NCERT Solutions for Miscellaneous Exercise Chapter 5 Class 11 - Complex Numbers and Quadratic Equations

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NCERT Solutions for Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations Miscellaneous Exercise- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations Miscellaneous Exercise- NCERT solutions for Class 11 Maths Chapter 5 miscellaneous exercise summarises questions from all topics of the NCERT chapter complex numbers and quadratic equations. Class 11 Maths Chapter 5 miscellaneous exercise solutions present questions from basics of complex numbers, algebra of complex numbers, modulus and argument of complex numbers and conjugate of the complex numbers. A few more topics covered in the miscellaneous exercise chapter 5 Class 11 are the polar representation of complex numbers and solutions of quadratic equations with negative discriminant. Along with the NCERT syllabus Class 11 Maths chapter 5 miscellaneous solutions, the following exercises and their solutions are also available. Students can utilise these for their preparation.

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NCERT Solutions for Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations Miscellaneous Exercise- Download Free PDF
NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations Miscellaneous Exercise
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**As per the CBSE Syllabus 2023-24, this class 11 chapter 5 maths miscellaneous solutions is now designated as Chapter 4.

NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations Miscellaneous Exercise

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Access Complex Numbers And Quadratic Equations Class 11 Chapter 5- Miscellaneous Exercise

Question:1 Evaluate $\small \left [ i^{18}+\left ( \frac{1}{i} \right )^2^5\right ]^3$ .

Answer:

The given problem is
$\small \left [ i^{18}+\left ( \frac{1}{i} \right )^2^5\right ]^3$
Now, we will reduce it into
$\small \left [ i^{18}+\left ( \frac{1}{i} \right )^2^5\right ]^3= \left [ (i^4)^4.i^2+\frac{1}{(i^4)^6.i} \right ]^3$
$=\left [ 1^4.(-1)+\frac{1}{1^6.i} \right ]^3$ $(\because i^4 = 1, i^2 = -1 )$
$= \left [ -1+\frac{1}{i} \right ]^3$
$= \left [ -1+\frac{1}{i} \times \frac{i}{i}\right ]^3$
$= \left [ -1+\frac{i}{i^2} \right ]^3$
$= \left [ -1+\frac{i}{-1} \right ]^3 = \left [ -1-i \right ]^3$
Now,
$-(1+i)^3=-(1^3+i^3+3.1^2.i+3.1.i^2)$ $(using \ (a+b)^3=a^3+b^3+3.a^2.b+3.a.b^2)$
$= -(1 - i +3i+3(-1))$ $(\because i^3=-i , i^2 = -1)$
$= -(1 - i +3i-3)= -(-2+2i)$
$=2-2i$
Therefore, answer is $2-2i$

Question:2 For any two complex numbers $\small z_1$ and $\small z_2$ , prove that $\small Re (z_1z_2)=Re\hspace {1mm}z_1\hspace {1mm}Re\hspace {1mm}z_2-Imz_1\hspace {1mm}Imz_2$

Answer:

Let two complex numbers are
$z_1=x_1+iy_1$
$z_2=x_2+iy_2$
Now,
$z_1.z_2=(x_1+iy_1).(x_2+iy_2)$
$=x_1x_2+ix_1y_2+iy_1x_2+i^2y_1y_2$
$=x_1x_2+ix_1y_2+iy_1x_2-y_1y_2$ $(\because i^2 = -1)$
$=x_1x_2-y_1y_2+i(x_1y_2+y_1x_2)$
$Re(z_1z_2)= x_1x_2-y_1y_2$
$=Re(z_1z_2)-Im(z_1z_2)$

Hence proved

Question:3 Reduce $\small \left ( \frac{1}{1-4i}-\frac{2}{1+i} \right )\left ( \frac{3-4i}{5+i} \right )$ to the standard form.

Answer:

Given problem is
$\small \left ( \frac{1}{1-4i}-\frac{2}{1+i} \right )\left ( \frac{3-4i}{5+i} \right )$
Now, we will reduce it into

$\small \left ( \frac{1}{1-4i}-\frac{2}{1+i} \right )\left ( \frac{3-4i}{5+i} \right ) = \left ( \frac{(1+i)-2(1-4i)}{(1+i)(1-4i)} \right )\left ( \frac{3-4i}{5+i} \right )$
$=\left ( \frac{1+i-2+8i}{1-4i+i-4i^2} \right )\left ( \frac{3-4i}{5+i} \right )$
$=\left ( \frac{-1+9i}{1-3i-4(-1)} \right )\left ( \frac{3-4i}{5+i} \right )$
$=\left ( \frac{-1+9i}{5-3i} \right )\left ( \frac{3-4i}{5+i} \right )$
$=\left ( \frac{-3+4i+27i-36i^2}{25+5i-15i-3i^2} \right )= \left ( \frac{-3+31i+36}{25-10i+3} \right )= \frac{33+31i}{28-10i}= \frac{33+31i}{2(14-5i)}$

Now, multiply numerator an denominator by $(14+5i)$
$\Rightarrow \frac{33+31i}{2(14-5i)}\times \frac{14+5i}{14+5i}$
$\Rightarrow \frac{462+165i+434i+155i^2}{2(14^2-(5i)^2)}$ $(using \ (a-b)(a+b)=a^2-b^2)$
$\Rightarrow \frac{462+599i-155}{2(196-25i^2)}$
$\Rightarrow \frac{307+599i}{2(196+25)}= \frac{307+599i}{2\times 221}= \frac{307+599i}{442}= \frac{307}{442}+i\frac{599}{442}$

Therefore, answer is $\frac{307}{442}+i\frac{599}{442}$

Question:4 If $\small x-iy=\sqrt{\frac{a-ib}{c-id}}$ , prove that $\small (x^2+y^2)^2=\frac{a^2+b^2}{c^2+d^2}.$

Answer:

the given problem is

$\small x-iy=\sqrt{\frac{a-ib}{c-id}}$
Now, multiply the numerator and denominator by

$\sqrt{c+id}$
$x-iy = \sqrt{\frac{a-ib}{c-id}\times \frac{c+id}{c+id}}$
$= \sqrt{\frac{(ac+bd)+i(ad-bc)}{c^2-i^2d^2}}= \sqrt{\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}}$
Now, square both the sides
$(x-iy)^2=\left ( \sqrt{\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}} \right )^2$
$=\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}$
$x^2-y^2-2ixy=\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}$
On comparing the real and imaginary part, we obtain

$x^2-y^2 = \frac{ac+bd}{c^2+d^2} \ \ and \ \ -2xy = \frac{ad-bc}{c^2+d^2} \ \ \ -(i)$

Now,
$(x^2+y^2)^2= (x^2-y^2)^2+4x^2y^2$
$= \left ( \frac{ac+bd}{c^2+d^2} \right )^2+\left ( \frac{ad-bc}{c^2+d^2} \right )^2 \ \ \ \ (using \ (i))$
$=\frac{a^2c^2+b^2d^2+2acbd+a^2d^2+b^2c^2-2adbc}{(c^2+d^2)^2}$
$=\frac{a^2c^2+b^2d^2+a^2d^2+b^2c^2}{(c^2+d^2)^2}$
$=\frac{a^2(c^2+d^2)+b^2(c^2+d^2)}{(c^2+d^2)^2}$
$=\frac{(a^2+b^2)(c^2+d^2)}{(c^2+d^2)^2}$
$=\frac{(a^2+b^2)}{(c^2+d^2)}$

Hence proved

Question:5(i) Convert the following in the polar form:

$\small \frac{1+7i}{(2-i)^2}$

Answer:

Let
$z =\small \frac{1+7i}{(2-i)^2} = \frac{1+7i}{4+i^2-4i}= \frac{1+7i}{4-1-4i}= \frac{1+7i}{3-4i}$

Now, multiply the numerator and denominator by $3+4i$
$\Rightarrow z = \frac{1+7i}{3-4i}\times \frac{3+4i}{3+4i}= \frac{3+4i+21i+28i^2}{3^2+4^2}= \frac{-25+25i}{25}= -1+i$
Now,
let
$r\cos\theta = -1 \ \ and \ \ r\sin \theta = 1$
On squaring both and then add
$r^2(\cos^2\theta+\sin^2\theta)= (-1)^2+1^2$
$r^2=2$
$r = \sqrt2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because r > 0)$
Now,
$\sqrt2 \cos \theta = -1 \ \ and \ \ \sqrt2\sin \theta = 1$
$\cos \theta = -\frac{1}{\sqrt2} \ \ and \ \ \sin \theta = \frac{1}{\sqrt2}$
Since the value of $\cos \theta$ is negative and $\sin \theta$ is positive this is the case in II quadrant
Therefore,
$\theta = \pi - \frac{\pi}{4}= \frac{3\pi}{4} \ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ II \ quadrant)$
$z = r\cos \theta + ir\sin \theta$
$=\sqrt2\cos \frac{3\pi}{4} + i\sqrt2\sin \frac{3\pi}{4}$
$=\sqrt2\left ( \cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4} \right )$
Therefore, the required polar form is

$\sqrt2\left ( \cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4} \right )$

Question:5(ii) Convert the following in the polar form:

$\small \frac{1+3i}{1-2i}$

Answer:

Let
$z =\frac{1+3i}{1-2i}$

Now, multiply the numerator and denominator by $1+2i$
$\Rightarrow z =\frac{1+3i}{1-2i} \times \frac{1+2i}{i+2i}= \frac{1+2i+3i-6}{1+4}= \frac{-5+5i}{5}=-1+i$
Now,
let
$r\cos\theta = -1 \ \ and \ \ r\sin \theta = 1$
On squaring both and then add
$r^2(\cos^2\theta+\sin^2\theta)= (-1)^2+1^2$
$r^2=2$
$r = \sqrt2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because r > 0)$
Now,
$\sqrt2 \cos \theta = -1 \ \ and \ \ \sqrt2\sin \theta = 1$
$\cos \theta = -\frac{1}{\sqrt2} \ \ and \ \ \sin \theta = \frac{1}{\sqrt2}$
Since the value of $\cos \theta$ is negative and $\sin \theta$ is positive this is the case in II quadrant
Therefore,
$\theta = \pi - \frac{\pi}{4}= \frac{3\pi}{4} \ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ II \ quadrant)$
$z = r\cos \theta + ir\sin \theta$
$=\sqrt2\cos \frac{3\pi}{4} + i\sqrt2\sin \frac{3\pi}{4}$
$=\sqrt2\left ( \cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4} \right )$
Therefore, the required polar form is

$\sqrt2\left ( \cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4} \right )$

Question:6 Solve each of the equation: $\small 3x^2-4x+\frac{20}{3}=0$

Answer:

Given equation is
$\small 3x^2-4x+\frac{20}{3}=0$
Now, we know that the roots of the quadratic equation are given by the formula
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
In this case the value of

$a=3,b=-4 \ and \ c= \frac{20}{3}$
Therefore,
$\frac{-(-4)\pm \sqrt{(-4)^2-4.3.\frac{20}{3}}}{2.3}= \frac{4\pm\sqrt{16-80}}{6} = \frac{4\pm\sqrt{-64}}{6}$ $=\frac{4\pm8i}{6}= \frac{2}{3}\pm i\frac{4}{3}$
Therefore, the solutions of requires equation are

$\frac{2}{3}\pm i\frac{4}{3}$

Question:7 Solve each of the equation: $\small x^2-2x+\frac{3}{2}=0$

Answer:

Given equation is
$\small x^2-2x+\frac{3}{2}=0$
Now, we know that the roots of the quadratic equation are given by the formula
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
In this case the value of $a=1,b=-2 \ and \ c= \frac{3}{2}$
Therefore,
$\frac{-(-2)\pm \sqrt{(-2)^2-4.1.\frac{3}{2}}}{2.1}= \frac{2\pm\sqrt{4-6}}{2} = \frac{2\pm\sqrt{-2}}{2}$ $=\frac{2\pm i\sqrt2}{2}=1\pm i\frac{\sqrt2}{2}$
Therefore, the solutions of requires equation are

$1\pm i\frac{\sqrt2}{2}$

Question:8 Solve each of the equation: $\small 27x^2-10x+1=0$ .

Answer:

Given equation is
$\small 27x^2-10x+1=0$
Now, we know that the roots of the quadratic equation are given by the formula
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
In this case the value of $a=27,b=-10 \ and \ c= 1$
Therefore,
$\frac{-(-10)\pm \sqrt{(-10)^2-4.27.1}}{2.27}= \frac{10\pm\sqrt{100-108}}{54} = \frac{10\pm\sqrt{-8}}{54}$ $=\frac{10\pm i2\sqrt2}{54}=\frac{5}{27}\pm i\frac{\sqrt2}{27}$
Therefore, the solutions of requires equation are $\frac{5}{27}\pm i\frac{\sqrt2}{27}$

Question:9 Solve each of the equation: $\small 21x^2-28x+10=0$

Answer:

Given equation is
$\small 21x^2-28x+10=0$
Now, we know that the roots of the quadratic equation are given by the formula
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
In this case the value of $a=21,b=-28 \ and \ c= 10$
Therefore,
$\frac{-(-28)\pm \sqrt{(-28)^2-4.21.10}}{2.21}= \frac{28\pm\sqrt{784-840}}{42} = \frac{28\pm\sqrt{-56}}{42}$ $=\frac{28\pm i2\sqrt{14}}{42}=\frac{2}{3}\pm i\frac{\sqrt{14}}{21}$
Therefore, the solutions of requires equation are

$\frac{2}{3}\pm i\frac{\sqrt{14}}{21}$

Question:10 If $\small z_1=2-i, z_2=1+i$ , find $\small \left |\frac{z_1+z_2+1}{z_1-z_2+1} \right |$ .

Answer:

It is given that
$\small z_1=2-i, z_2=1+i$
Then,
$\small \left |\frac{z_1+z_2+1}{z_1-z_2+1} \right | =\left | \frac{2-i+1+i+1}{2-i-1-i+1} \right | = \left | \frac{4}{2(1-i)} \right |= \left | \frac{2}{(1-i)} \right |$
Now, multiply the numerator and denominator by $1+i$
$\Rightarrow \left | \frac{2}{(1-i)} \times \frac{1+i}{1+i} \right |=\left |\frac{2(1+i)}{1^2-i^2} \right |=\left | \frac{2(1+i)}{1+1} \right |= \left| 1+i \right |$
Now,
$|1+i| = \sqrt{1^2+1^2}=\sqrt{1+1}=\sqrt{2}$
Therefore, the value of

$\small \left |\frac{z_1+z_2+1}{z_1-z_2+1} \right |$ is $\sqrt{2}$

Question:11 If $\small a+ib=\frac{(x+i)^2}{2x^2+1}$ , prove that $\small a^2+b^2=\frac{(x^2+1)^2}{(2x^2+1)^2}$ .

Answer:

It is given that
$\small a+ib=\frac{(x+i)^2}{2x^2+1}$
Now, we will reduce it into

$\small a+ib=\frac{(x+i)^2}{2x^2+1} = \frac{x^2+i^2+2xi}{2x^2+1}=\frac{x^2-1+2xi}{2x^2+1}=\frac{x^2-1}{2x^2+1}+i\frac{2x}{2x^2+1}$
On comparing real and imaginary part. we will get
$a=\frac{x^2-1}{2x^2+1}\ and \ b=\frac{2x}{2x^2+1}$
Now,
$a^2+b^2=\left ( \frac{x^2-1}{2x^2+1} \right )^2+\left ( \frac{2x}{2x^2+1} \right )^2$
$= \frac{x^4+1-2x^2+4x^2}{(2x^2+1)^2}$
$= \frac{x^4+1+2x^2}{(2x^2+1)^2}$
$= \frac{(x^2+1)^2}{(2x^2+1)^2}$
Hence proved

Question:12(i) Let $\small z_1=2-i,z_2=-2+i.$ Find

$\small Re\left ( \frac{z_1z_2}{\bar{z_1}} \right )$

Answer:

It is given that
$\small z_1=2-i \ and \ z_2=-2+i$
Now,
$z_1z_2= (2-i)(-2+i)= -4+2i+2i-i^2=-4+4i+1= -3+4i$
And
$\bar z_1 = 2+i$
Now,
$\frac{z_1z_2}{\bar z_1}= \frac{-3+4i}{2+i}= \frac{-3+4i}{2+i}\times \frac{2-i}{2-i}= \frac{-6+3i+8i-4i^2}{2^2-i^2}= \frac{-6+11i+4}{4+1}$ $= \frac{-2+11i}{5}= -\frac{2}{5}+i\frac{11}{5}$
Now,
$Re\left ( \frac{z_1z_2}{z_1} \right )= -\frac{2}{5}$
Therefore, the answer is

$-\frac{2}{5}$

Question:12(ii) Let $\small z_1=2-i,z_2=-2+i.$ Find

$\small Im\left ( \frac{1}{z_1\bar{z_1}} \right )$

Answer:

It is given that
$z_1= 2-i$
Therefore,
$\bar z_1= 2+i$
NOw,
$z_1\bar z_1= (2-i)(2+i)= 2^2-i^2=4+1=5$ $(using \ (a-b)(a+b)= a^2-b^2)$
Now,
$\frac{1}{z_1\bar z_1}= \frac{1}{5}$
Therefore,
$Im\left ( \frac{1}{z_1\bar z_1} \right )= 0$
Therefore, the answer is 0

Question:13 Find the modulus and argument of the complex number $\small \frac{1+2i}{1-3i}$ .

Answer:

Let
$z = \small \frac{1+2i}{1-3i}$
Now, multiply the numerator and denominator by $(1+3i)$

$\Rightarrow z=\frac{1+2i}{1-3i}\times \frac{1+3i}{1+3i}= \frac{1+3i+2i+6i^2}{1^2-(3i)^2}= \frac{1+5i-6}{1-9i^2}= \frac{-5+5i}{10}$ $= -\frac{1}{2}+i\frac{1}{2}$
Therefore,
$r\cos \theta= -\frac{1}{2} \ \ and \ \ r\sin \theta =\frac{1}{2}$
Square and add both the sides
$r^2\cos^2\theta +r^2\sin^2\theta= \left ( -\frac{1}{2} \right )^2+\left ( \frac{1}{2} \right )^2$
$r^2(\cos^2\theta +\sin^2\theta)= \left ( \frac{1}{4} \right )+\left ( \frac{1}{4} \right )$
$r^2= \frac{1}{2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because \sin^2\theta +\cos^2\theta = 1)$
$r = \frac{1}{\sqrt2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because r > 0)$
Therefore, the modulus is $\frac{1}{\sqrt2}$
Now,
$\frac{1}{\sqrt2} \cos\theta = -\frac{1}{2} \ \ and \ \ \frac{1}{\sqrt2} \sin\theta = \frac{1}{2}$
$\cos\theta = -\frac{1}{\sqrt2} \ \ and \ \ \sin\theta = \frac{1}{\sqrt2}$
Since the value of $\cos\theta$ is negative and the value of $\sin\theta$ is positive and we know that it is the case in II quadrant
Therefore,
Argument $=\left ( \pi-\frac{\pi}{4} \right )= \frac{3\pi}{4}$
Therefore, Argument and modulus are $\frac{3\pi}{4} \ \ and \ \ \frac{1}{\sqrt2}$ respectively

Question:14 Find the real numbers x andy if $\small (x-iy)(3+5i)$ is the conjugate of $\small -6-24i$ .

Answer:

Let
$z = \small (x-iy)(3+5i) = 3x+5xi-3yi-5yi^2= 3x+5y+i(5x-3y)$
Therefore,
$\bar z = (3x+5y)-i(5x-3y) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$
Now, it is given that
$\bar z = -6-24i \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$
Compare (i) and (ii) we will get
$(3x+5y)-i(5x-3y) = -6-24i$
On comparing real and imaginary part. we will get
$3x+5y=-6 \ \ \ and \ \ \ 5x-3y = 24$
On solving these we will get
$x = 3 \ \ \ and \ \ \ y =- 3$

Therefore, the value of x and y are 3 and -3 respectively

Question:15 Find the modulus of $\small \frac{1+i}{1-i}-\frac{1-i}{1+i}$ .

Answer:

Let
$z =\small \frac{1+i}{1-i}-\frac{1-i}{1+i}$
Now, we will reduce it into
$z =\small \frac{1+i}{1-i}-\frac{1-i}{1+i} = \frac{(1+i)^2-(1-i)^2}{(1+i)(1-i)}= \frac{1^2+i^2+2i-1^2-i^2+2i}{1^2-i^2}$ $= \frac{4i}{1+1}= \frac{4i}{2}=2i$
Now,
$r\cos\theta = 0 \ \ and \ \ r\sin \theta = 2$
square and add both the sides. we will get,
$r^2\cos^2\theta+r^2\sin^2 \theta = 0^2+2^2$
$r^2(\cos^2\theta+\sin^2 \theta) = 4$
$r^2 = 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because \cos^2\theta+\sin^2 \theta = 1)$
$r = 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because r > 0)$

Therefore, modulus of

$\small \frac{1+i}{1-i}-\frac{1-i}{1+i}$ is 2

Question:16 If $\small (x+iy)^3=u+iv$ , then show that $\small \frac{u}{x}+\frac{v}{y}=4 (x^2-y^2).$

Answer:

it is given that
$\small (x+iy)^3=u+iv$
Now, expand the Left-hand side
$x^3+(iy)^3+3.(x)^2.iy+3.x.(iy)^2= u + iv$
$x^3+i^3y^3+3x^2iy+3xi^2y^2= u + iv$
$x^3-iy^3+3x^2iy-3xy^2= u + iv$ $(\because i^3 = -i \ \ and \ \ i^2 = -1)$
$x^3-3xy^2+i(3x^2y-y^3)= u + iv$
On comparing real and imaginary part. we will get,
$u = x^3-3xy^2 \ \ \ and \ \ \ v = 3x^2y-y^3$
Now,
$\frac{u}{x}+\frac{v}{y}= \frac{x(x^2-3y^2)}{x}+\frac{y(3x^2-y^2)}{y}$
$= x^2-3y^2+3x^2-y^2$
$= 4x^2-4y^2$
$= 4(x^2-y^2)$
Hence proved

Question:17 If $\small \alpha$ and $\small \beta$ are different complex numbers with $\small |\beta|=1$ , then find $\small \left | \frac{\beta -\alpha }{1-\bar{\alpha }\beta } \right |$ .

Answer:

Let
$\alpha = a+ib$ and $\beta = x+iy$
It is given that
$\small |\beta|=1\Rightarrow \sqrt{x^2+y^2} = 1\Rightarrow x^2+y^2 = 1$
and
$\small \bar \alpha = a-ib$
Now,
$\small \left | \frac{\beta -\alpha }{1-\bar{\alpha }\beta } \right | = \left | \frac{(x+iy)-(a+ib)}{1-(a-ib)(x+iy)} \right | = \left | \frac{(x-a)+i(y-b)}{1-(ax+iay-ibx-i^2yb)} \right |$
$\small = \left | \frac{(x-a)+i(y-b)}{(1-ax-yb)-i(bx-ay)} \right |$
$\small = \frac{\sqrt{(x-a)^2+(y-b)^2}}{\sqrt{(1-ax-yb)^2+(bx-ay)^2}}$
$\small = \frac{\sqrt{x^2+a^2-2xa+y^2+b^2-yb}}{\sqrt{1+a^2x^2+b^2y^2-2ax+2abxy-by+b^2x^2+a^2y^2-2abxy}}$
$\small = \frac{\sqrt{(x^2+y^2)+a^2-2xa+b^2-yb}}{\sqrt{1+a^2(x^2+y^2)+b^2(x^2+y^2)-2ax+2abxy-by-2abxy}}$
$\small = \frac{\sqrt{1+a^2-2xa+b^2-yb}}{\sqrt{1+a^2+b^2-2ax-by}}$ $\small (\because x^2+y^2 = 1 \ given)$
$\small =1$

Therefore, value of $\small \left | \frac{\beta -\alpha }{1-\bar{\alpha }\beta } \right |$ is 1

Question:18 Find the number of non-zero integral solutions of the equation $\small |1-i|^x=2^x$ .

Answer:

Given problem is
$\small |1-i|^x=2^x$
Now,
$( \sqrt{1^2+(-1)^2 })^x=2^x$
$( \sqrt{1+1 })^x=2^x$
$\left ( \sqrt{2 }\right )^x=2^x$
$2^{\frac{x}{2}}= 2^x$
$\frac{x}{2}=x$
$\frac{x}{2}=0$
x = 0 is the only possible solution to the given problem

Therefore, there are 0 number of non-zero integral solutions of the equation $\small |1-i|^x=2^x$

Question:19 If $\small (a+ib)(c+id)(e+if)(g+ih)=A+iB,$ then show that $\small (a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2)=A^2+B^2$

Answer:

It is given that
$\small (a+ib)(c+id)(e+if)(g+ih)=A+iB,$
Now, take mod on both sides
$\left | (a+ib)(c+id)(e+if)(g+ih) \right |= \left | A+iB \right |$
$|(a+ib)||(c+id)||(e+if)||(g+ih)|= \left | A+iB \right |$ $(\because |z_1z_2|=|z_1||z_2|)$
$(\sqrt{a^2+b^2})(\sqrt{c^2+d^2})(\sqrt{e^2+f^2})(\sqrt{g^2+h^2})= (\sqrt{A^2+B^2})$
Square both the sides. we will get

$({a^2+b^2})({c^2+d^2})({e^2+f^2})({g^2+h^2})= (A^2+B^2)$

Hence proved

Question:20 If $\small \left ( \frac{1+i}{1-i} \right )^m=1,$ then find the least positive integral value of $\small m$ .

Answer:

Let
$z = \left ( \frac{1+i}{1-i} \right )^m$
Now, multiply both numerator and denominator by $(1+i)$
We will get,
$z = \left ( \frac{1+i}{1-i}\times \frac{1+i}{1+i} \right )^m$
$= \left ( \frac{(1+i)^2}{1^2-i^2} \right )^m$
$= \left ( \frac{1^2+i^2+2i}{1+1} \right )^m$
$= \left ( \frac{1-1+2i}{2} \right )^m$ $(\because i^2 = -1)$
$= \left ( \frac{2i}{2} \right )^m$
$= i^m$
We know that $i^4 = 1$
Therefore, the least positive integral value of $\small m$ is 4

Topics Covered in Miscellaneous Exercise Class 11 Chapter 5

Miscellaneous Exercise Topics in NCERT Solutions for Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations:

Complex Numbers
Algebraic Operations with Complex Numbers
Modulus and Conjugate of Complex Numbers
Argand Plane and Polar Representation
Quadratic Equations

Benefits of NCERT Solutions for Class 11 Maths Chapter 5 Miscellaneous Exercise

NCERT syllabus Class 11 Maths chapter 5 miscellaneous exercise solutions can be accessed easily and downloaded for future reference
Solving miscellaneous exercise chapter 5 Class 11 helps students to make sure that whether they have familiarised them with the concepts discussed in the chapter.
Class 11 Maths chapter 5 miscellaneous solutions also helps in the preparation for competitive exams like JEE Main.

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Key Features of Class 11 Maths Ch 5 Miscellaneous Exercise Solutions

Comprehensive Coverage: The miscellaneous exercise class 11 chapter 5 solutions comprehensively address all 20 questions from the Miscellaneous Exercise in Chapter 5, ensuring a thorough understanding of complex numbers and quadratic equations.
Conceptual Clarity: The class 11 chapter 5 maths miscellaneous solutions provide clear explanations, step-by-step procedures, and valuable insights to help students grasp challenging mathematical concepts effectively.
Relevance to Exams: Designed in alignment with the Class 11 syllabus, the Class 11 maths miscellaneous exercise chapter 5 solutions serve as an invaluable resource for exam preparation, offering a strategic approach to solving miscellaneous exercise questions.
Preparation for Higher Studies: The covered concepts serve as a robust foundation for advanced studies in mathematics and engineering, with a focus on real-world applications of complex numbers.
Competitive Exam Readiness: The class 11 maths ch 5 miscellaneous exercise solutions aid in preparing for competitive exams, such as JEE Main, by reinforcing key mathematical principles and enhancing problem-solving skills.
User-Friendly Access: The class 11 chapter 5 miscellaneous exercise solutions are easily accessible, providing students with the convenience of free PDF downloads for future reference, study, and revision.
Verification of Understanding: Students can use the class 11 chapter 5 maths miscellaneous solutions solutions to verify their understanding of Chapter 5 concepts, ensuring a strong foundation in complex numbers and quadratic equations.

NCERT Solutions of Class 11 Subject Wise

Subject Wise NCERT Exampler Solutions

Frequently Asked Questions (FAQs)

1. Who introduced the symbol i for root(-1)?

Euler introduced symbol i for root(-1) first time

2. Who represented a+ib as ordered pair (a, b)?

W,R. Hamilton

3. Find Z1.Z2 if Z1=a+ib and Z2=c+id

Z1Z2=(ac-bd)+i(ad+bc)

4. Is the statement “all real numbers are complex numbers” true?

Yes, the statement is true. Example: 1 can be written as 1+0i

5. What is the argument of (1+i)/(1-i)

The argument of (1+i)/(1-i) is pi/2

6. How many questions are given in the miscellaneous examples of Class 11 NCERT Maths chapter 5?

5 questions are solved in the miscellaneous examples of complex numbers and quadratic equations.

7. How many questions are given in NCERT solutions for Class 11 Maths chapter 5 miscellaneous exercise?

Twenty questions of miscellaneous exercise chapter 5 Class 11 are solved in the Class 11 Maths chapter 5 miscellaneous exercise solutions

8. Give the ordered pair of (1+i)/(1-i)?

(1+i)/(1-i)=i

So the ordered pair is (0,1)

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Solutions for Miscellaneous Exercise Chapter 5 Class 11 - Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations Miscellaneous Exercise- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations Miscellaneous Exercise

Question:1 Evaluate .

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Topics Covered in Miscellaneous Exercise Class 11 Chapter 5

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NCERT Solutions of Class 11 Subject Wise

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Question:1 Evaluate $\small \left [ i^{18}+\left ( \frac{1}{i} \right )^2^5\right ]^3$ .