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Have you ever faced a situation where solving an equation led to the square root of a negative number? Do you know how engineers, physicists, and mathematicians deal with this situation? Yes, this is where Complex Numbers come in! This chapter discusses the idea of Complex numbers and quadratic equations. A complex number has both real and imaginary parts. The imaginary parts are represented by i(iota). Students will carry out basic algebraic operations on complex numbers in the same way as on real numbers, such as addition, subtraction, multiplication, and division. The quadratic equations, or second-degree order equations, educate one on how to determine the roots of these equations through the discriminant and quadratic formulas. This article on NCERT solutions for class 11 maths will help students to understand all the concepts more thoroughly.
JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | Unacademy
Suggested: JEE Main: high scoring chapters | Past 10 year's papers
The NCERT solutions for Class 11 complex numbers and quadratic equations deliver step-by-step processes to solve exercises and simplify complicated matters. In addition to textbook exercises, students are recommended to solve Class 11 Maths Chapter 4 Question Answer to reinforce learning and ensure they are well prepared for exams. Through the use of resources such as the NCERT Exemplar solutions for Class 11 Maths, students can be able to solve more complex problems and enhance their knowledge of both complex numbers and quadratic equations. Students need to refer to the NCERT Class 10 Maths books to gain more knowledge. With consistent practice and a clear understanding of these concepts, students can approach both theoretical and applied mathematics confidently. For syllabus, notes, and PDF, refer to this link: NCERT.
Class 10 Maths chapter 4 solutions Exercise: 4.1 Page number: 82-83 Total questions: 14 |
Question 1: Express the given complex number in the form a+ib: (5i)(-3i/5)?
Answer:
Given
Question 2: Express each of the complex number in the form
Answer:
We know that
Now, we will reduce
Now, in the form of
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
Now, in the form of
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, the answer is
Class 10 Maths chapter 4 solutions Miscellaneous Exercise Page number: 85-86 Total questions: 14 |
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
Answer:
Let two complex numbers are
Now,
Hence proved
Question 3: Reduce
Answer:
Given problem is
Now, we will reduce it into
Now, multiply numerator an denominator by
Therefore, answer is
Question 4: If
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: If
Answer:
It is given that
Then,
Now, multiply the numerator and denominator by
Now,
Therefore, the value of
Question 6: If
Answer:
It is given that
Now, we will reduce it into
On comparing real and imaginary part. we will get
Now,
Hence proved
Question 7 (i): Let
Answer:
It is given that
Now,
And
Now,
Now,
Therefore, the answer is
Question 7 (ii): Let
Answer:
It is given that
Therefore,
NOw,
Now,
Therefore,
Therefore, the answer is 0.
Question 8: Find the real numbers x and y if
Answer:
Let
Therefore,
Now, it is given that
Comparing (i) and (ii), we will get
On comparing the real and imaginary part, we will get
On solving these we will get
Therefore, the values of x and y are 3 and -3 respectively
Question 9: Find the modulus of
Answer:
Let
Now, we will reduce it into
Now,
square and add both the sides. we will get,
Therefore, the modulus of
Question 10: If
Answer:
it is given that
Now, expand the Left-hand side
On comparing real and imaginary part. we will get,
Now,
Hence proved
Question 11: If
Answer:
Let
It is given that
and
Now,
Therefore, value of
Question 12: 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 13: If
Answer:
It is given that
Now, take mod on both sides
Square both sides, we will get
Hence proved
Question 14: If
Answer:
Let
Now, multiply both numerator and denominator by
We will get,
We know that
Therefore, the least positive integral value of
Also, read,
Question:
If
Solution:
We have,
Now if
Then
Hence, the correct answer is 11.
The topics discussed in the NCERT Solutions for class 11, chapter 4, Complex Numbers and Quadratic Equations are:
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.
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.
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.
Here are some approaches that students can use to solve complex numbers and quadratic equation problems.
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The links below allow students to access all the Maths solutions from the NCERT book.
Also, read,
Given below are the subject-wise NCERT solutions of class 11 NCERT:
NCERT solutions for class 11 biology |
NCERT solutions for class 11 maths |
NCERT solutions for class 11 chemistry |
NCERT solutions for class 11 physics |
Here are some useful links for NCERT books and the NCERT syllabus for class 11:
Addition/Subtraction: Combine real & imaginary parts separately.
Multiplication: Distribute like algebra, using
Division: Multiply by the conjugate to simplify.
Modulus: Distance from origin,
Conjugate: Flip the sign of
If the discriminant
Definition: If
Its Uses:
Expressed as:
Where:
-
-
Useful for multiplication, division, and powers of complex numbers.
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