Prove that: i 21 - 1 i 46 2 = 2 i

We have to prove that i 21 - 1 i 46 2 = 2 i . Consider L.H.S = i 21 - 1 i 46 2 = i 4 × 5 + 1 - i - 4 × 12 + 2 2 = i 1 - i 2 2 ...... { ∵ i 4 n = 1 , i 4 n + 1 = i , i 4 n + 2 = i 2 and i 4 n + 3 = i 3 } = i + 1 2 = i 2 + 1 + 2 i ...... a + b 2 = a 2 + b 2 + 2 a b = - 1 + 1 + 2 i = 2 i ⇒ L.H.S = R.H.S Hence, proved.

Prove that: i 18 + 1 i 25 3 = 2 1 - i

We have to prove that i 18 + 1 i 25 3 = 2 1 - i . Consider L.H.S = i 18 + 1 i 25 3 = { i 18 + i - 25 } 3 = i 4 × 4 + 2 + i 4 × ( - 7 ) + 3 3 = i 2 + i 3 3 = { - 1 - i } 3 = ( - 1 ) 3 - ( i 3 ) + 3 i ( - 1 - i ) ... { a - b 3 = a 3 - b 3 - 3 a b a - b } = - 1 + i - 3 i - 3 i 2 = - 1 - 2 i + 3 = 2 1 - i Thus L.H.S = R.H.S Hence proved.

Prove that: 1 - i n 1 - 1 i n = 2 n for all values of n ∈ N .

We have to prove that 1 - i n 1 - 1 i n = 2 n for all n ∈ N . Consider L.H.S = 1 - i n 1 - 1 i n = ( 1 - i ) n ( 1 - i - 1 ) n = 1 - i n 1 - i 4 × ( - 1 ) + 3 n = 1 - i n 1 - i 3 n = 1 - i n 1 + i n . . . . . { ∵ i 3 = - i } = 1 - i 1 + i n . . . . { ∵ a n b n = a b n } = 1 - i 2 n = 2 n L.H.S = R.H.S Hence, proved.

Prove that: - 16 + 3 - 25 + - 36 - - 625 = 0

We have to prove that - 16 + 3 - 25 + - 36 - - 625 = 0 . Consider L.H.S = - 16 + 3 - 25 + - 36 - - 625 = 16 i + 3 25 i + 36 i - 625 i ..... { ∵ i = - 1 } = 4 i + 15 i + 6 i - 25 i = 0 L.H.S = R.H.S Hence proved.

Prove that 1 + i 2 + i 4 + i 6 + i 8 + … + i 20 = 1

We have to prove that 1 + i 2 + i 4 + i 6 + i 8 + … . + i 20 = 1 . Consider L.H.S = 1 + i 2 + i 4 + i 6 + i 8 + … . + i 20 = ∑ n = 0 n = 10 i 2 n = ∑ n = 0 n = 10 ( i 2 ) n = ∑ n = 0 n = 10 ( - 1 ) n = 1 + ∑ n = 1 n = 10 ( - 1 ) n = 1 + - 1 ( 1 - ( - 1 ) 10 ) 1 - ( - 1 ) ...... { ∑ n = 1 n = 10 ( - 1 ) n is a geometric series with common ratio - 1 } In this summation, power of n will be odd for 5 values and even for 5 values and 0 once. So we get, L.H.S = 1 = R.H.S Hence proved.

Prove that: ∑ n = 1 13 i n + i n + 1 = - 1 + i , n ∈ N .

We have to prove that ∑ n = 1 13 i n + i n + 1 = - 1 + i , n ∈ N . L.H.S = ∑ n = 1 13 i n + i n + 1 = i 1 + i 2 + i 3 + i 4 + i 5 + i 6 + … . + i 13 + i 14 = i 1 + i 3 + . . i 13 + i 2 + i 4 + . . + i 12 = i - 1 - i + 1 + i - 1 . . . + i - 1 All the terms will get cancelled out consecutively except the first term and last term. So L.H.S = i - 1 Thus, L.H.S = R.H.S Hence proved.

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Senior Secondary School Mathematics FOR CLASS 11>Algebra>Chapter 5 - Complex Numbers and Quadratic Equations>EXERCISE 5A

R S Aggarwal Solutions for Exercise 1: EXERCISE 5A

Author:R S Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 1: EXERCISE 5A

Attempt the free practice questions from Exercise 1: EXERCISE 5A with hints and solutions to strengthen your understanding. Senior Secondary School Mathematics FOR CLASS 11 solutions are prepared by Experienced Embibe Experts.