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

Edited By Ramraj Saini | Updated on Sep 21, 2023 11:22 PM IST

Complex Numbers and Quadratic Equations Class 11 Questions And Answers

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations are discussed here. These NCERT solutions are created by expert team at Careers360 keeping in mind of latest syllabus of CBSE 2023-24. 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 this class 11 maths chapter 5 question answer, you will learn to solve equations like x²+ 2 = 0.

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Complex Numbers and Quadratic Equations Class 11 Questions And Answers
Complex Numbers and Quadratic Equations Class 11 Questions And Answers PDF Free Download
Complex Numbers and Quadratic Equations Class 11 Solutions - Important Formulae
Complex Numbers and Quadratic Equations Class 11 NCERT Solutions (Intext Questions and Exercise)
Highlights Of NCERT Class 11 Chapter 5 Complex Numbers And Quadratic Equations
Complex Numbers And Quadratic Equations Exercise Wise Solutions
NCERT Solutions For Class 11 Mathematics - Chapter Wise
Benefits of NCERT Class 11 Maths ch 5 Question Answer
NCERT Solutions For Class 11 - Subject wise
NCERT Books and NCERT Syllabus

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

The CBSE Syllabus for 2023 includes the chapter Complex Numbers and Quadratic Equations, which comprises significant mathematical theorems and formulae. The NCERT textbook provides ample practice problems to cover all these concepts, facilitating students' comprehension of advanced concepts in the future. To aid in this process, Careers360 offers detailed solutions for all complex numbers class 11 problems in the textbook. These complex numbers and quadratic equations class 11 solutions are particularly beneficial for students who aim to pass their exams with last-minute preparations. However, the primary focus of NCERT solutions for class 11 is to master concepts and develop a deeper understanding of the subject.

Complex Numbers and Quadratic Equations Class 11 Questions And Answers PDF Free Download

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Complex Numbers and Quadratic Equations Class 11 Solutions - Important Formulae

Complex Numbers:

A complex number is expressed as a + bi, where 'a' and 'b' are real numbers, and 'i' is the imaginary unit.

Imaginary Numbers: The square root of a negative real number is called an imaginary number, represented as √-1 = i.

Equality of Complex Number: Two complex numbers z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂ are equal if x₁ = x₂ and y₁ = y₂.

Algebra of Complex Numbers:

Addition: (z₁+ z₂) = (x₁ + x₂) + i(y₁ + y₂)

Subtraction: (z₁ - z₂) = (x₁ - x₂) + i(y₁ - y₂)

Multiplication: (z₁ * z₂) = (x₁x₂ - y₁y₂) + i(x₁y₂ + x₂y₁)

Division: (z₁ / z₂) = [(x₁x₂ + y₁y₂) + i(x₂y₁ - x₁y₂)] / (x₂² + y₂²), where z₂ ≠ 0.

Conjugate of Complex Number: The conjugate of a complex number z = x + iy is represented as z¯ = x - iy.

Modulus of a Complex Number: |z| = √(x² + y²)

Argument of a Complex Number: The angle made by the line joining the point z to the origin, with the positive X-axis in an anti-clockwise sense is called the argument (arg) of the complex number.

Principal Value of Argument:

When x > 0 and y > 0 ⇒ arg(z) = θ
When x < 0 and y > 0 ⇒ arg(z) = π - θ
When x < 0 and y < 0 ⇒ arg(z) = -(π - θ)
When x > 0 and y < 0 ⇒ arg(z) = -θ

Polar Form of a Complex Number: z = |z| (cosθ + isinθ), where θ = arg(z).

The general polar form of z is z = |z| [cos(2nπ + θ) + isin(2nπ + θ)], where n is an integer.

Free download NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations for CBSE Exam.

Complex Numbers and Quadratic Equations Class 11 NCERT Solutions (Intext Questions and Exercise)

Class 11 maths chapter 5 NCERT solutions - 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 $0-i\frac{7\sqrt2}{2}$

Class 11 maths chapter 5 ncert solutions - 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 $\cos \frac{\pi}{2} +i\sin\frac{\pi}{2}$

Class 11 maths chapter 5 ncert solutions - 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

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

Complex numbers and quadratic equations class 11 solutions - 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

Highlights Of NCERT Class 11 Chapter 5 Complex Numbers And Quadratic Equations

The NCERT Solutions for Chapter 5 - Complex Numbers and Quadratic Equations in Class 11 Maths consists of 3 exercises and a miscellaneous exercise to provide enough practice problems for the students to comprehend all the concepts. The PDF of NCERT Solutions for Class 11 Maths includes detailed explanations of the following topics and sub-topics:

5.1 Introduction

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This section of ch 5 maths class 11 covers quadratic equations with no real solutions and their solutions using complex numbers. The solution to the quadratic equation ax^2 + bx + c = 0 has been derived when the discriminant D = b^2 - 4ac is less than 0.

5.2 Complex Numbers It defines complex numbers and explains the real and imaginary parts of complex numbers with examples.
5.3 Algebra of Complex Numbers This section covers the basic BODMAS operations on complex numbers i.e:

5.3.1 Addition of two complex numbers

5.3.2 Difference of two complex numbers

5.3.3 Multiplication of two complex numbers

5.3.4 Division of two complex number

5.3.5 Power of i

5.3.6 The square roots of a negative real number

5.3.7 Identities

After completing these exercises of complex no class 11 , students will have a better understanding of the fundamental BODMAS operations on complex numbers, as well as their properties, the power of i, square roots of negative real numbers, and complex number identitie

5.4 The Modulus and the Conjugate of a Complex Number This section provides detailed explanations of the modulus and conjugate of a complex number with solved examples.
5.5 Argand Plane and Polar Representation This section explains the complex plane or Argand plane and polar representation of complex numbers, including how to write ordered pairs for complex numbers.

A complex number is a number expressed in the form of a + ib where a and b are real numbers. The real part of the complex number is denoted by "a" while the imaginary part is denoted by "b".
If z1 = a + ib and z2 = c + id are two complex numbers, then their addition is (a + c) + i(b + d) and their product is (ac - bd) + i(ad + bc).
For any non-zero complex number z = a + ib (where a and b are not equal to zero), there exists a unique complex number called the multiplicative inverse of z, denoted by 1/z or z^(-1).
For any integer k, i^(4k) equals 1, i^(4k+1) equals i, i^(4k+2) equals -1, and i^(4k+3) equals -i.
The polar form of the complex number z = x + iy is represented as r(cosθ + i sinθ), where r is the modulus or absolute value of the complex number, and θ is the argument or angle made by the complex number with the positive x-axis.
A polynomial equation of degree n has exactly n roots, which may or may not be complex.

Complex Numbers And Quadratic Equations Exercise Wise Solutions

NCERT Solutions For Class 11 Mathematics - Chapter Wise

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

Benefits of NCERT Class 11 Maths ch 5 Question Answer

Clarity of concepts: The NCERT Solutions for maths chapter 5 class 11 provide clear explanations and examples that help students to understand the concepts easily. The solutions are designed to make the learning process easier and more enjoyable.

Practice: The class 11 complex numbers and quadratic equations NCERT solutions come with a wide range of practice problems that help students to practice and master the concepts covered in the chapter. The more problems they solve, the better they become at the topic.

Exam preparation: NCERT Solutions for ch 5 maths class 11 are designed to help students prepare for their exams. The solutions provide a comprehensive overview of the chapter, including all the important topics and subtopics.

NCERT Solutions For Class 11 - Subject wise

NCERT solutions for class 11 biology

NCERT solutions for class 11 maths

NCERT solutions for class 11 chemistry

NCERT solutions for class 11 physics

Complex Numbers And Quadratic Equation Class 11 Chapter-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.

So a complex number of the form a+ ib can be represented as $r(cos\theta +i sin\theta)$

and

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

and the above representation is known as the polar form of a complex number. T he polar form of the complex number makes the problem very easy to solve. There are many problems in the 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.

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Frequently Asked Questions (FAQs)

1. What are important topics of the chapter Complex Numbers and Quadratic Equations ?

Class 11 maths NCERT solutions chapter 5 includes topics such as Complex numbers, algebra of complex numbers, modulus and the conjugate of a complex number, argand plane and polar representation, and quadratic equations are the important topics in this chapter. to get command in these topics students can practice problems from complex numbers class 11 pdf.

2. How does the NCERT solutions are helpful ?

NCERT solutions are highly beneficial for students as they provide well-structured and comprehensive solutions to the textbook questions. These solutions are designed by subject matter experts and cover all the important topics and concepts in a clear and concise manner. Students can practice complex numbers class 11 solutions to get good hold on the concepts of chapter 5 class 11.

3. What are the most difficult chapters in the class 11 maths ?

Most of the students consider permutation and combination, trigonometry as the most difficult chapters in class 11 maths but with the rigorous practice students can get command on them also.

4. How many chapters are there in CBSE class 11 maths ?

There are 16 chapters starting from set to probability in the CBSE class 11 maths.

5. Does CBSE provides the solutions of NCERT class 11 maths ?

No, CBSE doesn’t provided NCERT solutions for any class or subject.

6. Where can I find the complete solutions of NCERT for class 11 maths ?

Here you will get the detailed NCERT solutions for class 11 maths by clicking on the link.

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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) $200 \, \, J - 500 \, \, J$
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Option 3) $zero\;$	Option 4) $K/4$

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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

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

Complex Numbers and Quadratic Equations Class 11 Questions And Answers

Complex Numbers and Quadratic Equations Class 11 Questions And Answers PDF Free Download

Complex Numbers and Quadratic Equations Class 11 Solutions - Important Formulae

Complex Numbers and Quadratic Equations Class 11 NCERT Solutions (Intext Questions and Exercise)

Highlights Of NCERT Class 11 Chapter 5 Complex Numbers And Quadratic Equations

Complex Numbers And Quadratic Equations Exercise Wise Solutions

NCERT Solutions For Class 11 Mathematics - Chapter Wise

Benefits of NCERT Class 11 Maths ch 5 Question Answer

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