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Class XI – NCERT NCERT – Maths Maths
Chapter 5 Complex Numbers and Quadratic Equations Exercise Exerc ise 5.1
Question 1:
3 Express the given complex number in the form a ib : 5i i 5 Solution 1: 3 3 5i i 5 i i 5 5
3i 2 3 1
i 2 1
=3
Question 2:
Express the given complex number in the form a ib :i9 i19 Solution 2:
i9 i19 i 421 i 44 3 2
4
i 4 i i 4 i 3 i 4 1, i 3 i
1 i 1 i
i i
0
Question 3:
Express the given complex number in the form a ib : i 39 Solution 3:
i
9
i 493 i4 i 3
39
9
1 i3
1 i
3
1
i
i 4 1 i3 i
1 i
i i
i
2
i
i i 1
Question 4:
i 2 1
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Class XI – NCERT NCERT – Maths Maths
Chapter 5 Complex Numbers and Quadratic Equations
Solution 4:
3 7 i7 i 7 i7 21 21i 7i 7i 2
21 28i 7 1
i 2 1
14 28i
Question 5:
Express the given complex number in the form a ib : 1 i 1 i6 . Solution 5:
1 i 1 i6 1 i 1 6i
2 7i
Question 6:
5 1 2 Express the given complex number in the form a ib : i 4 i 2 5 5 Solution 6: 5 1 2 5 i 5 4 i 2
1
2
5
i 4 i
5 5 2 1 2 5 4 i 5 5 2
21 i 5 10 19 21 i
19
5
10
Question 7:
1 7 1 4 Express the given complex co mplex number in the form a ib : i 4 i i 3 3 3 3 Solution 7:
1 7 1 4 i 4 i i 3 3 3 3 1
7
1
4
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Class XI – NCERT NCERT – Maths Maths
17 3
i
Chapter 5 Complex Numbers and Quadratic Equations
5 3
Question 8:
Express the given complex number in the form a ib : 1 i
4
Solution 8: 2 1 i 1 i 4
12 i2 2i
2
2
2
1 1 2i
2i
2
2i 2i i 2 1
4i 2 4
Question 9: 3
1 Express the given complex number in the form a ib : 3i 3 Solution 9: 3
3
3 1 1 1 1 3 i 3 i 3 3 i 3 i 3 3 3 3 1 1 27i3 3i 3i 27 3
1 27 1
27 i i 9i 2
i 3 i
27i i 9
i 2 1
27 1
9 i 27 1 27 242 242 26i 27
Question 10:
3
1
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Chapter 5 Complex Numbers and Quadratic Equations
Class XI – NCERT NCERT – Maths Maths
3
3
1 1 3 2 i 1 2 i 3 3 3 i 3 i i 2 3 2 2 3 3 3
i3 i 8 2i 2 3 27 i 2i 2 8 4i 3 27 i 2 8 4i 3 27 22 107i 27 3
22 107 3
27
i 3 i i 2 1
i
Question 11: Find the multiplicative inverse of the co mplex number number 4 3i . Solution 11: Let z 4 3i Then, 2
2 z 4 3i and z 4 3 16 9 25
Therefore, the multiplicative inverse of 4 3i is given by z 1
z z
2
4 3i 25
4 25
3 2 25 5
i
Question 12:
Find the multiplicative inverse of the complex number Solution 12:
Let z 5 3i 2
Then, z 5 3i and z
5
2
32 5 9 14
Therefore, the multiplicative inverse of z
1
z 2
5 3i
5
3i
5 3i
5 3i
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Class XI – NCERT NCERT – Maths Maths
Chapter 5 Complex Numbers and Quadratic Equations
Find the multiplicative inverse of the complex number i Solution 13: Let z i 2
Then, z i and z 12 1 Therefore, the multiplicative inverse of i is given by z 1
z
i
i
2
z
1
Question 14: Express the following expression in the t he form of a ib .
3 i 5 3 i 5 3 2i 3 i 2 Solution 14:
3 i 5 3 i 5 3 2i 3 i 2 3 i 5 2
2
3 2i 3 2i
a b a b a 2 b 2
9 5i 2 2 2i
9 5 1 2 2i 95 i
2 2i 14i 2 2i 2
i
i 2 1
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Chapter 5 Complex Numbers and Quadratic Equations
Class XI – NCERT NCERT – Maths Maths
Exercise 5.2 Question 1:
Find the modulus and the argument ar gument of the complex number z 1 i 3 Solution 1:
z 1 i 3 Let r cos 1 and r sin 3 On squaring and adding, we obtain
r cos r sin 1 2
2
2
3
2
r 2 cos2 sin 2 1 3 cos2 sin 2 1 onventio entiona nall lly, y, r 0 Conv
r 2 4 r 4 2
Modulus = 2
3 2 cos 1 and 2 sin cos cos
1
and sin
3
2 2 Since both the values of sin and cos negative and sin and cos are negative in III quadrant, 2 Argument 3 3 Thus, the modulus and argument of the complex number
1 3i are 2 and
respectively.
Question 2:
Find the modulus and the argument ar gument of the complex number z 3 i Solution 2:
2 3
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Chapter 5 Complex Numbers and Quadratic Equations
Class XI – NCERT NCERT – Maths Maths
cos
3 2
and sin
1 2
5 6 6
[As lies in the II quadrant]
Thus, the modulus an argument of the t he complex number 3 i are 2 and
5 6
respectively.
Question 3: Convert the given complex number in polar form: 1 i Solution 3:
1 i Let r cos 1 and r sin 1 On squaring and adding, we obtain r 2 cos2 r 2 sin2 12 1
2
r 2 cos2 sin 2 1 1
r 2 2 r 2
Conventionally, r 0
cos 1 and 2 co cos cos
4
1 2
1 2 sin
and sin
1 2
lies in the the IVqua IVquadra drant As lie
1 i r cos i r sin 2 cos i 2 sin 2 cos i sin 4 4 4 4 This is the required polar form.
Question 4: Convert the given complex number in polar form:
1 i
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Chapter 5 Complex Numbers and Quadratic Equations
Class XI – NCERT NCERT – Maths Maths
cos cos
1
and 2 3 4 4 It can be written,
2 sin 1 [As lies in the II quadrant]
1 i r cos i r sin 2 cos
3 3 3 3 i 2 sin 2 cos i sin 4 4 4 4
This is the required polar form.
Question 5: Convert the given complex number in polar form:
1 i
Solution 5:
1 i Let r cos 1 and r sin 1
On squaring and adding, we obtain 2
2
r 2 cos2 r 2 sin2 1 1
r 2 cos2 sin2 1 1
r 2 2 r 2 2 cos 1 and cos cos
1 2
[Conventionally, r 0 ]
2 sin 1
and sin
1 2
3 4 4
1 i r cos i r sin 2 cos 3 3 2 cos i sin 4 4 This is the required polar form.
[As lies in the III quadrant]
3 4
i 2 sin
3 4
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Chapter 5 Complex Numbers and Quadratic Equations
Class XI – NCERT NCERT – Maths Maths
r 2 9 r 9 3 3 cos 3 and 3 sin 0 cos 1 and sin 0
[Conventionally, [Conventionally, r > 0]
3 r cos i r sin 3 cos i 3sin 3 cos i sin This is the required polar form.
Question 7:
Convert the given complex number in polar form:
3 i
Solution 7:
3 i Let r cos 3 and r sin 1 On squaring and adding, we obtain r 2 cos2 r 2 sin 2
3
2
12
r 2 cos2 sin 2 3 1
r 2 4 r 4 2
Conventionally, r 0
2 cos 3 and 2 sin 1
cos cos
6
3 2
and sin
1 2 [As lies in the I quadrant]
3 i r cos i r sin 2 cos This is the required polar form.
i 2 sin 2 cos i sin 6 6 6 6
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Chapter 5 Complex Numbers and Quadratic Equations
Class XI – NCERT NCERT – Maths Maths
r 2 1 r 1 1
Conventionally, r 0
cos 0 and sin 1
2
i r cos i r sin cos
i sin 2 2
This is the required polar form.
Exercise 5.3 Question 1:
Solve the equation equat ion x2 3 0 Solution 1:
The given quadratic equation is x2 3 0 On comparing the given equation with ax2 bx c 0, We obtain a 1, b 0, and c 3 Therefore, the discriminant of the given equation is
D b2 4ac 02 4 1 3 12 Therefore, the required solutions are
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Class XI – NCERT NCERT – Maths Maths
Chapter 5 Complex Numbers and Quadratic Equations
Question 3:
Solve the equation equat ion x2 3x 9 0 Solution 3:
The given quadratic equation is x2 3x 9 0 On comparing the given equation with wit h ax2 bx c 0 , We obtain a 1, b 3, and c 9 Therefore, the discriminant of the given equation is
D b2 4ac 32 41 9 9 36 27 Therefore, the required solutions are
b D 2a
3 27 2 1
3 3 3 2
3 3 3i 2
Question 4: Solve the equation equat ion x2 x 2 0 Solution 4: The given quadratic equation is x2 x 2 0 On comparing the given equation with ax2 bx c 0 , We obtain a 1, b 1, and c 2 Therefore, the discriminant of the given equation is 2 2 D b 4ac 1 4 1 2 1 8 7
Therefore, the required solutions are
1 i
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Class XI – NCERT NCERT – Maths Maths
Chapter 5 Complex Numbers and Quadratic Equations
Solution 6:
The given quadratic equation is x2 x 2 0 On comparing the given equation with ax2 bx c 0 , We obtain a 1, b 1, and c 2 Therefore, the discriminant of the given equation is 2
D b2 4ac 1 4 1 2 1 8 7 Therefore, the required solutions are
b D 2a
1 7 1 7i 2 1 2
1 i
Question 7:
Solve the equation equat ion
2 x2 x 2 0
Solution 7:
2 x2 x 2 0 On comparing the given equation with ax2 bx c 0 , The given quadratic equation is
We obtain a 2 , b 1, and c 2 Therefore, the discriminant of the given equation is
D b2 4ac 12 4 2 2 1 8 7 Therefore, the required solutions are
b D
1 7
1 7i
1 i
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Chapter 5 Complex Numbers and Quadratic Equations
Class XI – NCERT NCERT – Maths Maths
Solve the equation equat ion x2 x
1 2
0
Solution 9:
The given quadratic equation is x2 x This equation can also be written as
1 2
0
2 x2 2x 1 0
On comparing the given equation with ax2 bx c 0 , we obtain a 2, b 2, and c 1
Discriminant D b2 4ac
2
2
4
2 1 2 4
Therefore, the required solutions are
2 2 4 2 2 2 1 2 2 2a 2 2 2 2 2 2 2 2 1 i 1 i 2 2 b D
1
2 2 1 i 2
Question 10:
Solve the equation equat ion x2
x
1 0
2
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Class XI – NCERT NCERT – Maths Maths
Chapter 5 Complex Numbers and Quadratic Equations Miscellaneous Exercise
Question 1:
18 1 25 Evaluate: i i
3
Solution 1:
18 1 25 i i
3
1 i 442 461 i
3
4 1 4 2 i i 6 i4 i 1 i 2 i
3
i 4 1
1 i 1 i i i 1 2 i
3
3
3
1 i 3
3
3
1 1 i
i 2 1
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Class XI – NCERT NCERT – Maths Maths
Chapter 5 Complex Numbers and Quadratic Equations
i 2 1
x1x2 ix1 y2 iy1x2 y1 y2
x1x2 y1 y2 i x1 y2 y1x2 Re z1z2 x1x2 y1y2 Re z1z2 Re z1 Re z2 Im z1 Im z2
Hence, proved.
Question 3: 2 3 4i 1 Reduce to the standard form 1 4i 1 i 5 i Solution 3:
2 3 4i 1 i 2 1 4i 3 4i 1 1 4 i 1 i 5 i 1 4i 1 i 5 i
1 i 2 8i 3 4i
1 9i 3 4i
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Class XI – NCERT NCERT – Maths Maths
2
x iy
Chapter 5 Complex Numbers and Quadratic Equations
ac bd i ad bc
x2 y 2 2ixy
c2 d 2 ac bd i ad bc
c2 d 2 On comparing real and imaginary parts, we obtain ac bd ad bc x2 y 2 2 , 2 xy ......1 c d2 c2 d 2
x
2
2
2
y 2 x2 y2 4x2 y 2 2
ac bd ad bc 2 2 2 2 c d c d a2c2 b2 d 2 2acbd a 2d 2 b 2c 2 2adbc 2 2 2 c d
a2c2 b2 d 2 a2 d 2 b2 c2
c
2
d 2
2
[Using (1)]
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Class XI – NCERT NCERT – Maths Maths
Chapter 5 Complex Numbers and Quadratic Equations
Let r cos 1 and r sin 1
On squaring and adding, we obtain r 2 cos2 sin 2 1
r 2 cos2 sin 2 2
r 2 2 r 2 cos 1 an and 2 ssiin 1 2 co 1 1 and sin cos cos 2
2
3 4 4 z r cos i r sin
2 cos
3 4
i 2 sin
cos2 sin 2 1 [Conventionally, r 0 ]
[As lies in II quadrant]
3 3 2 cos i sin 4 4 4
3
This is the required polar form.
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Chapter 5 Complex Numbers and Quadratic Equations
Class XI – NCERT NCERT – Maths Maths
Solve the equation equat ion 3 x2 4x
20 3
0
Solution 6:
The given quadratic equation is 3 x2 4x
20
0 3 This equation can also be written as 9 x2 12x 20 0 On comparing this equation with ax2 bx c 0 , we obtain a 9, b 12 and c 20 Therefore, the discriminant of the given equation is 2
D b2 4ac 12 4 9 20 144 720 576 Therefore, the required solutions are
12 576 12 576 i 2a 2 9 18 12 24i 6 2 4i 2 4i 2 4 i b D
18
18
3
3
3
1 i
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Class XI – NCERT NCERT – Maths Maths
Chapter 5 Complex Numbers and Quadratic Equations
2
D b2 4ac 10 4 27 1 100 108 8 Therefore, the required solutions are
b D 2a
5 2i 27
10 8 10 2 2 i 2 27 54
5 27
2 27 27
1 i
i
Question 9:
Solve the equation equat ion 21 x2 28x 10 0 Solution 9:
The given quadratic equation is 21 x2 28x 10 0 On comparing this equation with ax2 bx c 0 , we obtain a 21, b 28 and c 10 Therefore, the discriminant of the given equation is D b2 4ac 28
2
4 2110 7 78 84 840 56
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Class XI – NCERT NCERT – Maths Maths
Chapter 5 Complex Numbers and Quadratic Equations
1 i 12 12 2 Thus, the value of
z1 z2 1 z1 z2 1
is
2.
Question 11:
If a ib
x i 2
2
2
1
, prove that a 2 b2
x
2
1
2 2
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Class XI – NCERT NCERT – Maths Maths
Chapter 5 Complex Numbers and Quadratic Equations
Solution 12: z1 2 i, z2 2 i
(i) z1z2 2 i 2 i 4 2i 2i i 4 4i 1 3 4i 2
z1 2 i z1 z2 3 4i
z1
2i
On multiplying numerator and denominator by 2 i , we obtain
3 4i 2 i
6 3i 8i 4i 2
6 11i 4 1
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Chapter 5 Complex Numbers and Quadratic Equations
Class XI – NCERT NCERT – Maths Maths
1 2
cos
cos
1 2
1 2
and
1 2
sin
and sin
3 4 4
1 2
1 2 [As lies in the II quadrant]
Therefore, the modulus and argument of the given complex number are
1 2
and
3 4
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Chapter 5 Complex Numbers and Quadratic Equations
Class XI – NCERT NCERT – Maths Maths
2
1 i 1 i 1 i 1 i 1 i 1 i
1 i
1 i
2
1 i 2 2i 1 i 2 2i
12 12 4i
2i 2 1 i 1 i 1 i
1 i
2i 22 2
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Class XI – NCERT NCERT – Maths Maths
It is given that,
1
x2 y 2 1
x2 y2 1 .......i 1
x iy a ib 1 a ib x iy
x a i y b
Chapter 5 Complex Numbers and Quadratic Equations
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Class XI – NCERT NCERT – Maths Maths
2 x x 0
Chapter 5 Complex Numbers and Quadratic Equations
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Class XI – NCERT NCERT – Maths Maths
Chapter 5 Complex Numbers and Quadratic Equations
is some integer. m 4k where k is
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Thank You