Prove that ( 1 + i ) 4 × 1 + 1 i 4 = 16 .

Given L.H.S is ( 1 + i ) 4 × i + 1 i 4 and R.H.S is 16 Since, L.H.S is integral power of 1 + i 2 ( 1 + i ) 2 = 1 + 2 i + i 2 ( 1 + i ) 2 = 1 + 2 i - 1 ∵ i 2 = - 1 ( 1 + i ) 2 = 2 i . . . ( 1 ) Now, simplify the L.H.S ( 1 + i ) 4 × 1 + 1 i 4 ( 1 + i ) 4 × 1 + 1 i 4 = ( 1 + i ) 4 × i + 1 i 4 ⇒ ( 1 + i ) 4 × 1 + 1 i 4 = ( 1 + i ) 8 i 4 ⇒ ( 1 + i ) 4 × 1 + 1 i 4 = ( 1 + i ) 8 1 ∵ i 4 = 1 ⇒ ( 1 + i ) 4 × 1 + 1 i 4 = ( 1 + i ) 2 4 Substitute the value of ( 1 + i ) 2 ⇒ ( 1 + i ) 4 × 1 + 1 i 4 = 2 i 4 ⇒ ( 1 + i ) 4 × 1 + 1 i 4 = 16 i 4 ⇒ ( 1 + i ) 4 × 1 + 1 i 4 = 16 ⇒ ( 1 + i ) 4 × 1 + 1 i 4 = R.H.S Hence, proved.

If a = - 1 + 3 i 2 , b = - 1 - 3 i 2 , show that a 2 = b and b 2 = a .

Given, a = - 1 + 3 i 2 and b = - 1 - 3 i 2 Squaring both side of the equation a = - 1 + 3 i 2 ∴ a 2 = - 1 + 3 i 2 2 ⇒ a 2 = ( - 1 ) 2 + 2 ( - 1 ) ( 3 i ) + ( 3 i ) 2 2 2 ⇒ a 2 = 1 - 2 3 i - 3 4 ∵ i 2 = - 1 ⇒ a 2 = - 2 - 2 3 i 4 ⇒ a 2 = - 1 - 3 i 2 ⇒ a 2 = b Now, squaring both side of the equation b = - 1 - 3 i 2 ∴ b 2 = - 1 - 3 i b 2 ⇒ b 2 = ( - 1 ) 2 + 2 ( - 1 ) ( - 3 i ) + ( - 3 i ) 2 2 2 ⇒ b 2 = 1 + 2 ( 3 i ) - 3 4 ⇒ b 2 = - 2 + 2 ( 3 i ) 4 ⇒ b 2 = - 1 + 3 i 2 ⇒ b 2 = a Hence proved

If x + i y = ( a + i b ) 3 , show that x a + y b = 4 a 2 - b 2 .

Given that x + i y = ( a + i b ) 3 To prove: x a + y b = 4 a 2 - b 2 ⇒ x + i y = a 3 + 3 a 2 ( i b ) + 3 a ( i b ) 2 + ( i b ) 3 ⇒ x + i y = a 3 + 3 a 2 i b + 3 a b 2 i 2 + b 3 i 3 ⇒ x + i y = a 3 + 3 a 2 i b - 3 a b 2 - b 3 i ∴ x + y i = a 3 - 3 a b 2 + 3 a 2 b - b 3 i Equating the real and imaginary parts separately, we get x = a 3 - 3 a b 2 and y = 3 a 2 b - b 3 ⇒ x = a a 2 - 3 b 2 and y = b 3 a 2 - b 2 ⇒ x a = a 2 - 3 b 2 and y b = 3 a 2 - b 2 Now, addition of x a and y b will give ⇒ x a + y b = a 2 - 3 b 2 + 3 a 2 - b 2 = 4 a 2 - 4 b 2 ∴ x a + y b = 4 a 2 - b 2 Hence, Proved.

If a + 3 i 2 + i b = 1 - i , show that 5 a - 7 b = 0 .

Give that a + 3 i 2 + i b = 1 - i ⇒ a + 3 i = ( 2 + i b ) ( 1 - i ) ⇒ a + 3 i = 2 - 2 i + b i - b i 2 ⇒ a + 3 i = 2 - 2 i + b i + b ∵ i 2 = - 1 ∴ a + 3 i = ( b + 2 ) + ( b - 2 ) i Equating the real and imaginary parts separately, we get, a = b + 2 . . . 1 3 = b - 2 . . . 2 Solving 1 and 2 From 2 , we get b = 5 Substituting b = 5 in 1 , we get a = 7 Now put the value of a , b in 5 a - 7 b ⇒ 5 a - 7 b = 5 ( 7 ) - 7 ( 5 ) ⇒ 5 a - 7 b = 0 Hence, proved.