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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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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:
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.
Class 11 maths chapter 5 NCERT solutions - Exercise: 5.1
Question:1 Express each of the complex number in the form .
Answer:
On solving
we will get
Now, in the form of we can write it as
Question:2 Express each of the complex number in the form .
Answer:
We know that
Now, we will reduce into
Now, in the form of we can write it as
Therefore, the answer is
Question:3 Express each of the complex number in the form a+ib.
Answer:
We know that
Now, we will reduce into
Now, in the form of we can write it as
Therefore, the answer is
Question:4 Express each of the complex number in the form a+ib.
Answer:
Given problem is
Now, we will reduce it into
Therefore, the answer is
Question:5 Express each of the complex number in the form .
Answer:
Given problem is
Now, we will reduce it into
Therefore, the answer is
Question:6 Express each of the complex number in the form .
Answer:
Given problem is
Now, we will reduce it into
Therefore, the answer is
Question:7 Express each of the complex number in the form .
Answer:
Given problem is
Now, we will reduce it into
Therefore, the answer is
Question:8 Express each of the complex number in the form .
Answer:
The given problem is
Now, we will reduce it into
Therefore, the answer is
Question:9 Express each of the complex number in the form .
Answer:
Given problem is
Now, we will reduce it into
Therefore, the answer is
Question:10 Express each of the complex number in the form .
Answer:
Given problem is
Now, we will reduce it into
Therefore, the answer is
Question:11 Find the multiplicative inverse of each of the complex numbers.
Answer:
Let
Then,
And
Now, the multiplicative inverse is given by
Therefore, the multiplicative inverse is
Question:12 Find the multiplicative inverse of each of the complex numbers.
Answer:
Let
Then,
And
Now, the multiplicative inverse is given by
Therefore, the multiplicative inverse is
Question:13 Find the multiplicative inverse of each of the complex numbers.
Answer:
Let
Then,
And
Now, the multiplicative inverse is given by
Therefore, the multiplicative inverse is
Question:14 Express the following expression in the form of
Answer:
Given problem is
Now, we will reduce it into
Therefore, answer is
Class 11 maths chapter 5 ncert solutions - Exercise: 5.2
Question:1 Find the modulus and the arguments of each of the complex numbers.
Answer:
Given the problem is
Now, let
Square and add both the sides
Therefore, the modulus is 2
Now,
Since, both the values of is negative and we know that they are negative in III quadrant
Therefore,
Argument =
Therefore, the argument is
Question:2 Find the modulus and the arguments of each of the complex numbers.
Answer:
Given the problem is
Now, let
Square and add both the sides
Therefore, the modulus is 2
Now,
Since values of is negative and value is positive and we know that this is the case in II quadrant
Therefore,
Argument =
Therefore, the argument is
Question:3 Convert each of the complex numbers in the polar form:
Answer:
Given problem is
Now, let
Square and add both the sides
Therefore, the modulus is
Now,
Since values of is negative and value is positive and we know that this is the case in the IV quadrant
Therefore,
Therefore,
Therefore, the required polar form is
Question:4 Convert each of the complex numbers in the polar form:
Answer:
Given the problem is
Now, let
Square and add both the sides
Therefore, the modulus is
Now,
Since values of is negative and value is positive and we know that this is the case in II quadrant
Therefore,
Therefore,
Therefore, the required polar form is
Question:5 Convert each of the complex numbers in the polar form:
Answer:
Given problem is
Now, let
Square and add both the sides
&nbsnbsp;
Therefore, the modulus is
Now,
Since values of both and is negative and we know that this is the case in III quadrant
Therefore,
Therefore,
Therefore, the required polar form is
Question:6 Convert each of the complex numbers in the polar form:
Answer:
Given problem is
Now, let
Square and add both the sides
Therefore, the modulus is 3
Now,
Since values of is negative and is Positive and we know that this is the case in II quadrant
Therefore,
Therefore,
Therefore, the required polar form is
Question:7 Convert each of the complex numbers in the polar form:
Answer:
Given problem is
Now, let
Square and add both the sides
Therefore, the modulus is 2
Now,
Since values of Both and is Positive and we know that this is the case in I quadrant
Therefore,
Therefore,
Therefore, the required polar form is
Question:8 Convert each of the complex numbers in the polar form:
Answer:
Given problem is
Now, let
Square and add both the sides
Therefore, the modulus is 1
Now,
Since values of Both and is Positive and we know that this is the case in I quadrant
Therefore,
Therefore,
Therefore, the required polar form is
Class 11 maths chapter 5 ncert solutions - Exercise: 5.3
Question:1 Solve each of the following equations:
Answer:
Given equation is
Now, we know that the roots of the quadratic equation is given by the formula
In this case value of a = 1 , b = 0 and c = 3
Therefore,
Therefore, the solutions of requires equation are
Question:2 Solve each of the following equations:
Answer:
Given equation is
Now, we know that the roots of the quadratic equation are given by the formula
In this case value of a = 2 , b = 1 and c = 1
Therefore,
Therefore, the solutions of requires equation are
Question:3 Solve each of the following equations:
Answer:
Given equation is
Now, we know that the roots of the quadratic equation are given by the formula
In this case value of a = 1 , b = 3 and c = 9
Therefore,
Therefore, the solutions of requires equation are
Question:4 Solve each of the following equations:
Answer:
Given equation is
Now, we know that the roots of the quadratic equation is given by the formula
In this case value of a = -1 , b = 1 and c = -2
Therefore,
Therefore, the solutions of equation are
Question:5 Solve each of the following equations:
Answer:
Given equation is
Now, we know that the roots of the quadratic equation are given by the formula
In this case value of a = 1 , b = 3 and c = 5
Therefore,
Therefore, the solutions of the equation are
Question:6 Solve each of the following equations:
Answer:
Given equation is
Now, we know that the roots of the quadratic equation are given by the formula
In this case value of a = 1 , b = -1 and c = 2
Therefore,
Therefore, the solutions of equation are
Question:7 Solve each of the following equations:
Answer:
Given equation is
Now, we know that the roots of the quadratic equation is given by the formula
In this case the value of
Therefore,
Therefore, the solutions of the equation are
Question:8 Solve each of the following equations:
Answer:
Given equation is
Now, we know that the roots of the quadratic equation are given by the formula
In this case the value of
Therefore,
Therefore, the solutions of the equation are
Question:9 Solve each of the following equations:
Answer:
Given equation is
Now, we know that the roots of the quadratic equation is given by the formula
In this case the value of
Therefore,
Therefore, the solutions of the equation are
Question:10 Solve each of the following equations:
Answer:
Given equation is
Now, we know that the roots of the quadratic equation are given by the formula
In this case the value of
Therefore,
Therefore, the solutions of the equation are
Complex numbers and quadratic equations class 11 solutions - Miscellaneous Exercise
Question:1 Evaluate .
Answer:
The given problem is
Now, we will reduce it into
Now,
Therefore, answer is
Question:2 For any two complex numbers and , prove that
Answer:
Let two complex numbers are
Now,
Hence proved
Question:3 Reduce to the standard form.
Answer:
Given problem is
Now, we will reduce it into
Now, multiply numerator an denominator by
Therefore, answer is
Question:4 If , prove that
Answer:
the given problem is
Now, multiply the numerator and denominator by
Now, square both the sides
On comparing the real and imaginary part, we obtain
Now,
Hence proved
Question:5(i) Convert the following in the polar form:
Answer:
Let
Now, multiply the numerator and denominator by
Now,
let
On squaring both and then add
Now,
Since the value of is negative and is positive this is the case in II quadrant
Therefore,
Therefore, the required polar form is
Question:5(ii) Convert the following in the polar form:
Answer:
Let
Now, multiply the numerator and denominator by
Now,
let
On squaring both and then add
Now,
Since the value of is negative and is positive this is the case in II quadrant
Therefore,
Therefore, the required polar form is
Question:6 Solve each of the equation:
Answer:
Given equation is
Now, we know that the roots of the quadratic equation are given by the formula
In this case the value of
Therefore,
Therefore, the solutions of requires equation are
Question:7 Solve each of the equation:
Answer:
Given equation is
Now, we know that the roots of the quadratic equation are given by the formula
In this case the value of
Therefore,
Therefore, the solutions of requires equation are
Question:8 Solve each of the equation: .
Answer:
Given equation is
Now, we know that the roots of the quadratic equation are given by the formula
In this case the value of
Therefore,
Therefore, the solutions of requires equation are
Question:9 Solve each of the equation:
Answer:
Given equation is
Now, we know that the roots of the quadratic equation are given by the formula
In this case the value of
Therefore,
Therefore, the solutions of requires equation are
Question:10 If , find .
Answer:
It is given that
Then,
Now, multiply the numerator and denominator by
Now,
Therefore, the value of
is
Question:11 If , prove that .
Answer:
It is given that
Now, we will reduce it into
On comparing real and imaginary part. we will get
Now,
Hence proved
Question:12(ii) Let Find
Answer:
It is given that
Therefore,
NOw,
Now,
Therefore,
Therefore, the answer is 0
Question:13 Find the modulus and argument of the complex number .
Answer:
Let
Now, multiply the numerator and denominator by
Therefore,
Square and add both the sides
Therefore, the modulus is
Now,
Since the value of is negative and the value of is positive and we know that it is the case in II quadrant
Therefore,
Argument
Therefore, Argument and modulus are respectively
Question:14 Find the real numbers x andy if is the conjugate of .
Answer:
Let
Therefore,
Now, it is given that
Compare (i) and (ii) we will get
On comparing real and imaginary part. we will get
On solving these we will get
Therefore, the value of x and y are 3 and -3 respectively
Question:15 Find the modulus of .
Answer:
Let
Now, we will reduce it into
Now,
square and add both the sides. we will get,
Therefore, modulus of
is 2
Question:16 If , then show that
Answer:
it is given that
Now, expand the Left-hand side
On comparing real and imaginary part. we will get,
Now,
Hence proved
Question:17 If and are different complex numbers with , then find .
Answer:
Let
and
It is given that
and
Now,
Therefore, value of is 1
Question:18 Find the number of non-zero integral solutions of the equation .
Answer:
Given problem is
Now,
x = 0 is the only possible solution to the given problem
Therefore, there are 0 number of non-zero integral solutions of the equation
Question:19 If then show that
Answer:
It is given that
Now, take mod on both sides
Square both the sides. we will get
Hence proved
Question:20 If then find the least positive integral value of .
Answer:
Let
Now, multiply both numerator and denominator by
We will get,
We know that
Therefore, the least positive integral value of is 4
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:
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.
5.3.1 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
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 |
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 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 and
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
and
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.
Happy Reading !!!
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.
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.
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.
There are 16 chapters starting from set to probability in the CBSE class 11 maths.
No, CBSE doesn’t provided NCERT solutions for any class or subject.
Here you will get the detailed NCERT solutions for class 11 maths by clicking on the link.
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