Complex Numbers Class 11 OP Malhotra Exe-9A ISC Maths Solutions Ch-9 Solutions. In this article you would learn about Integral Powers of i. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.

Complex Numbers Class 11 OP Malhotra Exe-9A ISC Maths Solutions Ch-9

Board	ISC
Publications	S Chand
Subject	Maths
Class	11th
Chapter-9	Complex Numbers
Writer	O.P. Malhotra
Exe-9(A)	Integral Powers of i.

Exercise- 9A

Complex Number Class 11 OP Malhotra Exe-9A Solution.

Que-1: 3i . 2

Sol: 3i . 2
= 6i

Que-2: i (- i)

Sol: i (- i)
= – i²
= – (- 1)
= 1 + 0i

Que-3: – i (- i)

Sol: – i (- i)
= i²
= – 1

Que-4: 5i (- 8i)

Sol: 5i (- 8i)
= – 40i²
= – 40 (- 1)
= 40

Que-5: 20i / 4

Sol: 20i / 5
= 4i

Que-6: √-25

Sol: √-25
= √(5×5) √-1
= 5i

Que-7: √-8

Sol: √-8
√(2×2) √-2
= 2√2 i

Que-8: √(-1/3)

Sol: √(-1/3)
= √(1/3)i
= √{(1/3)×(3/3)} i
= √3/3 i

Que-9: (1/2) √(-3/4)

Sol: (1/2) √(-3/4)
= (1/2) (1/2) √-3i
= √3/4 i

Que-10: 6 / -i

Sol: 6/−i = (6/−i) × (i/i)
= 6i/−i²
= 6i/ −(−1)
= 6i

Que-11: √-144

Sol: √(12×12) √-1
= 12 i

Que-12: x / i

Sol: x/i
= (x/i) × (i/i)
= ix/i²
= – i x

Simplify

Que-13: i¹³

Sol: i¹³  = i¹² i
= (i⁴)³ i  = 1³ i
= i [∵ i⁴ = 1]

Que-14: i²⁸

Sol: i²⁸ = (i⁴)⁷
= 1 [∵ i⁴ = (i²)² = (- 1)² = 1]

Que-15: i¹⁸

Sol:  i¹⁸ = (i⁴)⁴ i²
= 1⁴ x (- 1)
= – 1

Que-16: i²³

Sol: i²³ = (i⁴)⁵ i³
= 1⁵ x i² x i
= 1 x (- 1) x i
= – i

Que-17: √-4 + √-16 – √-25

Sol: 2i + 4i – 5i
= 6i – 5i = i

Que-18: √-20 + √-12

Sol: 2√5i + 2√3i
= 2(√+√3) i

Que-19: -√(-7/4)  – √(-1/7)

Sol: -(√7i)/2 – (i/√7)
= {-i(7+2)}/2√7
= (-9i/2√7) (√7/√7)
= -9√7i / 14

Que-20: √-2 / √-8

Sol: √2i / 2√2
= 1/2

Que-21: 1/i + 1/i² + 1/i³ + 1/i^4

Sol: (1/i) + (1/−1) + (1/−i) + 1
= (1/i) − 1 − (1/i) + 1 = 0

Que-22: 1/i − 1/i² + 1/i³ − 1/i^4

Sol: (1/i) + 1 − (1/i) − 1
= 0 [∵ i² = – 1 and i4 = 1]

Que-23: i + 2i² + 3i³ + i⁴

Sol: i + 2i² + 3i³ + i^4
= i + 2 (- 1) + 3i (- 1) + 1
= i – 2 – 3i + 1
= – 2i – 1

Que-24: [i^(18) + (1/i)^(25)]³

Sol: [i^(18) + (1/i)^(25)]³

= [-1 – i]³ = – (1+i)³
= – [1+i³+3i(1+i)]
= – [1-i+3i-3]
= – (-2+2i)
= 2 (1-i)

Que-25: √(-x/4) + √(-x/16) + √(-x/64),  where x is a positive real number.

Sol: √(-x/4) + √(-x/16) + √(-x/64)

Que-26: √-5x^8 – √-20x^8 + √-45x^8, where x is a positive real number.

Sol: √-5x^8 – √-20x^8 + √-45x^8
= √5x^4 i – 2√5x^4 i + 3√5x^4 i
= 2√5 i x^4

Que-27: If i = √-1 prove the following :
(x + 1 + i) (x + 1 – i) (x – 1 – i) = x^4 + 4.

Sol: L.H.S = (x + 1 + i) (x + 1 – i) (x – 1 + i) (x – 1 – i)
= [(x + 1)² – i²] [(x – 1)² – i²]
= [x² + 2x + 1 + 1] [x² – 2x + 1 + 1]
= (x² + 2x + 2) (x² – 2x + 2)
= (x² + 2 + 2x) (x² + 2 – 2x)
= (x² + 2)² – (2x)²
= x⁴ + 4x² + 4 – 4x²
= x⁴ + 4 = R.H.S

Que-28: Find the value of (1-i)^n [1 – (1/i)]^n, where n is a positive integer.

Sol: (1-i)^n [1 – (1/i)]^n
= (1-i)^n [1 – (i^n)/i]^n
= (1-i)^n (1+i³)^n
= [(1-i)(1+i)]^n
= (1-i²)^n
= 2^n

–: End of Complex Number Class 11 OP Malhotra Exe-9A ISC Math Ch-9 Solution :–

Return to :-OP Malhotra ISC Class-11 S Chand Publication Maths Solutions

Thanks

Please share with your friends