Show that: 1 i - 1 i 2 + 1 i 3 - 1 i 4 = 0

To Prove: 1 i - 1 i 2 + 1 i 3 - 1 i 4 = 0 L.H.S. = 1 i - 1 i 2 + 1 i 3 - 1 i 4 = 1 i - 1 - 1 + 1 i 2 · i - 1 i 2 2 We know that i 2 = - 1 = 1 i + 1 + 1 - 1 · i - 1 - 1 2 = 1 i + 1 - 1 i - 1 = 0 = R.H.S. Hence, Proved.

Given, - 81 We know that for any two real numbers a and b , a × b = a b is true only when at least one of a and b is zero or positive. Now, - 81 = - 1 × 81 = - 1 × 81 = i × 9 = 9 i Hence, - 81 = 9 i

Evaluate: 4 - 4 + 5 - 9 - 3 - 16

Given, 4 - 4 + 5 - 9 - 3 - 16 We know that - a = i a , where a is any positive real number. ⇒ 4 - 4 + 5 - 9 - 3 - 16 = 4 × 4 i + 5 9 i - 3 16 i = 4 × 2 i + 5 × 3 i - 3 × 4 i = 8 i + 15 i - 12 i = 23 i - 12 i = 11 i Hence, 4 - 4 + 5 - 9 - 3 - 16 = 11 i

Simplify and express the result in the form a + i b ( 3 + - 16 ) - ( 4 - - 9 )

Given, ( 3 + - 16 ) - ( 4 - - 9 ) = 3 + - 1 × 16 - 4 + - 1 × 9 = 3 + - 1 × 16 - 4 + - 1 × 9 = 3 + i × 4 - 4 + i × 3 = 3 + 4 i - 4 + 3 i = 3 - 4 + i 4 + 3 = - 1 + i 7 Hence, ( 3 + - 16 ) - ( 4 - - 9 ) = - 1 + 7 i