get app

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.

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

5.6 Quadratic Equations

The NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations is given below:

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

Question:1 Express each of the complex number in the form $a+ib$ .

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

Answer:

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
   $= 3+0i$

Question:2 Express each of the complex number in the form $a+ib$ .

$i^9+i^1^9$

Answer:

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 $o+io$

Question:3 Express each of the complex number in the form a+ib.

$i^{-39}$

Answer:

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 $o+i1$

Question:4 Express each of the complex number in the form a+ib.

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

Answer:

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 $14+i28$

Question:5 Express each of the complex number in the form $a+ib$ .

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

Answer:

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   $2-7i$

Question:6 Express each of the complex number in the form $a+ib$ .

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

Answer:

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   $-\frac{19}{5}-i\frac{21}{10}$

Question:7 Express each of the complex number in the form $a+ib$ .

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

Answer:

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   $\frac{17}{3}+i\frac{5}{3}$

Question:8 Express each of the complex number in the form $a+ib$ .

$(1-i)^4$

Answer:

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 $-4+i0$

Question:9 Express each of the complex number in the form $a+ib$ .

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

Answer:

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

$\left ( \frac{1}{3}+3i \right )^3=\left ( \frac{1}{3} \right )^3+(3i)^3+3.\left ( \frac{1}{3} \right )^2.3i+3.\frac{1}{3}.(3i)^2$    $(using \ (a+b)^3=a^3+b^3+3a^2b+3ab^2)$
   $= \frac{1}{27}+27i^3+i + 9i^2$
$= \frac{1}{27}+27(-i)+i + 9(-1)$ $(\because i^3=-i \ and \ i^2 = -1)$
   $=\frac{1}{27}-27i+i-9$
   $=\frac{1-243}{27}-26i$
   $=-\frac{242}{27}-26i$

Therefore, the answer is

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

Question:10 Express each of the complex number in the form $a+ib$ .

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

Answer:

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   $-\frac{22}{3}-i\frac{107}{27}$

Question:11 Find the multiplicative inverse of each of the complex numbers.

$4-3i$

Answer:

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

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

Question:12 Find the multiplicative inverse of each of the complex numbers.

$\sqrt{5}+3i$

Answer:

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 $\frac{\sqrt5}{14}-i\frac{3}{14}$

Question:13 Find the multiplicative inverse of each of the complex numbers.

$-i$

Answer:

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 $0+i1$

Question:14 Express the following expression in the form of $a+ib:$

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

Answer:

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

Question:1 Find the modulus and the arguments of each of the complex numbers.

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

Answer:

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

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

Question:2 Find the modulus and the arguments of each of the complex numbers.

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

Answer:

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

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

Question:3 Convert each of the complex numbers in the polar form:

$1-i$

Answer:

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

Question:4 Convert each of the complex numbers in the polar form:

$-1+i$

Answer:

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

Question:5 Convert each of the complex numbers in the polar form:

$-1-i$

Answer:

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

Question:6 Convert each of the complex numbers in the polar form:

$-3$

Answer:

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 $3\left ( \cos \pi +i\sin\pi \right )$

Question:7 Convert each of the complex numbers in the polar form:

$\sqrt{3}+i$

Answer:

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 $2\left ( \cos \frac{\pi}{6} +i\sin\frac{\pi}{6} \right )$

Question:8 Convert each of the complex numbers in the polar form:

$i$

Answer:

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

Question:1 Solve each of the following equations: $x^2+3=0$

Answer:

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 $\pm\sqrt3i$

Question:2 Solve each of the following equations: $2x^2+x+1=0$

Answer:

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

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

Question:3 Solve each of the following equations: $x^2+3x+9=0$

Answer:

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

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

Question:4 Solve each of the following equations: $-x^2+x-2=0$

Answer:

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

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

Question:5 Solve each of the following equations: $x^2+3x+5=0$

Answer:

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 $\frac{-3\pm\sqrt{11}i}{2}$

Question:6 Solve each of the following equations: $x^2-x+2=0$

Answer:

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

Question:7 Solve each of the following equations: $\sqrt{2}x^2+x+\sqrt{2}=0$

Answer:

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

Question:8 Solve each of the following equations: $\sqrt{3}x^2-\sqrt{2}x+3\sqrt{3}=0$

Answer:

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 $\frac{\sqrt2\pm\sqrt{34}i}{2\sqrt3}$

Question:9 Solve each of the following equations: $x^2+x+\frac{1}{\sqrt{2}}=0$

Answer:

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:

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

Answer:

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

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

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

Some important point to remember:

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.

Happy Reading !!!

Recently Asked Questions

Q.
Express each of the complex number in the form a+ib.

Q : 4 $3(7+7i)+i(7+7i)$

Q.
Convert each of the complex numbers in the polar form:

Q : 8 $i$

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

View All CBSE 11 Class Question

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series

View All

Exams

Articles

Questions

Exams

Colleges

Predictors

Resources

Quick links

Exams

Colleges

Resources

Colleges

Resources

Exams

Exams

Colleges

Predictor

Resources

Exams

Predictors

Resources

Upcoming Events

Exams

Ranking

Products & Resources

NCERT Solutions

Exams

Country

Resources

Exams

Colleges

Predictors

Resources

Exams

Colleges

Predictors

Resources

Engineering Preparation

Government Jobs

Medical Preparation

NCERT Solutions

Exams

Colleges

Predictors

Resources

Resources

Exams

Colleges

Predictors

Exams

Predictors

Colleges

Resources

Exams

Colleges

Resources

Exams

Resources

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

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

The NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations is given below:

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

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

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

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

NCERT solutions for class 11 mathematics

NCERT solutions for class 11- Subject wise

Some important point to remember:

As mentioned in the first paragraph and

Recently Asked Questions

Related Articles

Ask Question

Register to post Answer

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