2 }{ \theta } -i\sin ^{ 2 }{ \theta } \Rightarrow 2\sin { \theta } .\cos { \theta } +i\left( \cos ^{ 2 }{ \theta } -\sin ^{ 2 }{ \theta } \right)$

$2\sin { \theta } .\cos { \theta } =\sin { 2\theta } ;\cos ^{ 2 }{ \theta } -\sin ^{ 2 }{ \theta } =\cos { 2\theta } \Rightarrow \sin { 2\theta } +i\cos { 2\theta }$

modulus

$=\sqrt { \sin ^{ 2 }{ 2\theta } +\cos ^{ 2 }{ 2\theta } } =\sqrt { 1 } =1$

none of these

$\cfrac{\pi}{3}$ and

$\cfrac{\pi}{4}$ are the arguments of

${z}_{1}$ and

${\overline { z } }_{ 2 }$ , then the value of arg

$({z}_{1}{z}_{2})$ is

0%
$\cfrac{5\pi}{12}$
0%
$\cfrac{\pi}{12}$
0%
$\cfrac{7\pi}{12}$
0%
None of these

Explanation

$arg(z_1.z_2)=arg.(z_1)+arg(z_2)$

$=\dfrac{\pi}{3}+\left(\dfrac{-\pi}{4}\right)$

$\dfrac{\pi}{12}$

$n\in N,\ { \left( \dfrac { 1+i }{ \sqrt { 2 } } \right) }^{ 8n }+{ \left( \dfrac { 1-i }{ \sqrt { 2 } } \right) }^{ 8n }=$

0%
$0$
0%
$1$
0%
$2$
0%
$-2$

Explanation

We have,

${{\left( \dfrac{1+i}{\sqrt{2}} \right)}^{8n}}+{{\left( \dfrac{1-i}{\sqrt{2}} \right)}^{8n}}$

${{\left[ {{\left( \dfrac{1+i}{\sqrt{2}} \right)}^{2}} \right]}^{4n}}+{{\left[ {{\left( \dfrac{1-i}{\sqrt{2}} \right)}^{2}} \right]}^{4n}}$

$={{\left[ \dfrac{{{1}^{2}}+{{i}^{2}}+2i}{2} \right]}^{4n}}+{{\left[ \dfrac{{{1}^{2}}+{{i}^{2}}-2i}{2} \right]}^{4n}}$

$={{\left[ \dfrac{1-1+2i}{2} \right]}^{4n}}+{{\left[ \dfrac{1-1-2i}{2} \right]}^{4n}}$

$={{\left( {{i}^{4}} \right)}^{n}}+{{\left( {{\left( -i \right)}^{4}} \right)}^{n}}$

$={{1}^{n}}+{{1}^{n}}$

$=1+1$

$=2$

Hence, this is the answer.

The roots of the equation

${x}^{\sqrt{x}}={(\sqrt{x})}^{x}$ are

0%
$0$ and $1$
0%
$0$ and $4$
0%
$1$ and $4$
0%
$0,1$ and $4$

Explanation

$x^{\sqrt{x}}=(\sqrt{x})^{x}$

$x^{\sqrt{x}}=x^{x/2}$

$\Rightarrow \sqrt{x}=\dfrac{x}{2}$ or

$x=1$

$\Rightarrow 4x=x^{2}$

$\Rightarrow (x)(x-4)=0$

$\therefore x=4$ or

$x=0$

$\therefore$ But

$x\neq 0$ because of definition of exponent

$\therefore x= 1$ or

$4$

$\left(\dfrac{1+\cos \dfrac{\pi}{8}+i\sin \dfrac{\pi}{8}}{1+\cos \dfrac{\pi}{8}-i\sin \dfrac{\pi}{8}}\right)^{8}=$ ?

0%
$1+i$
0%
$1-i$
0%
$1$
0%
$-1$

The complex number

$z$ satisfies

$z+|z|=2+8i$ . The value of

$|z|$ is

0%
$10$
0%
$13$
0%
$17$
0%
$23$

Explanation

$Assume,z=a+ib,where \ i=\sqrt { -1 }$

$Then,z+|z|=a+ib+\sqrt { { a }^{ 2 }+{ b }^{ 2 } } =2+8i$

$Thus,b=8.$

$And\quad a+\sqrt { { a }^{ 2 }+64 } =2$

${ a }^{ 2 }+64={ (2−a) }^{ 2 }$

${ a }^{ 2 }+64=4−4a+{ a }^{ 2 }$

$a=−15$

$Thus\quad z=−15+8i$

$Thus|z|=\sqrt { { (-15) }^{ 2 }+{ 8 }^{ 2 } } =17$

For a complex number $z$ , the minimum value of $\left| z \right| + \left| {z - 1} \right|$ is

0%
1
0%
2
0%
3
0%
none of these

The value of $\sum\limits_{n = 1}^{13} {\left( {{i^n} + {i^{n + 1}}} \right)}$ , where $i = \sqrt { - 1}$ equals:

0%
$i$
0%
$i - 1$
0%
$- i$
0%
0

The modulus of the complex quantity

$(2-3i)(-1+7i)$ .

0%
$5\sqrt{13}$
0%
$5\sqrt{26}$
0%
$13\sqrt{5}$
0%
$26\sqrt{5}$

Explanation

The given complex quantity is

$(2-3i)(-1+7i)$ .

Let

$z_{1}=2-3i$ and

$z_{2}=-1+7i$

Therefore,

$|z_{1}|=\sqrt{2^2+(-3)^2}=\sqrt{13}$

$|z_{2}|=\sqrt{(-1)^2+7^2}=\sqrt{50}=5\sqrt{2}$

Thus, required modulus =

$|z_{1}z_{2}|=\sqrt{13} \times 5\sqrt{2}=5\sqrt{26}$ .

If A and B be two complex numbers satisfying

$\dfrac{A}{B}+\dfrac{B}{A}=1$ . Then the two points represented by A and B and the origin form the vertices of

0%
An equilateral triangle
0%
An isosceles triangle which is not equilateral
0%
An isosceles triangle which is not right angled
0%
A right angled triangle

Explanation

Given A and B are 2 complex numbers and

$\dfrac{A}{B}+\dfrac{B}{A} -1$

Let

$A = z_1$

$B =z_2$

$z_3 = 0$

$\dfrac{z_1}{z_2}+\dfrac{z_{2}}{z_1} = 1 \Rightarrow z_{1}^{2}+z_{2}^{2} =z_{1}z_{2}$

$\Rightarrow z_{1}^{2}+z_{2}^{2} + z_{3}^{2} = z_1z_2 + z_2 z_3 + z_3 z_1$

This means

$z_1, z_2, z_3$ from an equilateral triangle

1155189_1110682_ans_18f985bb4bf14ed6a7b5c254df7b11f3.jpg

$3+2\ i\ \sin \theta$ will be real, if

$\theta=$

0%
$2n \pi$
0%
$n \pi +\pi/2$
0%
$n\pi$
0%
$none\ of\ these$

Explanation

$3+2i\sin \theta$ will be real if imaginary term is zero

i.e

$2\sin\theta=0$

$\Rightarrow \sin \theta=0$

$\Rightarrow \theta=n\pi$

$z_{1},\ z_{2}$ are two complex numbers such that

$arg\left( { z }_{ 1 }+{ z }_{ 2 } \right) =0$ and

$Im\left( { z }_{ 1 }{ z }_{ 2 } \right) =0$ , then

0%
$z_{1}=-z_{2}$
0%
$z_{1}=z_{2}$
0%
$z_{1}=\bar { { z }_{ 2 } }$
0%
$none\ of\ these$

Explanation

Let the complex numbers be,

$z_1=a+ib,z_2=x+iy$

$arg(z_1+z_2)=arg(a+x+i(b+y))=0$

$\tan^{-1}\left ( \dfrac{b+y}{a+x} \right )=0$

$\dfrac{b+y}{a+x}=0$

$b+y=0$ .....(1)

$Im(z_1z_2)=Im(ax-by+i(ay+bx))=0$

$(ay+bx)=0$

from (1),

$ay=xy\Rightarrow a=x$

$\Rightarrow b=-y,a=x$

$\therefore z_1=x-iy,z_2=x+iy$

Hence proved,

$z_1=\bar{z}_2$

The argument of the complex number

$\sin \dfrac {6\pi}{5}+i\left(1+\cos \dfrac {6\pi}{5}\right)$ is

0%
$\dfrac {6\pi}{5}$
0%
$\dfrac {5\pi}{6}$
0%
$\dfrac {9\pi}{10}$
0%
$\dfrac {2\pi}{5}$

Explanation

$\sin{\dfrac{6\pi}{5}}+i\left(1+\cos{\dfrac{6\pi}{5}}\right)$ is equal to

$\sin{\left(\dfrac{\pi}{5}\right)}+i\left(1-\cos{\left(\dfrac{\pi}{5}\right)}\right)$

Lies in the second quardent of complex plane and its argument is

$arg \left(n+iy\right)=\pi-{\tan}^{-1}{\left|\dfrac{y}{x}\right|}$

$=\pi-{\tan}^{-1}{\left|\dfrac{1-\cos{\left(\dfrac{\pi}{5}\right)}}{ \sin{\left(\dfrac{\pi}{5}\right)}}\right|}$

$=\pi-{\tan}^{-1}{\left|\dfrac{2{\sin}^{2}{\left(\dfrac{\pi}{10}\right)}}{ 2\sin{\left(\dfrac{\pi}{10}\right)\cos{\left(\dfrac{\pi}{10}\right)}}}\right|}$

$=\pi-{\tan}^{1}{\left|\dfrac{\sin{\left(\dfrac{\pi}{10}\right)}}{\cos{\left(\dfrac{\pi}{10}\right)}}\right|}$

$=\pi-{\tan}^{-1}{\left|\tan{\left(\dfrac{\pi}{10}\right)}\right|}$

$=\pi-{\tan}^{-1}{\left(\tan{\left(\dfrac{\pi}{10}\right)}\right)}$

$=\pi-\dfrac{\pi}{10}$

$=\dfrac{9\pi}{10}$

Let

$a, b, c\epsilon R_{0}$ and

$1$ be a root of the equation

$ax^{2} + bx + c = 0$ , then the equation

$4ax^{2} + 3bx + 2c = 0$ has

0%
Imaginary roots
0%
Real and equal roots
0%
Real and unequal roots
0%
Rational roots

If z is a complex number such that

$|z|\ge 2$ , then the minimumm value of

$\left|z+\dfrac{1}{2}\right|$ :

0%
is equal to $\dfrac{5}{2}$
0%
lies in the interval $(1,2)$
0%
is strictly greater then $\dfrac{5}{2}$
0%
is strictly greater than $\dfrac{3}{2}$ but less than $\dfrac{5}{2}$

The complex no.

$\dfrac{1+2i}{1-i}$ lies in which quadrant of the complex plane

0%
first
0%
second
0%
third
0%
fourth

$z \neq 0$ , then

$\overset{100}{\underset{0}{\int}}arg(-|z|)dx =$

0%
$0$
0%
Not defined
0%
$100$
0%
$100\pi$

Explanation

Let z is a complex number. Then,

⇒ z=x+iy and

∣z∣=

$(X^2+ Y^2)^{1/2}$

∣z∣ is a distance of a complex number from the origin which is always real number and positive.

We know, in complex number line x−axis= Real number and y−axis= imaginary number.

If ∣z∣ lies of real axis then it will make

$0^0$ angle with respect to the origin.

∴ arg∣z∣=0 Then,

⇒

$arg(-∣z∣)=\pi$

$\int^{100} _0 arg(-∣z∣)dx=\int^{100} _0 \pi dx=100\pi$

hence D is the correct answer

$z$ is purely real and

$Re(z)<0$ , then

$arg(x)$ is

0%
$0$
0%
${\pi}$
0%
$-{\pi}$
0%
$\dfrac{\pi}{2}$

Explanation

Given

$z$ is purely real is

$z = a + i(0)$ for some

$a \in R$

$\Rightarrow z = a$

given

$Re(z) < 0$

$\Rightarrow a < 0$

$\therefore z$ can be represented as Ref. image

$\therefore Arg (z) = \pi$

1427394_1113864_ans_30c9a64de0634ea898d1e1370158bca9.png

$Z$ is a complex number such that

$|z| \ge 2$ ,
then the minimum value of

$\left|z + \dfrac{1}{2}\right|$

0%
Is equal to $\dfrac{5}{2}$
0%
Lies in the interval $(1, 2)$
0%
Is strictly grater than $\dfrac{5}{2}$
0%
Is strictly greater than $\dfrac{3}{2}$ but less than $\dfrac{5}{2}$

Explanation

R.E.F image

graph of

$|z|\geq 2$

everything outside the

circle and on the circle is

$|z|\geq 2$

Minimum distance bet

$^{n}$

$|z+1/2|$

$\Rightarrow$ distance

$AB$

By using distance formula

$\Rightarrow \dfrac 32$

$\therefore$ Option B is correct

1181346_1134501_ans_d6590265571e4db4af8ff1e9a0f0f07b.png

$|z|=1$ and

$|\omega -1| =1$ where

$z, \omega \in C$ , then the largest set of values of

$|2z - 1|^2 + | 2\omega -1|^2$ equals

0%
$[1, 9]$
0%
$[2, 6]$
0%
$[2, 12]$
0%
$[2, 18]$

Explanation

$|z|=1$ implies that Z in Q circle with radius with a center (0,0) in the complex plane.

$|w-1|=1$ implies that it in a circle with the radius unit, center

$(1,0)$ in the complex plane.

$|2z-1|^2+|2w-1|^2=4\left(\left|z-\dfrac{1}{2}\right|^2+\left|w-\dfrac{1}{2}\right|^2\right)$

It means it is the range of

$4$ times the sum of the square of the distance of points from

$(0.5,)$ from the respective circles of low of the z,w

The point

$(0.5,0)$ in symmetric with the both circles on the center of a radius on the X-axis so both of them get equal ranges so it simply becomes

$8$ times of annual circle range.

The min or max possible values of the distance from the point become the distance b/w the point and the vertices of the diameter on choice it lies that is

$\rightarrow \dfrac{1}{2},\dfrac{3}{2}$

$=8\times \left[\left(\dfrac{1}{2}\right)^2,\left(\dfrac{3}{2}\right)^2\right]=[2,18]$

2117502_1132651_ans_28e18deb27384e3283fccfdf8d12839f.png

If for complex number

$z_{1}and z_{2}arg(z_{1})-arg(z_{2})=0then \mid z_{ 1}-z_{2}\mid$ is equal to:

0%
$\mid z_{1}+z_{2}\mid$
0%
$\mid z_{1}\mid +\mid z_{2}\mid$
0%
$\parallel z_{1}\mid -\mid z_{2}\parallel$
0%
0

$z_{1}$ and

$z_{2}$ are two non zero complex numbers such that

$|z_{1}+z_{2}|=|z_{1}|+|z_{2}|$ then

$arg\ z_{1}$ -

$arg\ z_{2}$ is equal to

0%
$-\pi$
0%
$\dfrac {\pi}{2}$
0%
$-\dfrac {\pi}{2}$
0%
$0$

Explanation

A] Let

$z_{1} = cos\theta _{1}+i\,sin\theta _{1}$

$z_{2} = cos\theta _{2}+i\,sin\theta _{2}$

$z_{1}+z_{2} = (cos\theta _{1}+cos\theta _{2})+i(sin\theta _{1}+sin\theta _{2})$

$|z_{1}+z_{2}| = |z_{1}|+|z_{2}|$

$= \sqrt{(cos\theta _{1}+cos\theta _{2})^{2}+(sin\theta _{1}+sin\theta _{2})^{2}} = 1+1$

$= 2(1+cos(\theta _{1}-\theta _{2})) = 4$ [squaring]

$= cos(\theta _{1}-\theta _{2}) = 1$

$= \theta _{1}-\theta _{2} = 0$ [as

$cos0^{\circ} = 1$ ]

$= Arg z_{1}-Argz_{2} = 0$

1179877_1145563_ans_a73007a5ead048e0bbb57a610b8397bf.jpg

Argument and modules of

$[\dfrac{1+i}{1-i}]^{2\pi i}$ are respectively.................

0%
$\dfrac{-\pi}{2}$ and $1$
0%
$\dfrac{\pi}{2}$ and $\sqrt{2}$
0%
$0$ and $\sqrt{2}$
0%
$\dfrac{\pi}{2}$ and $1$

Explanation

$(\dfrac{1+i}{1-i})^{2\pi i}$

let

$2 = (\dfrac{1+i}{1-i})^{2\pi i}$

Rationalizing the above, we get

$= \dfrac{(1+i)^2}{(1)^2-(i)^2} = \dfrac{1+(i)^2+2i}{1-(-1)} = \dfrac{1-1+2i}{1+1} = \dfrac{2i}{2} = i$

$z = 0+i = 1$

here x =0 ,y = 1

modules of

$2 = \sqrt{x^2 + y^2 } = \sqrt{0+1^2} = 1$

modules of z = 1

$z =i ,$ As per the question,

$(i)^{2\pi i }$

$z = r(cos\theta + sin\theta)$

$\Rightarrow (r(cos\theta + 1sin \theta))^{2\pi i}$

$= [r(cos(2\pi i))+\theta +i sin (2\pi i )\theta ]$

$(i)^{2\pi i} = (r)^{2\pi i}$

$\Rightarrow r = i$

$r^2 cos^2 \theta = 0 ; r^2 sin^2 \theta = 1$

$tan \theta = \dfrac{\pi}{2}$

$0+1 = r^2 cos^2\theta +r sin^2 \theta$

$1 = r^2(cos^2 \theta + sin^2 \theta)$

$1 = r^2 \Rightarrow r = 1$

argument of

$z = \dfrac{\pi}{2}$

Option (D) is correct.

1169585_1140014_ans_7099f96ac0f2472c82e25114b401bbaa.jpeg

In Argand diagram, O, P, Q represents the origin, $z$ and $z+iz$
respectively. then $\angle OPQ =$

0%
$\dfrac{\pi }{4}$
0%
$\dfrac{\pi }{6}$
0%
$\dfrac{\pi }{2}$
0%
$\dfrac{\pi }{3}$

Explanation

$z=x+iy$

$z+iz=(x+iy)+i(x+iy)$

$=x+iy+xi-y$

$=(x-y)+(x+y)i$

Slope of line

$PQ = \dfrac{x+y-y}{x-y-x} = \dfrac{-x}{y}=m_{1}$

Slope of

$OP = \dfrac{y}{x}=m_{2}$

$m_{1}m_{2}=-1$

lines are perpendicular.

$\angle OPQ=90^{\circ}=\dfrac{\pi }{2}$

C is correct.

1068412_1156428_ans_221d53268b954a5e9e528480a8d52c2a.png

What is the modulus of following complex number:

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

0%
4
0%
5
0%
2
0%
3

$\theta$ and

$\phi$ are the roots of the equation

$8x^{2}+22x+5=0$ , then

0%
both $\sin^{-1}{\theta}$ and $\sin^{-1}{\phi}$ are real
0%
both $\sec^{-1}{\theta}$ and $\sec^{-1}{\phi}$ are real
0%
both $\tan^{-1}{\theta}$ and $\tan^{-1}{\phi}$ are real
0%
None of these

Explanation

Solving given quadratic eq., we get

$8x^2+22x+5=0$

$\implies (4x+1)(2x+5)=0$

$\implies x=\dfrac{-1}{4}, \dfrac{-5}{2}$

Hence these values are not in the range of

$sec^{-1}x$ and

$sin^{-1}x$

So,

$tan^{-1}\theta$ and

$tan^{-1}\phi$ are real

If root of the equation

${ \left( q-r \right) x }^{ 2 }+\left( r-p \right) x+\left( p-q \right) =0$ are equal, then

$p,q,r$ are in

0%
$AP$
0%
$GP$
0%
$HP$
0%
$None\ of\ these$

Explanation

$(q-r)x^2+(r-p)x+(p-q)=0$

If roots are equal,

$D=0$

$\therefore (r-p)^2=4(p-q)(q-r)$

$\therefore r^2+p^2-2rp=4(p-q)(q-r)$

$\therefore r^2+p^2-2rp=4(pq-pr-q^2+2r)$

$\therefore r^2+p^2-2rp=4p2-4pr-4q^2+4qr$

$\therefore (r+p)^2=4pq-4q^2+4qr$

$\therefore (r+p)^2=4(pq-q^2+qr)$

$\therefore 2q=p+r$

Hence

$p, q, r$ are in

$A.P$ .

The integral values of

$'a'$ for which the equation

$\cos ^{2}x-(a^{2}+a+5)|\cos x|+(a^{3}+3a^{2}+2a+6)=0$ has real solution(s)

0%
$-3$
0%
$-2$
0%
$-1$
0%
$0$

Explanation

$Let$

$\left| \cos { x } \right| =t$

$\therefore { t }^{ 2 }-\left( { a }^{ 2 }+a+5 \right) t+\left( { a }^{ 3 }+3{ a }^{ 2 }+2a+6 \right) =0$

$For$

$a=-3$

${ t }^{ 2 }-11=0$

$t=0$ Or

$t=11$

$\therefore \cos { x } =0$ or

$\cos { x } =11$

$But$

$\cos { x } \le 1$

$\cos { x } =0$

$For$

$a=-2$

${ t }^{ 2 }-7t+6=0$

$(t-1)(t-6)=0$

$t=1$ Or

$t=0$

$\therefore \cos { x } =1$ Or

$\cos { x } =6$

$But$

$\cos { x } \le 1$

$\therefore \cos { x } =1$

$Similarly\space checking\space for\space rest\space two\space options, the\space case\space is\space not\space possible$

$So\space both\space option\space A\space and\space B\space are\space true.$

$x_{r}=\cos(\dfrac{\pi}{3^{r}})+i\sin(\dfrac{\pi}{3^{r}})$ , then

$x_{1}x_{2}x_{3}$ .... upto

$infinity=i$ .

0%
True
0%
False

$z=(3+7i)(p+iq)$ where

$p,q\in I-\left\{ 0 \right\}$ , is purely imaginary then minimum value of

${ \left| z \right| }^{ 2 }$ is

0%
$0$
0%
$58$
0%
$\dfrac{3364}{3}$
0%
$3364$

The figure formed by four points

$1+0i;-1+0i,3+4i$ and

$\cfrac{25}{-3-4i}$ on the argand plane is

0%
parallelogram but not a rectangle
0%
a trapesium which is not equilateral
0%
cyclic quadrilateral
0%
none of these

Explanation

$1+0i, -1+0i, 3+4i, \dfrac{25}{-3-4i}$

$\dfrac{25}{-3-4i}\dfrac{(-3+4i)}{(-3+4i)}$

$=\dfrac{25(-3+4i)}{25}=-3+4i$ .

1059275_1159885_ans_ba2730beb1db49deb35e0ebada0c1630.jpg

The amplitude of

$\cfrac { 1+\sqrt { 3i } }{ \sqrt { 3 } +1 }$ is

0%
$\cfrac { \pi }{ 3 }$
0%
$-\cfrac { \pi }{ 3 }$
0%
$\cfrac { \pi }{ 6 }$
0%
$-\cfrac { \pi }{ 6 }$

Explanation

$\displaystyle \frac{1}{\sqrt{3}+1}+\frac i{\sqrt{3}}{\sqrt{3}+1}$

$\displaystyle \tan^{-1}\left ( \frac{\sqrt{3}/\sqrt{3}+1}{1/\sqrt{3}+1} \right )$

$=\tan^{-1}(\sqrt{3})$

$=\dfrac{\pi}{3}$

$Z = \frac{{1 - \sqrt 3 i}}{{1 + \sqrt 3 i}}$ then find

$arg(z).$

0%
$- \dfrac{2\pi }{3}$
0%
$\dfrac{2\pi }{5}$
0%
$\dfrac{\pi }{3}$
0%
$\dfrac{2\pi }{3}$

Explanation

Consider the following question.

$Z=\dfrac{1-\sqrt{3}i}{1+\sqrt{3i}}$

$=\dfrac{\left( 1-\sqrt{3}i \right)}{\left( 1+\sqrt{3i} \right)}\times \dfrac{\left( 1-\sqrt{3}i \right)}{\left( 1-\sqrt{3}i \right)}$

$=\dfrac{{{\left( 1-\sqrt{3}i \right)}^{2}}}{1+3}$

$=\dfrac{1+3{{i}^{2}}-2\sqrt{3}}{4}$

$=\dfrac{1-3-2\sqrt{3}}{4}$

$=\dfrac{-1-i\sqrt{3}}{2}$

$=-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}$

$\arg \left( -\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2} \right)=-\pi +\dfrac{\pi }{3}$

$=\dfrac{2\pi }{3}$

Hence, this is the required answer.

$\left| {z + 2 - i} \right| = 5$ then the maximum value of

$\left| {3z + 9 - 7i} \right|$ is

0%
18
0%
19
0%
20
0%
8

The set of the possible values of

$a$ for which the expression

$x^{2}-ax+1-2a^{2}$ is always positive is

$(\alpha, \beta)$ , then

0%
$\alpha-2\beta=2$
0%
$\alpha+2\beta=2$
0%
$\beta+2\alpha=2$
0%
$\beta-2\alpha=2$

Explanation

$(x^2-ax-2a^2+1)$ is always positive for

$ax=(\alpha , \beta)$

$\therefore \ D < O$

$\therefore \ a^2-4(1) (1-2a^2) < 0$

$\therefore \ a^2-4+8a^2 < 0$

$\therefore \ 9a^2-4 < 0$

$\therefore \ a^2 < \dfrac {4}{3}$

$\therefore \ a \in \left (\dfrac {-2}{3},\ \dfrac {2}{3}\right)$

$\therefore \ \alpha =2/3$

$\beta =-2/3$

$\therefore \ \alpha /2\beta =\dfrac {2}{3}+\dfrac {4}{3}=2$

$\left|z\right|=1$ and

$\varpi=\dfrac{z-1}{z+1}$ , where

$z\neq-1$ , then

$Re\left(\varpi\right)$ is

0%
$0$
0%
$-\dfrac{1}{|z+1|^{2}}$
0%
$\dfrac{1}{\sqrt{2}}{\left|z+1\right|^{2}}$
0%
$\dfrac{\sqrt{2}}{|z+1|^{2}}$

$a,b,c$ are distinct, positive in

$H.P$ , then quadratic equation

${ ax }^{ 2 }+2bx+c=0$ has

0%
Two non-real roots such that their sum is real
0%
Two distinct real roots
0%
Two non real conjugate roots
0%
Two equal real roots

Explanation

Given that

$a,b,c$ are distinct and real roots in HP.

Thus we can say

$\dfrac {1}{a}, \dfrac {1}{b}, \dfrac {1}{c}$ are in AP.

$\therefore \dfrac {1}{b}=\dfrac {\dfrac {1}{a}+\dfrac {1}{c}}{2}=\dfrac {a+c}{2ac}$

Thus

$b=\dfrac {2ac}{a+c}$ ....(i)

Now given quadratic equation is

$ax^2+2bx+c=0$

We know discriminant

$D=b^2-4ac$

$=(2b)^2-4\times a \times c$

$=4b^2-4ac$

$=4\left[\dfrac {4a^2c^2}{(a+c)^2}\right]-4ac$

$=\dfrac {16a^2c^2}{(a+c)^2}-4ac$

$=4ac\left(\dfrac {4ac}{(a+c)^2}-1\right)$

$=4ac\left[\dfrac {4ac-(a+c)^2}{(a+c)^2}\right]$

$=4ac\left [\dfrac {-a^2-c^2+2ac}{(a+c)^2}\right]$

$=-4ac\left [\dfrac {(a-c)^2}{(a+c)^2}\right]$

Given that

$a,b,c$ are positive, but overall result is negative.

Thus

$D<0$

We can say roots are imaginary.

The root of the equation

${\left| x \right|^2} + \left| x \right| - 6 = 0$ are -

0%
only one real number
0%
real and sum = $1 \, or \, -1$
0%
real and sum = $0$
0%
real and product = $0$

Explanation

For

$x>0$

${\left|x\right|}^{2}+\left|x\right|-6=0$ becomes

${x}^{2}+x-6=0$

${x}^{2}+3x-2x-6=0$

$x\left(x+3\right)-2\left(x+3\right)=0$

$\left(x+3\right)\left(x-2\right)=0$

$\Rightarrow\,x=-3,2$

Roots are real and sum

$=-3+2=-1$

For

$x<0$

${\left|x\right|}^{2}+\left|x\right|-6=0$ becomes

${x}^{2}-x-6=0$

${x}^{2}-3x+2x-6=0$

$x\left(x-3\right)+2\left(x-3\right)=0$

$\left(x-3\right)\left(x+2\right)=0$

$\Rightarrow\,x=3,-2$

Roots are real and sum

$=3-2=1$

Hence real and sum

$=-1\,or \,1$

$z=(3+7i)(p+iq)$ , where

$p,q\in I-\left\{ 0 \right\}$ , is a purely imaginary, then minimum value of

${ \left| z \right| }^{ 2 }$ is

0%
$0$
0%
$58$
0%
$\cfrac{3364}{3}$
0%
$3364$

Explanation

$z=(3+7i)(p+iq) \quad \dots (1)$

$z=(3p-7q)+i(3q+7p)$

Now,

$z$ is purely imaginary

$3p-7q=0$

$\cfrac{p}{q}=\cfrac{7}{3}$

$\cfrac{p}{q}+i=\cfrac{7}{3}+i$

Therefore,

$\cfrac{(p+qi)}{q}=\cfrac{7+3i}{3}$

$p+iq=7+3i \quad \dots (2)$

Substitute

$(2)$ in

$(1)$

$z=21+9i+49i-21$

Thus,

$z=58i$

$\left| { z }^{ 2 } \right| =3364$

$z=\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}i$ then

$z\bar{z}$ is

0%
$-1$
0%
$0$
0%
$1$
0%
$2$

Explanation

Conjugate of

$\dfrac{\sqrt{3}+i}{2}=\dfrac{\sqrt{3}-i}{2}$

$\therefore\,\bar{z}=\dfrac{\sqrt{3}-i}{2}$

Now,

$z\bar{z}=\left(\dfrac{\sqrt{3}+i}{2}\right)\left(\dfrac{\sqrt{3}-i}{2}\right)=\dfrac{3-{i}^{2}}{4}=\dfrac{3+1}{4}=\dfrac{4}{4}=1$

Solve

$i^{57}+\dfrac{1}{i^{125}}$

0%
$0$
0%
$2i$
0%
$-2i$
0%
$2$

Explanation

We know that

${i}^{2}=-1$

${i}^{3}={i}^{2}.i=-i$

${i}^{4}={i}^{2}\times {i}^{2}=-1\times -1=1$

Thus,

${i}^{57}={\left({i}^{4}\right)}^{14}\times i={1}^{14}\times i=i$

${i}^{125}={\left({i}^{4}\right)}^{31}\times i={1}^{4}\times i=i$

$\dfrac{1}{{i}^{125}}=\dfrac{1}{i}$

$\dfrac{1}{i}=\dfrac{1}{i}\times\dfrac{i}{i}=\dfrac{i}{{i}^{2}}=-i$

Now

${i}^{57}+\dfrac{1}{{i}^{125}}=i+\left(-i\right)=i-i=0$

$(y^{2}+1)^{2}-y^{2}=0$ has .

0%
No real roots
0%
One real root
0%
Two real roots
0%
Three real roots
0%
None of these

Explanation

$({ y }^{ 2 }+1)^{ 2 }-{ y }^{ 2 }=0\\ \Rightarrow { y }^{ 4 }+1+2{ y }^{ 2 }-{ y }^{ 2 }=0\\ \Rightarrow { y }^{ 4 }+{ y }^{ 2 }+1=0\\ D={ b }^{ 2 }-4ac\\ =1-4\times 1\times 1\\ =-3\\ \because D<0$

$\therefore$ No real roots

$z_{1} and z_{2} are on straight line$

$\left| \frac { 1 } { 2 } \left( z _ { 1 } + z _ { 2 } \right) + \sqrt { z _ { 1 } z _ { 2 } } \right| + \left| \frac { 1 } { 2 } \left( z _ { 1 } + z _ { 2 } \right) - \sqrt { z _ { 1 } z _ { 2 } } \right| =$

0%
$\left| z _ { 1 } + z _ { 2 } \right|$
0%
$\left| z _ { 1 } - z _ { 2 } \right|$
0%
$\left| z _ { 1 } \right| + \left| z _ { 2 } \right|$
0%
$\left| z _ { 1 } \right| - \left| z _ { 2 } \right|$

The value of

$2x^{4}+5x^{3}+7x^{2}-x+41$ , when

$x=-2-\sqrt{3i}$ is:

0%
-4
0%
4
0%
-6
0%
6

Explanation

$2x^{4}+5x^{3}+4x^{2}-x+41$ ---(1)

when

$x = -2-\sqrt{3}i = -(2+\sqrt{3}i)$ ---(2)

$x^{2} = (2+\sqrt{3}i)^{2}$

$= 4+3(i)^{2}+4\sqrt{3}i$

$=4-3+4\sqrt{3}!$

$(\because i^{2} = -1)$

$x^{2} = 1+4\sqrt{3}i$ ---(3)

$x^{3} = x^{2}.x$

$= (1+4\sqrt{3}i)[-(2+\sqrt{3}i)]$

$= -(2+\sqrt{3}i+8\sqrt{3}i+12!^{2})$

$= -(2-12+9\sqrt{3}i)$

$x^{3} = 10-9\sqrt{3}i$ ---(4)

$x^{4} = (x^{2})^{2}$

$= (1+4\sqrt{3}i)^{2}$

$= 1+48!^{2}+8i\sqrt{3}$

$= 1-48+8i\sqrt{3}$

$(i^{2} = -1)$

$x^{4} = -47+8!^{3}$ ---(5)

From (1),(2),(3),(4) and (5)

$2(-47+8i\sqrt{3})+5(10-9\sqrt{3}i)+7(1+4\sqrt{3}i)-(-2\sqrt{3}i)+41$

$-94+16i\sqrt{3}+50-45\sqrt{3}i+7+28\sqrt{3}i+2+\sqrt{3}i+41$

$= 6$ d)

1112264_1200633_ans_9aa025ae23a8492fb71450047d74f29c.jpg

The argument of the complex number

$\sin \dfrac{{6\pi }}{5} + i\left( {1 + \cos \dfrac{{6\pi }}{5}} \right)$ is

0%
$\dfrac{{6\pi }}{5}$
0%
$\dfrac{{5\pi }}{5}$
0%
$\dfrac{{9\pi }}{10}$
0%
$\dfrac{{7\pi }}{10}$

Explanation

Given that

$\sin { \cfrac { 6\pi }{ 5 } } +i\left( 1+\cos { \cfrac { 6\pi }{ 5 } } \right)$

Notice the point

$\sin { \left( \cfrac { 6\pi }{ 5 } \right) } +i\left( 1+\cos { \cfrac { 6\pi }{ 5 } } \right)$ or

$-\sin { \left( \cfrac { 6\pi }{ 5 } \right) } +i\left( 1-\cos { \cfrac { \pi }{ 5 } } \right)$ lies in the second quadrant of complex plane hence its argument is given as

$arg\left( z \right) =\pi -\tan ^{ -1 }{ \left| y/x \right| } \quad$ (

$\because z=x+iy$ )

$\left( \forall x<0,y\ge 0 \right)$

$arg(z)=\pi -\tan ^{ -1 }{ \left| \cfrac { 1-\cos { \cfrac { \pi }{ 5 } } }{ \sin { \cfrac { \pi }{ 5 } } } \right| }$

$=\pi -\tan ^{ -1 }{ \left| \cfrac { 2\sin ^{ 2 }{ \cfrac { \pi }{ 10 } } }{ 2\sin { \cfrac { \pi }{ 10 } } \cos { \cfrac { \pi }{ 10 } } } \right| } =\pi -\tan ^{ -1 }{ \left| \cfrac { \sin { \cfrac { \pi }{ 10 } } }{ \cos { \cfrac { \pi }{ 10 } } } \right| }$

$=\pi -\tan ^{ -1 }{ \left| \tan { \cfrac { \pi }{ 10 } } \right| } \left( \because \tan ^{ -1 }{ \cfrac { \pi }{ 10 } } >1 \right)$

$=\pi -\cfrac { \pi }{ 10 } \left( \because -\cfrac { \pi }{ 2 } \le \tan ^{ -1 }{ x } \le \cfrac { \pi }{ 2 } \right)$

$arg(z)=\cfrac{9\pi}{10}$

$Re(\dfrac{z+2i}{z+4})=0$ then z lies on a circle with center:

0%
(-2,-1)
0%
(-2,1)
0%
(2,-1)
0%
(2,1)

Explanation

Let

$z=x+ iy$

Then

$w=\dfrac{z+2i}{z+4}=\dfrac{x+i(2+y)}{(x+4)+iy}$

Rationalizing we get:

$w=\dfrac{z+2i}{z+4}=\dfrac{x+i(2+y)}{(x+4)-iy}\times \dfrac{x+4-iy}{x+4+iy}=\dfrac{x^2+4x+y^2+2y+i(xy+2x+4y+xy+8)}{(x+4)^2+y^2}$

Since real part of

$w$ is 0, hence

$(x+2)^2+(y+1)^2-5=0$

Hence centre of circle is

$(-2,-1)$

If the six solutions of

$x^6 = -64$ are written in the form

$a + bi$ , where

$a$ and

$b$ are real, then the product those solution with

$a < 0$ , is

0%
$4$
0%
$8$
0%
$16$
0%
$64$

CBSE Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 8 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 8 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.