# Complex Numbers And Quadratic Equations Chapter 5 Class 11 Maths NCERT PDF

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### Chapter 5: Complex Numbers and Quadratic Equations

5.1 Introduction In earlier classes, we have studied linear equations in one and two variables and quadratic equations in one variable.

We have seen that the equation x2 + 1 = 0 has no real solution as x2 + 1 = 0 gives x2 = – 1 and square of every real number is non-negative.

So, we need to extend the real number system to a larger system so that we can find the solution of the equation x2 = – 1.

In fact, the main objective is to solve the equation ax2 + bx + c = 0, where D = b2 – 4ac < 0, which is not possible in the system of real numbers.

#### 5.3 Algebra of Complex Numbers

In this Section, we shall develop the algebra of complex numbers.

#### 5.3.1 Addition of two complex numbers Let z1

= a + ib and z2
= c + id be any two
complex numbers. Then, the sum z1

• z2
is defined as follows:
z1
• z2
= (a + c) + i (b + d), which is again a complex number.
For example, (2 + i3) + (– 6 +i5) = (2 – 6) + i (3 + 5) = – 4 + i 8

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

Express each of the complex numbers given in Exercises 1 to 10 in the form of a + ib.

1. (5i) (-3/5i)

Solution:

(5i) (-3/5i) = 5 x (-3/5) x i2

= -3 x -1 [i2 = -1]

= 3

Hence,

(5i) (-3/5i) = 3 + i0

2. i9 + i19

Solution:

i9 + i19 = (i2)4. i + (i2)9. i

= (-1)4 . i + (-1)9 .i

= 1 x i + -1 x i

= i – i

= 0

Hence,

i9 + i19 = 0 + i0

3. i-39

Solution:

i-39 = 1/ i39 = 1/ i4 x 9 + 3 = 1/ (19 x i3) = 1/ i3 = 1/ (-i) [i4 = 1, i3 = -I and i2 = -1]

Now, multiplying the numerator and denominator by i we get

i-39 = 1 x i / (-i x i)

= i/ 1 = i

Hence,

i-39 = 0 + i

4. 3(7 + i7) + i(7 + i7)

Solution:

3(7 + i7) + i(7 + i7) = 21 + i21 + i7 + i7

= 21 + i28 – 7 [i2 = -1]

= 14 + i28

Hence,

3(7 + i7) + i(7 + i7) = 14 + i28

5. (1 – i) – (–1 + i6)

Solution:

(1 – i) – (–1 + i6) = 1 – i + 1 – i6

= 2 – i7

Hence,

(1 – i) – (–1 + i6) = 2 – i7

6.

Solution:

7.

Solution:

8. (1 – i)4

Solution:

(1 – i)= [(1 – i)2]2

= [1 + i2 – 2i]2

= [1 – 1 – 2i]2 [i= -1]

= (-2i)2

= 4(-1)

= -4

Hence, (1 – i)4 = -4 + 0i

9. (1/3 + 3i)3

Solution:

Hence, (1/3 + 3i)3 = -242/27 – 26i

10. (-2 – 1/3i)3

Solution:

Hence,

(-2 – 1/3i)3 = -22/3 – 107/27i