Previously** **you must have learnt about quadratic and linear equations in one or two variables. The chapter Quadratic Equations and Complex Numbers will advance your knowledge by introducing the concept of complex numbers. You will learn how to find the solutions of equations beyond the real number. So let’s get started and quickly go through this important topic of** class 11 maths syllabus**.

##### This Blog Includes:

## What are Complex Numbers?

Class 11 maths NCERT solutions state that there is a number root over -1. The number can also be written in the form of i^{2}=-1. It goes to show that it is a solution to the equation x^{2}+1=0. It explains a number in the form of **e+ **if where **e** and **f** are real numbers is referred to as a complex number.

For a given equation of the form x^{2}+1=0, there will be no real solution because x^{2}=-1 and the square of all real numbers are non-negative. Thus, we need to extend the real number system to a larger system for finding the solutions of a quadratic equation- x^{2}= -1. Our main objective is to find the solution for the equation ax^{2}+bx+c=0 where D<0 and equal to b^{2}-4ac that is not feasible in the system of real numbers.

The chapter illustrates that, a complex number of the form **q=e+jf**, then **e** is referred to as the real portion denoted by R en and **f** is referred to the imaginary portion denoted by I mn of the complex number n. Hence, if n is in the form of **n=4+i10** then **R en=4** and **Im n=10.** For instance, if there are two complex numbers **n1=e+jf **and **n2=l+km** are equal if **e=l** and **f=m.**

**Example 1**. If 4l + i(3l – m) = 3 + i (– 6), where l and m are real numbers, then find the values of l and m.

**Solution: **

We have 4l + i (3l – m) = 3 + i (–6) … (1)

Equating the real and the imaginary portions of (1),

we get 4l = 3, 3l – m = – 6, which, on solving simultaneously, give l=3/4 and m=33/4

*Therefore, the *answer is l=3/4 and m=33/4

Previously** **you must have learnt about quadratic and linear equations in one or two variables. The chapter Quadratic Equations and Complex Numbers will advance your knowledge by introducing the concept of complex numbers. You will learn how to find the solutions of equations beyond the real number. So let’s get started and quickly go through this important topic of** class 11 maths syllabus**.

##### This Blog Includes:

## What are Complex Numbers?

Class 11 maths NCERT solutions state that there is a number root over -1. The number can also be written in the form of i^{2}=-1. It goes to show that it is a solution to the equation x^{2}+1=0. It explains a number in the form of **e+ **if where **e** and **f** are real numbers is referred to as a complex number.

For a given equation of the form x^{2}+1=0, there will be no real solution because x^{2}=-1 and the square of all real numbers are non-negative. Thus, we need to extend the real number system to a larger system for finding the solutions of a quadratic equation- x^{2}= -1. Our main objective is to find the solution for the equation ax^{2}+bx+c=0 where D<0 and equal to b^{2}-4ac that is not feasible in the system of real numbers.

The chapter illustrates that, a complex number of the form **q=e+jf**, then **e** is referred to as the real portion denoted by R en and **f** is referred to the imaginary portion denoted by I mn of the complex number n. Hence, if n is in the form of **n=4+i10** then **R en=4** and **Im n=10.** For instance, if there are two complex numbers **n1=e+jf **and **n2=l+km** are equal if **e=l** and **f=m.**

**Example 1**. If 4l + i(3l – m) = 3 + i (– 6), where l and m are real numbers, then find the values of l and m. **Solution: **

We have 4l + i (3l – m) = 3 + i (–6) … (1)

Equating the real and the imaginary portions of (1),

we get 4l = 3, 3l – m = – 6, which, on solving simultaneously, give l=3/4 and m=33/4*Therefore, the *answer is l=3/4 and m=33/4

**Example 2:** Express the following in the form of a+bi: (-5i)(1/8i) (-i)(2i)( -1/8i)3**Solution :**

(i) (-5i)(1/8i)= -5/8i2= -5/8(-1)=5/8=5/8+i()

(ii) (-i)(2i)(-1/8i)3= 2* (1/8*8*8)*i5= 1/256(i2)2 -> i=(1/256)i

**Example:3 ** Express (5 – 3i) ^{3} in the form a + ib**Solution **

We have, (5 – 3i) ^{3} = 5 ^{3} – 3 × 5^{2} × (3i) + 3 × 5 (3i) ^{2} – (3i) ^{3} = 125 – 225i – 135 + 27i = – 10 – 198i.

**Take This Maths Quiz If You Consider Yourself Genius!**

## Addition of Two Complex Numbers

The next topic that that chapter complex numbers exemplifies is the addition of two complex numbers. Let **n1 = l + jm **and **n2 = p + iq **be any two complex numbers. Then, the sum n1 + n2 is defined as follows: **n1 + n2 = (l + p) + i (m + q)**, which is again a complex number.

For example, (4 + i6) + (– 7 +i9) = (4 – 7) + i (6 + 9) = – 3 + i 16.** **Here are some properties that satisfy addition of complex numbers-

**The Closure Law:**It states that when two complex numbers are added it results in the formation of a complex number i.e., n1 + n2 is a complex number for all complex numbers n1 and n2**The Commutative Law**: Suppose you are given two complex numbers n1 and n2 , then n1 + n2 = n2 + n1**The Associative Law:****Class 11 maths NCERT solutions**explains**The existence of Additive Identity**: Suppose you are given a complex number of the form 0 + i 0 (denoted as 0), referred to as the additive identity or the zero complex number, such that, for every complex number n, n + 0 = n**The Existence of Additive Inverse:****Class 11 maths NCERT solutions**explains

**Permutation and Combination Class 11**

**Vedic Maths** **Maths for Competitive Exams**

## Subtraction of Two Complex Numbers

Given any two complex numbers n1 and n2, the subtraction n1 – n2 is defined as follows:** n1 – n2 = n1 + (– n2 ).**

For example, (6 + 3i) – (2 – i) = (6 + 3i) + (– 2 + i ) = 4 + 4i and (2 – i) – (6 + 3i) = (2 – i) + ( – 6 – 3i) = – 4 – 4i

## Multiplication

Let **n1 = a + ib** and **n2 = c + id **be any two complex numbers. Then, the multiplication **n1 n2 **is defined as follows: **n1 n2 = (ac – bd) + i(ad + bc).**

For example, (3 + i5) (2 + i6) = (3 × 2 – 5 × 6) + i(3 × 6 + 5 × 2) = – 24 + i28. The chapter also** **elaborates that** **when you have to multiply complex numbers it has to abide by the undermentioned properties.

**Closure law:**When you multiply two complex numbers a resulting number is also a complex number, the multiplication n1 n2 is a complex number for all complex numbers n1 and n2**Commutative law:**For any two complex numbers n1 and n2, n1 n2 = n2 n1**The associative law:**For any three complex numbers n1 , n2 , n3 , (n1 n2 ) n3 = n1 (n2 n3 )**Existence of multiplicative identity:**Suppose you are given a complex number of form 1 + i 0 (denoted as 1), referred to the multiplicative identity such that n.1 = n, for every complex number n**Existence of multiplicative inverse:**Suppose you are given non zero complex number of the form n = a + ib or a + bi(a ≠ 0, b ≠ 0), we have the complex number 22 22 a –b i ab ab + + + (denoted by 1 n or n–1 ), referred the multiplicative inverse of n such that 1 n. 1 n = (the multiplicative identity.

Class 11 maths NCERT solutions explains that** **the distributive law: For any three complex numbers n1 , n2 , n3 ,

(a) n1 (n2 + n3 ) = n1 n2 + n1 n3

(b) (n1 + n2 ) n3 = n1 n3 + n2 n3

## Argand Plane and Polar Representation

Similar to each ordered pair of real numbers (n1, n2), we will have a point unique in the N1N2 plane along with its opposite which will refer to a set of lines mutually perpendicular referred to as the n1-axis and the n2-axis. The complex number n1 + in2 which corresponds to the ordered pair (n1, n2) can be represented geometrically as the unique point P(n1, n2) in the N1N2-plane and vice-versa. Some complex numbers such as 4+81, – 4 + 6, 0 + 1i, 4 + 0i, – 10 –2i and 1 – 2i which correspond to the ordered pairs (4, 8), ( – 4, 6), (0, 1), (4, 0), ( –10, –2), and (1, – 2). Class 11 maths NCERT solutions states that** **when a complex number is contained in a plane then each of the points is called the complex or the Argand plane.

Thus, we hope that through these notes on complex numbers, we have helped you in understanding one of the most essential concepts of class 11 mathematics. For expert advice related to upcoming crucial decisions, get in touch with our **Leverage Edu** experts. Book an e-meeting.