Commerce Exam > Commerce Notes > Mathematics (Maths) Class 11 > RD Sharma Class 11 Solutions Chapter - Complex Numbers

RD Sharma Class 11 Solutions Chapter - Complex Numbers

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 Page 1


13. Complex Numbers
Exercise 13.1
1. Question
Evaluate the following:
(i) i
457
(ii) i
528
(iii) 
(iv) 
(v)
(vi) (i
77
 + i
70
 + i
87
 + i
414
 )
3
(vii) (vii) i
30
 + i
40
 + i
60
(viii) i
49
 + i
68
 + i
89
 + i
118
Answer
i. i
457
 = i 
(456 + 1)
= i
4(114)
 × i
= (1)
114
 × i = i since i
4
 = 1
ii. i
528
 = i
4(132)
= (1)
132
 =1 since i
4
 = 1
iii. 
 since i
4
 = 1
 = – 1 since i
2
 = – 1
iv. 
[since i
4
 = 1]
v. 
 = (i – i) = 0
[since ]
Page 2


13. Complex Numbers
Exercise 13.1
1. Question
Evaluate the following:
(i) i
457
(ii) i
528
(iii) 
(iv) 
(v)
(vi) (i
77
 + i
70
 + i
87
 + i
414
 )
3
(vii) (vii) i
30
 + i
40
 + i
60
(viii) i
49
 + i
68
 + i
89
 + i
118
Answer
i. i
457
 = i 
(456 + 1)
= i
4(114)
 × i
= (1)
114
 × i = i since i
4
 = 1
ii. i
528
 = i
4(132)
= (1)
132
 =1 since i
4
 = 1
iii. 
 since i
4
 = 1
 = – 1 since i
2
 = – 1
iv. 
[since i
4
 = 1]
v. 
 = (i – i) = 0
[since ]
vi. (i
77
 + i
70
 + i
87
 + i
414
 )
3
 = (i
(76 + 1)
 + i
(68 + 2)
 + i
(84 + 3)
 + i
(412 + 2)
 ) 
3
(i
77
 + i
70
 + i
87
 + i
414
 )
3
 = (i + i
2
 + i
3
 + i
2
 )
3
[since i
3
 = – i, i
2
 = – 1]
= (i + (– 1) + (– i) + (– 1))
3
 = (– 2)
3
(i
77
 + i
70
 + i
87
 + i
414
 )
3
 = –8
vii. i
30
 + i
40
 + i
60
 = i
(28 + 2) +
 i
40
 + i
60
= (i
4
)
7
 i
2
 + (i
4
)
10
 + (i
4
)
15
= i
2
 + 1
10
 + 1
15
 = – 1 + 1 + 1 = 1
viii. i
49
 + i
68
 + i
89
 + i
118
 = i
(48 + 1)
 + i
68
 + i
(88 + 1)
 + i
(116 + 2)
= (i
4
)
12
×i + (i
4
)
17
 + (i
4
)
11
×i + (i
4
)
29
×i
2
= i + 1 + i – 1 = 2i
2. Question
Show that 1 + i
10
 + i
20
 + i
30
 is a real number ?
Answer
1 + i
10
 + i
20
 + i
30
 = 1 + i
(8 + 2) +
 i
20
 + i
(28 + 2)
= 1 + (i
4
)
2
 × i
2
 + (i
4
)
5
 + (i
4
)
7
 × i
2
= 1 – 1 + 1 – 1 = 0
[ since i
4
 = 1, i
2
 = – 1]
Hence , 1 + i
10
 + i
20
 + i
30
 is a real number.
3 A. Question
Find the value of following expression:
i
49
 + i
68
 + i
89
 + i
110
Answer
i
49
 + i
68
 + i
89
 + i
110
 = i
(48 + 1)
 + i
68
 + i
(88 + 1)
 + i
(108 + 2)
= (i
4
)
12
 × i + (i
4
)
17
 + (i
4
)
11
 × i + (i
4
)
27
 × i
2
= i + 1 + i – 1 = 2i
[since i
4
 = 1, i
2
 = – 1]
i
49
 + i
68
 + i
89
 + i
110
 = 2i
3 B. Question
Find the value of following expression:
i
30
 + i
80
 + i
120
Answer
i
30
 + i
80
 + i
120
 = i
(28 + 2)
 + i
80
 + i
120
= (i
4
)
7
 × i
2
 + (i
4
)
20
 + (i
4
)
30
= – 1 + 1 + 1 = 1
Page 3


13. Complex Numbers
Exercise 13.1
1. Question
Evaluate the following:
(i) i
457
(ii) i
528
(iii) 
(iv) 
(v)
(vi) (i
77
 + i
70
 + i
87
 + i
414
 )
3
(vii) (vii) i
30
 + i
40
 + i
60
(viii) i
49
 + i
68
 + i
89
 + i
118
Answer
i. i
457
 = i 
(456 + 1)
= i
4(114)
 × i
= (1)
114
 × i = i since i
4
 = 1
ii. i
528
 = i
4(132)
= (1)
132
 =1 since i
4
 = 1
iii. 
 since i
4
 = 1
 = – 1 since i
2
 = – 1
iv. 
[since i
4
 = 1]
v. 
 = (i – i) = 0
[since ]
vi. (i
77
 + i
70
 + i
87
 + i
414
 )
3
 = (i
(76 + 1)
 + i
(68 + 2)
 + i
(84 + 3)
 + i
(412 + 2)
 ) 
3
(i
77
 + i
70
 + i
87
 + i
414
 )
3
 = (i + i
2
 + i
3
 + i
2
 )
3
[since i
3
 = – i, i
2
 = – 1]
= (i + (– 1) + (– i) + (– 1))
3
 = (– 2)
3
(i
77
 + i
70
 + i
87
 + i
414
 )
3
 = –8
vii. i
30
 + i
40
 + i
60
 = i
(28 + 2) +
 i
40
 + i
60
= (i
4
)
7
 i
2
 + (i
4
)
10
 + (i
4
)
15
= i
2
 + 1
10
 + 1
15
 = – 1 + 1 + 1 = 1
viii. i
49
 + i
68
 + i
89
 + i
118
 = i
(48 + 1)
 + i
68
 + i
(88 + 1)
 + i
(116 + 2)
= (i
4
)
12
×i + (i
4
)
17
 + (i
4
)
11
×i + (i
4
)
29
×i
2
= i + 1 + i – 1 = 2i
2. Question
Show that 1 + i
10
 + i
20
 + i
30
 is a real number ?
Answer
1 + i
10
 + i
20
 + i
30
 = 1 + i
(8 + 2) +
 i
20
 + i
(28 + 2)
= 1 + (i
4
)
2
 × i
2
 + (i
4
)
5
 + (i
4
)
7
 × i
2
= 1 – 1 + 1 – 1 = 0
[ since i
4
 = 1, i
2
 = – 1]
Hence , 1 + i
10
 + i
20
 + i
30
 is a real number.
3 A. Question
Find the value of following expression:
i
49
 + i
68
 + i
89
 + i
110
Answer
i
49
 + i
68
 + i
89
 + i
110
 = i
(48 + 1)
 + i
68
 + i
(88 + 1)
 + i
(108 + 2)
= (i
4
)
12
 × i + (i
4
)
17
 + (i
4
)
11
 × i + (i
4
)
27
 × i
2
= i + 1 + i – 1 = 2i
[since i
4
 = 1, i
2
 = – 1]
i
49
 + i
68
 + i
89
 + i
110
 = 2i
3 B. Question
Find the value of following expression:
i
30
 + i
80
 + i
120
Answer
i
30
 + i
80
 + i
120
 = i
(28 + 2)
 + i
80
 + i
120
= (i
4
)
7
 × i
2
 + (i
4
)
20
 + (i
4
)
30
= – 1 + 1 + 1 = 1
[since i
4
 = 1, i
2
 = – 1]
i
30
 + i
80
 + i
120
 = 1
3 C. Question
Find the value of following expression:
i + i
2
 + i
3
 + i
4
Answer
i + i
2
 + i
3
 + i
4
 = i + i
2
 + i
2
×i + i
4
= i – 1 + (– 1)×i + 1
since i
4
 = 1, i
2
 = – 1
= i – 1 – i + 1 = 0
3 D. Question
Find the value of following expression:
i
5
 + i
10
 + i
15
Answer
i
5
 + i
10
 + i
15
 = i
(4 + 1)
 + i
(8 + 2)
 + i
(12 + 3)
= (i
4
)
1
×i + (i
4
)
2
×i
2
 + (i
4
)
3
×i
3
= (i
4
)
1
×i + (i
4
)
2
×i
2
 + (i
4
)
3
×i
2
×i
= 1×i + 1×(– 1) + 1×(– 1)×i
= i – 1 – i = – 1
3 E. Question
Find the value of following expression:
Answer
                                    = i
10
                                         = i
8
i
2
                                   = (i
4
)
2
 i
2
 Since i
4
 = 1, i
2
 = -1           
 
                                = (1)
2
 (-1)                               
 = -1
                                   
3 F. Question
Find the value of following expression:
1 + i
2
 + i
4
 + i
6
 + i
8
 + ... + i
20
Answer
1 + i
2
 + i
4
 + i
6
 + i
8
 + ... + i
20
 = 1 + (– 1) + 1 + (– 1) + 1 + ... + 1
= 1
3 G. Question
Find the value of following expression:
(1 + i)
6
 + (1 – i)
3
Page 4


13. Complex Numbers
Exercise 13.1
1. Question
Evaluate the following:
(i) i
457
(ii) i
528
(iii) 
(iv) 
(v)
(vi) (i
77
 + i
70
 + i
87
 + i
414
 )
3
(vii) (vii) i
30
 + i
40
 + i
60
(viii) i
49
 + i
68
 + i
89
 + i
118
Answer
i. i
457
 = i 
(456 + 1)
= i
4(114)
 × i
= (1)
114
 × i = i since i
4
 = 1
ii. i
528
 = i
4(132)
= (1)
132
 =1 since i
4
 = 1
iii. 
 since i
4
 = 1
 = – 1 since i
2
 = – 1
iv. 
[since i
4
 = 1]
v. 
 = (i – i) = 0
[since ]
vi. (i
77
 + i
70
 + i
87
 + i
414
 )
3
 = (i
(76 + 1)
 + i
(68 + 2)
 + i
(84 + 3)
 + i
(412 + 2)
 ) 
3
(i
77
 + i
70
 + i
87
 + i
414
 )
3
 = (i + i
2
 + i
3
 + i
2
 )
3
[since i
3
 = – i, i
2
 = – 1]
= (i + (– 1) + (– i) + (– 1))
3
 = (– 2)
3
(i
77
 + i
70
 + i
87
 + i
414
 )
3
 = –8
vii. i
30
 + i
40
 + i
60
 = i
(28 + 2) +
 i
40
 + i
60
= (i
4
)
7
 i
2
 + (i
4
)
10
 + (i
4
)
15
= i
2
 + 1
10
 + 1
15
 = – 1 + 1 + 1 = 1
viii. i
49
 + i
68
 + i
89
 + i
118
 = i
(48 + 1)
 + i
68
 + i
(88 + 1)
 + i
(116 + 2)
= (i
4
)
12
×i + (i
4
)
17
 + (i
4
)
11
×i + (i
4
)
29
×i
2
= i + 1 + i – 1 = 2i
2. Question
Show that 1 + i
10
 + i
20
 + i
30
 is a real number ?
Answer
1 + i
10
 + i
20
 + i
30
 = 1 + i
(8 + 2) +
 i
20
 + i
(28 + 2)
= 1 + (i
4
)
2
 × i
2
 + (i
4
)
5
 + (i
4
)
7
 × i
2
= 1 – 1 + 1 – 1 = 0
[ since i
4
 = 1, i
2
 = – 1]
Hence , 1 + i
10
 + i
20
 + i
30
 is a real number.
3 A. Question
Find the value of following expression:
i
49
 + i
68
 + i
89
 + i
110
Answer
i
49
 + i
68
 + i
89
 + i
110
 = i
(48 + 1)
 + i
68
 + i
(88 + 1)
 + i
(108 + 2)
= (i
4
)
12
 × i + (i
4
)
17
 + (i
4
)
11
 × i + (i
4
)
27
 × i
2
= i + 1 + i – 1 = 2i
[since i
4
 = 1, i
2
 = – 1]
i
49
 + i
68
 + i
89
 + i
110
 = 2i
3 B. Question
Find the value of following expression:
i
30
 + i
80
 + i
120
Answer
i
30
 + i
80
 + i
120
 = i
(28 + 2)
 + i
80
 + i
120
= (i
4
)
7
 × i
2
 + (i
4
)
20
 + (i
4
)
30
= – 1 + 1 + 1 = 1
[since i
4
 = 1, i
2
 = – 1]
i
30
 + i
80
 + i
120
 = 1
3 C. Question
Find the value of following expression:
i + i
2
 + i
3
 + i
4
Answer
i + i
2
 + i
3
 + i
4
 = i + i
2
 + i
2
×i + i
4
= i – 1 + (– 1)×i + 1
since i
4
 = 1, i
2
 = – 1
= i – 1 – i + 1 = 0
3 D. Question
Find the value of following expression:
i
5
 + i
10
 + i
15
Answer
i
5
 + i
10
 + i
15
 = i
(4 + 1)
 + i
(8 + 2)
 + i
(12 + 3)
= (i
4
)
1
×i + (i
4
)
2
×i
2
 + (i
4
)
3
×i
3
= (i
4
)
1
×i + (i
4
)
2
×i
2
 + (i
4
)
3
×i
2
×i
= 1×i + 1×(– 1) + 1×(– 1)×i
= i – 1 – i = – 1
3 E. Question
Find the value of following expression:
Answer
                                    = i
10
                                         = i
8
i
2
                                   = (i
4
)
2
 i
2
 Since i
4
 = 1, i
2
 = -1           
 
                                = (1)
2
 (-1)                               
 = -1
                                   
3 F. Question
Find the value of following expression:
1 + i
2
 + i
4
 + i
6
 + i
8
 + ... + i
20
Answer
1 + i
2
 + i
4
 + i
6
 + i
8
 + ... + i
20
 = 1 + (– 1) + 1 + (– 1) + 1 + ... + 1
= 1
3 G. Question
Find the value of following expression:
(1 + i)
6
 + (1 – i)
3
Answer
(1 + i)
6
 + (1 – i) 
3
 = {(1 + i)
2
 }
3
 + (1 – i)
2
 (1 – i)
= {1 + i
2
 + 2i}
3
 + (1 + i
2 –
 2i)(1 – i)
= {1 – 1 + 2i}
3
 + (1 – 1 – 2i)(1 – i)
= (2i)
3
 + (– 2i)(1 – i)
= 8i
3
 + (– 2i) + 2i
2
[since i
3
 = – i, i
2
 = – 1]
= – 8i – 2i – 2
= – 10 i – 2
= – 2(1 + 5i)
Exercise 13.2
1 A. Question
Express the following complex numbers in the standard form a + i b :
(1 + i) (1 + 2i)
Answer
Given:
? a+ib = (1+i)(1+2i)
? a+ib = 1(1+2i)+i(1+2i)
? a+ib = 1+2i+i+2i
2
We know that i
2
=-1
? a+ib = 1+3i+2(-1)
? a+ib = 1+3i-2
? a+ib=-1+3i
? The values of a, b are -1, 3.
1 B. Question
Express the following complex numbers in the standard form a + i b :
Answer
Given:
? 
Multiplying and dividing with -2-i
? 
? 
We know that i
2
=-1
Page 5


13. Complex Numbers
Exercise 13.1
1. Question
Evaluate the following:
(i) i
457
(ii) i
528
(iii) 
(iv) 
(v)
(vi) (i
77
 + i
70
 + i
87
 + i
414
 )
3
(vii) (vii) i
30
 + i
40
 + i
60
(viii) i
49
 + i
68
 + i
89
 + i
118
Answer
i. i
457
 = i 
(456 + 1)
= i
4(114)
 × i
= (1)
114
 × i = i since i
4
 = 1
ii. i
528
 = i
4(132)
= (1)
132
 =1 since i
4
 = 1
iii. 
 since i
4
 = 1
 = – 1 since i
2
 = – 1
iv. 
[since i
4
 = 1]
v. 
 = (i – i) = 0
[since ]
vi. (i
77
 + i
70
 + i
87
 + i
414
 )
3
 = (i
(76 + 1)
 + i
(68 + 2)
 + i
(84 + 3)
 + i
(412 + 2)
 ) 
3
(i
77
 + i
70
 + i
87
 + i
414
 )
3
 = (i + i
2
 + i
3
 + i
2
 )
3
[since i
3
 = – i, i
2
 = – 1]
= (i + (– 1) + (– i) + (– 1))
3
 = (– 2)
3
(i
77
 + i
70
 + i
87
 + i
414
 )
3
 = –8
vii. i
30
 + i
40
 + i
60
 = i
(28 + 2) +
 i
40
 + i
60
= (i
4
)
7
 i
2
 + (i
4
)
10
 + (i
4
)
15
= i
2
 + 1
10
 + 1
15
 = – 1 + 1 + 1 = 1
viii. i
49
 + i
68
 + i
89
 + i
118
 = i
(48 + 1)
 + i
68
 + i
(88 + 1)
 + i
(116 + 2)
= (i
4
)
12
×i + (i
4
)
17
 + (i
4
)
11
×i + (i
4
)
29
×i
2
= i + 1 + i – 1 = 2i
2. Question
Show that 1 + i
10
 + i
20
 + i
30
 is a real number ?
Answer
1 + i
10
 + i
20
 + i
30
 = 1 + i
(8 + 2) +
 i
20
 + i
(28 + 2)
= 1 + (i
4
)
2
 × i
2
 + (i
4
)
5
 + (i
4
)
7
 × i
2
= 1 – 1 + 1 – 1 = 0
[ since i
4
 = 1, i
2
 = – 1]
Hence , 1 + i
10
 + i
20
 + i
30
 is a real number.
3 A. Question
Find the value of following expression:
i
49
 + i
68
 + i
89
 + i
110
Answer
i
49
 + i
68
 + i
89
 + i
110
 = i
(48 + 1)
 + i
68
 + i
(88 + 1)
 + i
(108 + 2)
= (i
4
)
12
 × i + (i
4
)
17
 + (i
4
)
11
 × i + (i
4
)
27
 × i
2
= i + 1 + i – 1 = 2i
[since i
4
 = 1, i
2
 = – 1]
i
49
 + i
68
 + i
89
 + i
110
 = 2i
3 B. Question
Find the value of following expression:
i
30
 + i
80
 + i
120
Answer
i
30
 + i
80
 + i
120
 = i
(28 + 2)
 + i
80
 + i
120
= (i
4
)
7
 × i
2
 + (i
4
)
20
 + (i
4
)
30
= – 1 + 1 + 1 = 1
[since i
4
 = 1, i
2
 = – 1]
i
30
 + i
80
 + i
120
 = 1
3 C. Question
Find the value of following expression:
i + i
2
 + i
3
 + i
4
Answer
i + i
2
 + i
3
 + i
4
 = i + i
2
 + i
2
×i + i
4
= i – 1 + (– 1)×i + 1
since i
4
 = 1, i
2
 = – 1
= i – 1 – i + 1 = 0
3 D. Question
Find the value of following expression:
i
5
 + i
10
 + i
15
Answer
i
5
 + i
10
 + i
15
 = i
(4 + 1)
 + i
(8 + 2)
 + i
(12 + 3)
= (i
4
)
1
×i + (i
4
)
2
×i
2
 + (i
4
)
3
×i
3
= (i
4
)
1
×i + (i
4
)
2
×i
2
 + (i
4
)
3
×i
2
×i
= 1×i + 1×(– 1) + 1×(– 1)×i
= i – 1 – i = – 1
3 E. Question
Find the value of following expression:
Answer
                                    = i
10
                                         = i
8
i
2
                                   = (i
4
)
2
 i
2
 Since i
4
 = 1, i
2
 = -1           
 
                                = (1)
2
 (-1)                               
 = -1
                                   
3 F. Question
Find the value of following expression:
1 + i
2
 + i
4
 + i
6
 + i
8
 + ... + i
20
Answer
1 + i
2
 + i
4
 + i
6
 + i
8
 + ... + i
20
 = 1 + (– 1) + 1 + (– 1) + 1 + ... + 1
= 1
3 G. Question
Find the value of following expression:
(1 + i)
6
 + (1 – i)
3
Answer
(1 + i)
6
 + (1 – i) 
3
 = {(1 + i)
2
 }
3
 + (1 – i)
2
 (1 – i)
= {1 + i
2
 + 2i}
3
 + (1 + i
2 –
 2i)(1 – i)
= {1 – 1 + 2i}
3
 + (1 – 1 – 2i)(1 – i)
= (2i)
3
 + (– 2i)(1 – i)
= 8i
3
 + (– 2i) + 2i
2
[since i
3
 = – i, i
2
 = – 1]
= – 8i – 2i – 2
= – 10 i – 2
= – 2(1 + 5i)
Exercise 13.2
1 A. Question
Express the following complex numbers in the standard form a + i b :
(1 + i) (1 + 2i)
Answer
Given:
? a+ib = (1+i)(1+2i)
? a+ib = 1(1+2i)+i(1+2i)
? a+ib = 1+2i+i+2i
2
We know that i
2
=-1
? a+ib = 1+3i+2(-1)
? a+ib = 1+3i-2
? a+ib=-1+3i
? The values of a, b are -1, 3.
1 B. Question
Express the following complex numbers in the standard form a + i b :
Answer
Given:
? 
Multiplying and dividing with -2-i
? 
? 
We know that i
2
=-1
? 
? 
? 
? The values of a, b are .
1 C. Question
Express the following complex numbers in the standard form a + i b :
Answer
Given:
? 
? 
We know that i
2
=-1
? 
? 
Multiplying and diving with 3-4i
? 
? 
? 
? 
? 
? The values of a, b is .
1 D. Question
Express the following complex numbers in the standard form a + i b :
Answer
Given:
? 
Multiplying and dividing by 1-i

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Document Description: RD Sharma Class 11 Solutions Chapter - Complex Numbers for Commerce 2026 is part of Mathematics (Maths) Class 11 preparation. The notes and questions for RD Sharma Class 11 Solutions Chapter - Complex Numbers have been prepared according to the Commerce exam syllabus. Information about RD Sharma Class 11 Solutions Chapter - Complex Numbers covers topics like and RD Sharma Class 11 Solutions Chapter - Complex Numbers Example, for Commerce 2026 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Class 11 Solutions Chapter - Complex Numbers.

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