# NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations: In the earlier classes you have studied the quadratic equations. You must have come across some equations like x2+2=0, x2=-2, for which there is no real solution. How to solve these quadratic equations? In NCERT solutions for class 11 maths chapter 5 complex numbers and quadratic equations, you will learn to solve equations like x2+2=0. This chapter will introduce you to a new term called i (iota), $i=\sqrt{-1}$. Using this you will solve the quadratic equation $ax^2+bx+c=0$ with $b^2-4ac<0$. This chapter is useful not only in solving quadratic equations but also in solving the alternating current circuits and in vector analysis. In CBSE NCERT solutions for class 11 maths chapter 5 complex numbers and quadratic equations, you will learn to solve quadratic equations that have imaginary roots. In this chapter, there are 32 questions in 3 exercises. All the questions are explained in solutions of NCERT for class 11 maths chapter 5 complex numbers and quadratic equations in a detailed manner. It will be very useful for you to understand the concepts. Check all NCERT solutions from class 6 to 12 to learn CBSE science and maths. There are three exercises and a miscellaneous exercise in this chapter which are explained below.

Exercise:5.1

Exercise:5.2

Exercise:5.3

Miscellaneous Exercise

## The main topics of the NCERT Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations are

5.1 Introduction

5.2 Complex Numbers

5.3 Algebra of Complex Numbers

5.4 The Modulus and the Conjugate of a Complex Number

5.5 Argand Plane and Polar Representation

## Solutions of NCERT for class 11 maths chapter 5 Complex Numbers and Quadratic Equations-Exercise: 5.1

$(5i)\left ( -\frac{3}{5} i\right )$

On solving

$(5i)\left ( -\frac{3}{5} i\right )$
we will get

$(5i)\left ( -\frac{3}{5} i\right ) = 5 \times (-\frac{3}{5})\times i \times i$
$= - 3 \times i^2$                                    $(\because i^2 = -1)$
$= - 3 \times -1$
$= 3$

Now, in the form of   $a+ib$   we can write it as
$= 3+0i$

$i^9+i^1^9$

We know that  $i^4 = 1$
Now, we will reduce   $i^9+i^{19}$   into

$i^9+i^1^9$$= (i^4)^2.i+(i^4)^3.i^3$
$= (1)^2.i+(1)^3.(-i)$                                                      $(\because i^4 = 1 , i^3 = -i\ and \ i^2 = -1)$
$=i-i = 0$
Now, in the form of  $a+ib$  we can write it as
$o+io$
Therefore, the answer is $o+io$

$i^{-39}$

We know that  $i^4 = 1$
Now, we will reduce   $i^{-39}$   into

$i^{-39}$ $= (i^{4})^{-9}.i^{-3}$
$= (1)^{-9}.(-i)^{-1}$                                                     $(\because i^4 = 1 , i^3 = -i)$
$= \frac{1}{-i}$
$= \frac{1}{-i} \times \frac{i}{i}$
$= \frac{i}{-i^2}$                                                                        $(\because i^2 = -1)$
$= \frac{i}{-(-1)}$
$=i$
Now, in the form of  $a+ib$  we can write it as
$o+i1$
Therefore, the answer is $o+i1$

$3(7+7i)+i(7+7i)$

Given problem is
$3(7+7i)+i(7+7i)$
Now, we will reduce it into

$3(7+7i)+i(7+7i)$  $= 21+21i+7i+7i^2$
$= 21+21i+7i+7(-1)$                             $(\because i^2 = -1)$
$= 21+21i+7i-7$
$=14+28i$

Therefore, the answer is $14+i28$

$(1-i)-(-1+6i)$

Given problem is
$(1-i)-(-1+6i)$
Now, we will reduce it into

$(1-i)-(-1+6i)$  $=1-i+1-6i$
$= 2-7i$

Therefore, the answer is  $2-7i$

$\left ( \frac{1}{5}+i\frac{2}{5} \right )-\left ( 4+i\frac{5}{2} \right )$

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

$\left ( \frac{1}{5}+i\frac{2}{5} \right )-\left ( 4+i\frac{5}{2} \right ) = \frac{1}{5}+i\frac{2}{5}-4-i\frac{5}{2}$
$= \frac{1-20}{5}+i\frac{(4-25)}{10}$
$= -\frac{19}{5}-i\frac{21}{10}$

Therefore, the answer is  $-\frac{19}{5}-i\frac{21}{10}$

$\left [ \left ( \frac{1}{3}+i\frac{7}{3} \right )+\left ( 4+i\frac{1}{3} \right ) \right ]-\left ( -\frac{4}{3}+i \right )$

Given problem is
$\left [ \left ( \frac{1}{3}+i\frac{7}{3} \right )+\left ( 4+i\frac{1}{3} \right ) \right ]-\left ( -\frac{4}{3}+i \right )$
Now, we will reduce it into

$\left [ \left ( \frac{1}{3}+i\frac{7}{3} \right )+\left ( 4+i\frac{1}{3} \right ) \right ]-\left ( -\frac{4}{3}+i \right ) = \frac{1}{3}+i\frac{7}{3} + 4+i\frac{1}{3} + \frac{4}{3}-i$
$=\frac{1+4+12}{3}+i\frac{(7+1-3)}{3}$
$=\frac{17}{3}+i\frac{5}{3}$

Therefore, the answer is  $\frac{17}{3}+i\frac{5}{3}$

$(1-i)^4$

The given problem is
$(1-i)^4$
Now, we will reduce it into

$(1-i)^4 = ((1-i)^2)^2$
$= (1^2+i^2-2.1.i)^2$                                                        $(using \ (a-b)^2= a^2+b^2-2ab)$

$=(1-1-2i)^2$                                                               $(\because i^2 = -1)$
$= (-2i)^2$
$= 4i^2$
$= -4$

Therefore, the answer is  $-4+i0$

$\left ( \frac{1}{3}+3i \right )^3$

Given problem is
$\left ( \frac{1}{3}+3i \right )^3$
Now, we will reduce it into

$\left ( \frac{1}{3}+3i \right )^3=\left ( \frac{1}{3} \right )^3+(3i)^3+3.\left ( \frac{1}{3} \right )^2.3i+3.\frac{1}{3}.(3i)^2$                $(using \ (a+b)^3=a^3+b^3+3a^2b+3ab^2)$
$= \frac{1}{27}+27i^3+i + 9i^2$

$= \frac{1}{27}+27(-i)+i + 9(-1)$                                                               $(\because i^3=-i \ and \ i^2 = -1)$
$=\frac{1}{27}-27i+i-9$
$=\frac{1-243}{27}-26i$
$=-\frac{242}{27}-26i$

$-\frac{242}{27}-26i$

$\left ( -2-\frac{1}{3}i \right )^3$

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

$\left ( -2-\frac{1}{3}i \right )^3=-\left ( (2)^3+\left ( \frac{1}{3}i \right )^3 +3.(2)^2\frac{1}{3}i+3.\left ( \frac{1}{3}i \right )^2.2 \right )$                $(using \ (a+b)^3=a^3+b^3+3a^2b+3ab^2)$
$=-\left ( 8+\frac{1}{27}i^3+3.4.\frac{1}{3}i+3.\frac{1}{9}i^2.2 \right )$

$=-\left ( 8+\frac{1}{27}(-i)+4i+\frac{2}{3}(-1) \right )$                                                               $(\because i^3=-i \ and \ i^2 = -1)$
$=-\left ( 8-\frac{1}{27}i+4i-\frac{2}{3} \right )$
$=-\left ( \frac{(-1+108)}{27}i+\frac{24-2}{3} \right )$
$=-\frac{22}{3}-i\frac{107}{27}$

Therefore, the answer is  $-\frac{22}{3}-i\frac{107}{27}$

$4-3i$

Let    $z = 4-3i$
Then,
$\bar z = 4+ 3i$
And
$|z|^2 = 4^2+(-3)^2 = 16+9 =25$
Now, the multiplicative inverse is given by
$z^{-1}= \frac{\bar z}{|z|^2}= \frac{4+3i}{25}= \frac{4}{25}+i\frac{3}{25}$

Therefore, the multiplicative inverse is

$\frac{4}{25}+i\frac{3}{25}$

$\sqrt{5}+3i$

Let    $z = \sqrt{5}+3i$
Then,
$\bar z = \sqrt{5}-3i$
And
$|z|^2 = (\sqrt5)^2+(3)^2 = 5+9 =14$
Now, the multiplicative inverse is given by
$z^{-1}= \frac{\bar z}{|z|^2}= \frac{\sqrt5-3i}{14}= \frac{\sqrt5}{14}-i\frac{3}{14}$

Therefore, the multiplicative inverse is   $\frac{\sqrt5}{14}-i\frac{3}{14}$

$-i$

Let    $z = -i$
Then,
$\bar z = i$
And
$|z|^2 = (0)^2+(1)^2 = 0+1 =1$
Now, the multiplicative inverse is given by
$z^{-1}= \frac{\bar z}{|z|^2}= \frac{i}{1}= 0+i$

Therefore, the multiplicative inverse is   $0+i1$

$\frac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$

Given problem is
$\frac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$
Now, we will reduce it into

$\frac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})} = \frac{3^2- (\sqrt5i)^2}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$                        $(using \ (a-b)(a+b)=a^2-b^2)$
$=\frac{9-5i^2}{\sqrt3+\sqrt2i-\sqrt3+\sqrt2i}$
$=\frac{9-5(-1)}{2\sqrt2i}$                                                           $(\because i^2 = -1)$
$=\frac{14}{2\sqrt2i}\times \frac{\sqrt2i}{\sqrt2i}$
$=\frac{7\sqrt2i}{2i^2}$
$=-\frac{7\sqrt2i}{2}$
Therefore, answer is $0-i\frac{7\sqrt2}{2}$

## Solutions of NCERT for class 11 maths chapter 5 Complex Numbers and Quadratic Equations-Exercise: 5.2

$z=-1-i\sqrt{3}$

Given the problem is
$z=-1-i\sqrt{3}$
Now, let
$r\cos \theta = - 1 \ \ \ and \ \ \ r\sin \theta = -\sqrt3$
Square and add both the sides
$r^2(\cos^2\theta +\sin^2\theta)= (-1)^2+(-\sqrt3)^2$                                                     $(\because \cos^2\theta +\sin^2\theta = 1)$
$r^2= 1+3$
$r^2 =4$
$r= 2$                                                                                                                         $(\because r > 0)$
Therefore, the modulus is 2
Now,
$2\cos \theta = -1 \ \ \ and \ \ \ 2\sin \theta = -\sqrt3$
$\cos \theta = -\frac{1}{2} \ \ \ and \ \ \ \sin \theta = -\frac{\sqrt3}{2}$
Since, both the values of   $\cos \theta \ and \ \sin \theta$  is negative and we know that they are negative in III quadrant
Therefore,
Argument = $-\left ( \pi - \frac{\pi}{3} \right )= - \frac{2\pi}{3}$
Therefore, the argument  is

$- \frac{2\pi}{3}$

$z=-\sqrt{3}+i$

Given the problem is
$z=-\sqrt{3}+i$
Now, let
$r\cos \theta = - \sqrt3 \ \ \ and \ \ \ r\sin \theta = 1$
Square and add both the sides
$r^2(\cos^2\theta +\sin^2\theta)= (-\sqrt3)^2+(1)^2$                                                     $(\because \cos^2\theta +\sin^2\theta = 1)$
$r^2= 1+3$
$r^2 =4$
$r= 2$                                                                                                                         $(\because r > 0)$
Therefore, the modulus is 2
Now,
$2\cos \theta = -\sqrt3 \ \ \ and \ \ \ 2\sin \theta = 1$
$\cos \theta = -\frac{\sqrt3}{2} \ \ \ and \ \ \ \sin \theta =\frac{1}{2}$
Since values of   $\cos \theta$  is negative and  value $\sin \theta$ is positive and  we know that this is the case in  II quadrant
Therefore,
Argument = $\left ( \pi - \frac{\pi}{6} \right )= \frac{5\pi}{6}$
Therefore, the argument  is

$\frac{5\pi}{6}$

$1-i$

Given problem is
$z=1-i$
Now, let
$r\cos \theta = 1 \ \ \ and \ \ \ r\sin \theta = -1$
Square and add both the sides
$r^2(\cos^2\theta +\sin^2\theta)= (1)^2+(-1)^2$                                                     $(\because \cos^2\theta +\sin^2\theta = 1)$
$r^2= 1+1$
$r^2 =2$
$r= \sqrt2$                                                                                                                         $(\because r > 0)$
Therefore, the modulus is $\sqrt2$
Now,
$\sqrt2\cos \theta = 1 \ \ \ and \ \ \ \sqrt2\sin \theta = -1$
$\cos \theta = \frac{1}{\sqrt2} \ \ \ and \ \ \ \sin \theta =-\frac{1}{\sqrt2}$
Since values of   $\sin \theta$  is negative and  value $\cos \theta$ is positive and  we know that this is the case in the IV quadrant
Therefore,
$\theta = -\frac{\pi}{4} \ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ IV \ quadrant)$
Therefore,
$1-i= r\cos \theta +ir\sin \theta$
$= \sqrt2\cos \left ( -\frac{\pi}{4} \right ) +i\sqrt2\sin \left ( -\frac{\pi}{4} \right )$
$= \sqrt2\left ( \cos \left ( -\frac{\pi}{4} \right ) +i\sin \left ( -\frac{\pi}{4} \right ) \right )$

Therefore, the required polar form is   $\sqrt2\left ( \cos \left ( -\frac{\pi}{4} \right ) +i\sin \left ( -\frac{\pi}{4} \right ) \right )$

$-1+i$

Given the problem is
$z=-1+i$
Now, let
$r\cos \theta = -1 \ \ \ and \ \ \ r\sin \theta = 1$
Square and add both the sides
$r^2(\cos^2\theta +\sin^2\theta)= (1)^2+(-1)^2$                                                     $(\because \cos^2\theta +\sin^2\theta = 1)$
$r^2= 1+1$
$r^2 =2$
$r= \sqrt2$                                                                                                                         $(\because r > 0)$
Therefore, the modulus is $\sqrt2$
Now,
$\sqrt2\cos \theta = -1 \ \ \ and \ \ \ \sqrt2\sin \theta = 1$
$\cos \theta = -\frac{1}{\sqrt2} \ \ \ and \ \ \ \sin \theta =\frac{1}{\sqrt2}$
Since values of   $\cos \theta$  is negative and  value $\sin \theta$ is positive and  we know that this is the case in  II quadrant
Therefore,
$\theta =\left ( \pi - \frac{\pi}{4} \right )= \frac{3\pi}{4}\ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ II \ quadrant)$
Therefore,
$-1+i= r\cos \theta +ir\sin \theta$
$= \sqrt2\cos \left ( \frac{3\pi}{4} \right ) +i\sqrt2\sin \left ( \frac{3\pi}{4} \right )$
$= \sqrt2\left ( \cos \left ( \frac{3\pi}{4} \right ) +i\sin \left ( \frac{3\pi}{4} \right ) \right )$

Therefore, the required polar form is   $\sqrt2\left ( \cos \left ( \frac{3\pi}{4} \right ) +i\sin \left ( \frac{3\pi}{4} \right ) \right )$

$-1-i$

Given problem is
$z=-1-i$
Now, let
$r\cos \theta = -1 \ \ \ and \ \ \ r\sin \theta = -1$
Square and add both the sides
$r^2(\cos^2\theta +\sin^2\theta)= (-1)^2+(-1)^2$                                                     $(\because \cos^2\theta +\sin^2\theta = 1)$
$r^2= 1+1$
$r^2 =2$
$r= \sqrt2$                                                          &nbsnbsp;                                                              $(\because r > 0)$
Therefore, the modulus is $\sqrt2$
Now,
$\sqrt2\cos \theta = -1 \ \ \ and \ \ \ \sqrt2\sin \theta = -1$
$\cos \theta = -\frac{1}{\sqrt2} \ \ \ and \ \ \ \sin \theta =-\frac{1}{\sqrt2}$
Since values of both  $\cos \theta$ and $\sin \theta$ is negative  and  we know that this is the case in  III quadrant
Therefore,
$\theta =-\left ( \pi - \frac{\pi}{4} \right )= -\frac{3\pi}{4}\ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ III \ quadrant)$
Therefore,
$-1-i= r\cos \theta +ir\sin \theta$
$= \sqrt2\cos \left ( -\frac{3\pi}{4} \right ) +i\sqrt2\sin \left (- \frac{3\pi}{4} \right )$
$= \sqrt2\left ( \cos \left ( -\frac{3\pi}{4} \right ) +i\sin \left ( -\frac{3\pi}{4} \right ) \right )$

Therefore, the required polar form is   $\sqrt2\left ( \cos \left (- \frac{3\pi}{4} \right ) +i\sin \left (- \frac{3\pi}{4} \right ) \right )$

$-3$

Given problem is
$z=-3$
Now, let
$r\cos \theta = -3 \ \ \ and \ \ \ r\sin \theta = 0$
Square and add both the sides
$r^2(\cos^2\theta +\sin^2\theta)= (-3)^2+(0)^2$                                                     $(\because \cos^2\theta +\sin^2\theta = 1)$
$r^2= 9+0$
$r^2 =9$
$r= 3$                                                                                                                         $(\because r > 0)$
Therefore, the modulus is 3
Now,
$3\cos \theta =- 3 \ \ \ and \ \ \ 3\sin \theta = 0$
$\cos \theta = -1\ \ \ and \ \ \ \sin \theta =0$
Since values of  $\cos \theta$ is negative and $\sin \theta$ is Positive  and  we know that this is the case in  II quadrant
Therefore,
$\theta =\pi\ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ II \ quadrant)$
Therefore,
$-3= r\cos \theta +ir\sin \theta$
$= 3\cos \left (\pi \right ) +i3\sin \left (\pi \right )$
$= 3\left ( \cos \pi +i\sin\pi \right )$

Therefore, the required polar form is   $3\left ( \cos \pi +i\sin\pi \right )$

$\sqrt{3}+i$

Given problem is
$z=\sqrt3+i$
Now, let
$r\cos \theta = \sqrt3 \ \ \ and \ \ \ r\sin \theta = 1$
Square and add both the sides
$r^2(\cos^2\theta +\sin^2\theta)= (\sqrt3)^2+(1)^2$                                                     $(\because \cos^2\theta +\sin^2\theta = 1)$
$r^2= 3+1$
$r^2 =4$
$r= 2$                                                                                                                         $(\because r > 0)$
Therefore, the modulus is 2
Now,
$2\cos \theta =\sqrt3 \ \ \ and \ \ \ 2\sin \theta = 1$
$\cos \theta = \frac{\sqrt3}{2}\ \ \ and \ \ \ \sin \theta =\frac{1}{2}$
Since values of Both $\cos \theta$  and $\sin \theta$ is Positive  and  we know that this is the case in  I quadrant
Therefore,
$\theta =\frac{\pi}{6}\ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ I \ quadrant)$
Therefore,
$\sqrt3+i= r\cos \theta +ir\sin \theta$
$= 2\cos \left (\frac{\pi}{6} \right ) +i2\sin \left (\frac{\pi}{6} \right )$
$= 2\left ( \cos \frac{\pi}{6} +i\sin\frac{\pi}{6} \right )$

Therefore, the required polar form is   $2\left ( \cos \frac{\pi}{6} +i\sin\frac{\pi}{6} \right )$

$i$

Given problem is
$z = i$
Now, let
$r\cos \theta = 0 \ \ \ and \ \ \ r\sin \theta = 1$
Square and add both the sides
$r^2(\cos^2\theta +\sin^2\theta)= (0)^2+(1)^2$                                                     $(\because \cos^2\theta +\sin^2\theta = 1)$
$r^2= 0+1$
$r^2 =1$
$r= 1$                                                                                                                         $(\because r > 0)$
Therefore, the modulus is 1
Now,
$1\cos \theta =0 \ \ \ and \ \ \ 1\sin \theta = 1$
$\cos \theta =0\ \ \ and \ \ \ \sin \theta =1$
Since values of Both $\cos \theta$  and $\sin \theta$ is Positive  and  we know that this is the case in  I quadrant
Therefore,
$\theta =\frac{\pi}{2}\ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ I \ quadrant)$
Therefore,
$i= r\cos \theta +ir\sin \theta$
$= 1\cos \left (\frac{\pi}{2} \right ) +i1\sin \left (\frac{\pi}{2} \right )$
$= \cos \frac{\pi}{2} +i\sin\frac{\pi}{2}$

Therefore, the required polar form is   $\cos \frac{\pi}{2} +i\sin\frac{\pi}{2}$

## CBSE NCERT solutions for class 11 maths chapter 5 Complex Numbers and Quadratic Equations-Exercise: 5.3

Given equation is
$x^2+3=0$
Now, we know that the roots of the quadratic equation is given by the formula
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
In this case value of a = 1 , b = 0 and c = 3
Therefore,
$\frac{-0\pm \sqrt{0^2-4.1.(3)}}{2.1}= \frac{\pm\sqrt{-12}}{2} = \frac{\pm2\sqrt3i}{2}=\pm\sqrt3i$
Therefore, the solutions of requires equation are   $\pm\sqrt3i$

Given equation is
$2x^2+x+1=0$
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 value of a = 2 , b = 1 and c = 1
Therefore,
$\frac{-1\pm \sqrt{1^2-4.2.1}}{2.2}= \frac{-1\pm\sqrt{1-8}}{4} = \frac{-1\pm\sqrt{-7}}{4}=\frac{-1\pm\sqrt7i}{4}$
Therefore, the solutions of requires equation are

$\frac{-1\pm\sqrt7i}{4}$

Given equation is
$x^2+3x+9=0$
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 value of a = 1 , b = 3 and c = 9
Therefore,
$\frac{-3\pm \sqrt{3^2-4.1.9}}{2.1}= \frac{-3\pm\sqrt{9-36}}{2} = \frac{-3\pm\sqrt{-27}}{2}=\frac{-3\pm3\sqrt3i}{2}$
Therefore, the solutions of requires equation are

$\frac{-3\pm3\sqrt3i}{2}$

Given equation is
$-x^2+x-2=0$
Now, we know that the roots of the quadratic equation is given by the formula
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
In this case value of a = -1 , b = 1 and c = -2
Therefore,
$\frac{-1\pm \sqrt{1^2-4.(-1).(-2)}}{2.(-1)}= \frac{-1\pm\sqrt{1-8}}{-2} = \frac{-1\pm\sqrt{-7}}{-2}=\frac{-1\pm\sqrt7i}{-2}$
Therefore, the solutions of equation are

$\frac{-1\pm\sqrt7i}{-2}$

Given equation is
$x^2+3x+5=0$
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 value of a = 1 , b = 3 and c = 5
Therefore,
$\frac{-3\pm \sqrt{3^2-4.1.5}}{2.1}= \frac{-3\pm\sqrt{9-20}}{2} = \frac{-3\pm\sqrt{-11}}{2}=\frac{-3\pm\sqrt{11}i}{2}$
Therefore, the solutions of the equation are   $\frac{-3\pm\sqrt{11}i}{2}$

Given equation is
$x^2-x+2=0$
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 value of a = 1 , b = -1 and c = 2
Therefore,
$\frac{-(-1)\pm \sqrt{(-1)^2-4.1.2}}{2.1}= \frac{1\pm\sqrt{1-8}}{2} = \frac{1\pm\sqrt{-7}}{2}=\frac{1\pm\sqrt{7}i}{2}$
Therefore, the solutions of equation are   $\frac{1\pm\sqrt{7}i}{2}$

Given equation is
$\sqrt{2}x^2+x+\sqrt{2}=0$
Now, we know that the roots of the quadratic equation is given by the formula
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
In this case the value of $a = \sqrt 2 , b =1 \ and \ c = \sqrt2$
Therefore,
$\frac{-1\pm \sqrt{1^2-4.\sqrt2.\sqrt2}}{2.\sqrt2}= \frac{-1\pm\sqrt{1-8}}{2\sqrt2} = \frac{-1\pm\sqrt{-7}}{2\sqrt2}=\frac{-1\pm\sqrt{7}i}{2\sqrt2}$
Therefore, the solutions of the equation are   $\frac{-1\pm\sqrt{7}i}{2\sqrt2}$

Given equation is
$\sqrt{3}x^2-\sqrt{2}x+3\sqrt{3}=0$
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 $a = \sqrt 3 , b =-\sqrt2 \ and \ c = 3\sqrt3$
Therefore,
$\frac{-(-\sqrt2)\pm \sqrt{(-\sqrt2)^2-4.\sqrt3.3\sqrt3}}{2.\sqrt3}= \frac{\sqrt2\pm\sqrt{2-36}}{2\sqrt3} = \frac{\sqrt2\pm\sqrt{-34}}{2\sqrt3}$$=\frac{\sqrt2\pm\sqrt{34}i}{2\sqrt3}$
Therefore, the solutions of the equation are   $\frac{\sqrt2\pm\sqrt{34}i}{2\sqrt3}$

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

$\frac{-1\pm\sqrt{(2\sqrt2-1)}i}{2}$

Question:10 Solve each of the following equations:

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

Given equation is
$x^2+\frac{x}{\sqrt{2}}+1=0$
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 $a = 1 , b =\frac{1}{\sqrt2} \ and \ c= 1$
Therefore,
$\frac{-\frac{1}{\sqrt2}\pm \sqrt{(\frac{1}{\sqrt2})^2-4.1.1}}{2.1}= \frac{-\frac{1}{\sqrt2}\pm\sqrt{\frac{1}{2}-4}}{2} = \frac{-\frac{1}{\sqrt2}\pm\sqrt{-\frac{7}{2}}}{2}$$=\frac{-1\pm\sqrt{7}i}{2\sqrt2}$
Therefore, the solutions of the equation are

$\frac{-1\pm\sqrt{7}i}{2\sqrt2}$

## NCERT solutions for class 11 maths chapter 5 Complex Numbers and Quadratic Equations-Miscellaneous Exercise

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 $2-2i$

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

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   $\frac{307}{442}+i\frac{599}{442}$

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:

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

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:

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

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

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

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

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

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

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

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    $\frac{5}{27}\pm i\frac{\sqrt2}{27}$

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

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

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  $\sqrt{2}$

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

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

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,
$Re\left ( \frac{z_1z_2}{z_1} \right )= -\frac{2}{5}$

$-\frac{2}{5}$

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

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,
$Im\left ( \frac{1}{z_1\bar z_1} \right )= 0$

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

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

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

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

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

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  $\small |1-i|^x=2^x$

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

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

## NCERT solutions for class 11 mathematics

 chapter-1 NCERT solutions for class 11 maths chapter 1 Sets chapter-2 Solutions of NCERT for class 11 chapter 2 Relations and Functions chapter-3 CBSE NCERT solutions for class 11 chapter 3 Trigonometric Functions chapter-4 NCERT solutions for class 11 chapter 4 Principle of Mathematical Induction chapter-5 NCERT solutions for class 11 maths chapter 5 Complex Numbers and Quadratic equations chapter-6 CBSE NCERT solutions for class 11 maths chapter 6 Linear Inequalities chapter-7 NCERT solutions for class 11 maths chapter 7 Permutation and Combinations chapter-8 Solutions of NCERT for class 11 maths chapter 8 Binomial Theorem chapter-9 CBSE NCERT solutions for class 11 maths chapter 9 Sequences and Series chapter-10 NCERT solutions for class 11 maths chapter 10 Straight Lines chapter-11 Solutions of NCERT for class 11 maths chapter 11 Conic Section chapter-12 CBSE NCERT solutions for class 11 maths chapter 12 Introduction to Three Dimensional Geometry chapter-13 NCERT solutions for class 11 maths chapter 13 Limits and Derivatives chapter-14 Solutions of NCERT for class 11 maths chapter 14 Mathematical Reasoning chapter-15 CBSE NCERT solutions for class 11 maths chapter 15 Statistics chapter-16 NCERT solutions for class 11 maths chapter 16 Probability

## NCERT solutions for class 11- Subject wise

 Solutions of NCERT for class 11 biology CBSE NCERT solutions for class 11 maths NCERT solutions for class 11 chemistry Solutions of NCERT for Class 11 physics

## As mentioned in the first paragraph $i=\sqrt{-1}$ and

$\\i^2={-1}\\i^3=-1\times i=-i\\i^4=-1\times-1=1$

and any number can be represented as a complex number of the form a+ ib where a is the real part and b is the imaginary part, for example, 1=1+0i. A complex number a+ib in the X-Y plane is represented as follows

Where is

$\\r=\sqrt{a^2+b^2}\\a=rcos\theta\\b=rsin\theta$

So a complex number of the form a+ ib can be represented as $r(cos\theta +i sin\theta)$ and the above representation is known as the polar form of a complex number. The polar form of the complex number makes the problem very easy to solve. There are many problems in the CBSE NCERT solutions for class 11 maths chapter 5 complex numbers and quadratic equations which are explained using the polar form of the complex number and some are solved using 2-D geometry.

So, NCERT solutions for class 11 maths chapter 5 complex numbers and quadratic equations can make learning easier for you so that you can score well.