If 1 + i 1 - i 3 - 1 - i 1 + i 3 = x + i y , find x and y .

Consider x + i y = 1 + i 1 - i 3 - 1 - i 1 + i 3 = 1 + i 1 - i × 1 + i 1 + i 3 - 1 - i 1 + i × 1 - i 1 - i 3 = ( 1 + i ) 2 ( 1 - i ) ( 1 + i ) 3 - ( 1 - i ) 2 ( 1 + i ) ( 1 - i ) 3 = ( ( 1 + i ) 2 ) 3 - ( ( 1 - i ) 2 ) 3 ( ( 1 - i ) ( 1 + i ) ) 3 = ( 1 + i 2 + 2 i ) 3 - ( 1 + i 2 - 2 i ) 3 ( 1 2 - i 2 ) 3 . . . . . . [ ( a - b ) ( a + b ) = a 2 - b 2 ] = ( 1 - 1 + 2 i ) 3 - ( 1 - 1 - 2 i ) 3 ( 1 - ( - 1 ) ) 3 = ( 2 i ) 3 - ( - 2 i ) 3 8 = 8 i 3 + 8 i 3 8 = 2 i 3 = - 2 i . Thus x + i y = 0 - 2 i . . . . . . ( i ) . Equating real and imaginary part on both the sides of ( i ) , we get x = 0 and y = - 2

If 1 + i 1 - i 3 - 1 - i 1 + i 3 = x + i y , find x and y .

Consider x + i y = 1 + i 1 - i 3 - 1 - i 1 + i 3 = 1 + i 1 - i × 1 + i 1 + i 3 - 1 - i 1 + i × 1 - i 1 - i 3 = ( 1 + i ) 2 ( 1 - i ) ( 1 + i ) 3 - ( 1 - i ) 2 ( 1 + i ) ( 1 - i ) 3 = ( ( 1 + i ) 2 ) 3 - ( ( 1 - i ) 2 ) 3 ( ( 1 - i ) ( 1 + i ) ) 3 = ( 1 + i 2 + 2 i ) 3 - ( 1 + i 2 - 2 i ) 3 ( 1 2 - i 2 ) 3 . . . . . . [ ( a - b ) ( a + b ) = a 2 - b 2 ] = ( 1 - 1 + 2 i ) 3 - ( 1 - 1 - 2 i ) 3 ( 1 - ( - 1 ) ) 3 = ( 2 i ) 3 - ( - 2 i ) 3 8 = 8 i 3 + 8 i 3 8 = 2 i 3 = - 2 i . Thus x + i y = 0 - 2 i . . . . . . ( i ) . Equating real and imaginary part on both the sides of ( i ) , we get x = 0 and y = - 2

If x + i y = a + i b a - i b , prove that x 2 + y 2 = 1 .

Consider the given equation, x + i y = a + i b a - i b = a + i b a - i b × a + i b a + i b = a 2 + ( 2 a b ) i + ( i b ) 2 ( a ) 2 - ( i b ) 2 . . . . . . ( a - b ) ( a + b ) = a 2 - b 2 = a 2 + ( 2 a b ) i + ( - 1 ) b 2 a 2 - ( - 1 ) b 2 . . . . . . i 2 = - 1 = a 2 + 2 i a b - b 2 a 2 + b 2 So, x + i y = a 2 - b 2 a 2 + b 2 + i 2 a b a 2 + b 2 . On comparing both the sides, we get x = a 2 - b 2 a 2 + b 2 and y = 2 a b a 2 + b 2 . Now consider x 2 + y 2 = a 2 - b 2 a 2 + b 2 2 + 2 a b a 2 + b 2 2 = 1 a 2 + b 2 2 a 2 - b 2 2 + ( 2 a b ) 2 = 1 a 2 + b 2 2 a 4 + b 4 - 2 a 2 b 2 + 4 a 2 b 2 = 1 a 2 + b 2 2 a 4 + b 4 + 2 a 2 b 2 = 1 a 2 + b 2 2 a 2 + b 2 2 = 1 = R.H.S

If ( a + i b ) = c + i c - i where c is real, prove that a 2 + b 2 = 1 and b a = 2 c c 2 - 1

Consider the given equation a + i b = c + i c - i a + i b = c + i c - i × c + i c + i = ( c + i ) 2 ( c ) 2 - ( i ) 2 . . . . . . ( a - b ) ( a + b ) = a 2 - b 2 = c 2 + 2 i c + i 2 c 2 - i 2 = c 2 + 2 i c + ( - 1 ) c 2 - ( - 1 ) . . . . . . i 2 = - 1 = c 2 + 2 i c - 1 c 2 + 1 a + i b = c 2 - 1 c 2 + 1 + i 2 c c 2 + 1 On comparing both the sides, we get a = c 2 - 1 c 2 + 1 and b = 2 c c 2 + 1 . Now, we have to prove that a 2 + b 2 = 1 Conisder a 2 + b 2 = c 2 - 1 c 2 + 1 2 + 2 c c 2 + 1 2 = 1 c 2 + 1 2 c 2 - 1 2 + ( 2 c ) 2 = 1 c 2 + 1 2 c 4 + 1 - 2 c 2 + 4 c 2 = 1 c 2 + 1 2 c 4 + 1 + 2 c 2 = 1 c 2 + 1 2 c 2 + 1 2 = 1 = R.H.S Now, we have to prove b a = 2 c c 2 - 1 Consider b a = 2 c c 2 + 1 c 2 - 1 c 2 + 1 = 2 c c 2 + 1 × c 2 + 1 c 2 - 1 = 2 c c 2 - 1 = R.H.S Hence proved.

Show that ( 1 - i ) n 1 - 1 i n = 2 n for all n ∈ N .

We have to show that ( 1 - i ) n 1 - 1 i n = 2 n . Consider ( 1 - i ) n 1 - 1 i n = ( 1 - i ) n 1 - 1 i × i i n = ( 1 - i ) n 1 - i i 2 n = ( 1 - i ) n 1 - i - 1 n . . . . . . ∵ i 2 = - 1 = ( 1 - i ) n ( 1 + i ) n = [ ( 1 - i ) ( 1 + i ) ] n = 1 - i 2 n . . . . . . ( a + b ) ( a - b ) = a 2 - b 2 = [ 1 - ( - 1 ) ] n . . . . . . ∵ i 2 = - 1 = 2 n Thus we have shown that ( 1 - i ) n 1 - 1 i n = 2 n for all n ∈ N .

If a = ( cos θ + i sin θ ) prove that 1 + a 1 - a = cot θ 2 i

Given that a = cos θ + i sin θ To prove: 1 + a 1 - a = cot θ 2 i Consider L.H.S = 1 + a 1 - a = 1 + cos θ + i sin θ 1 - ( cos θ + i sin θ ) = 1 + cos θ + i sin θ 1 - cos θ - i sin θ = 1 + cos θ + i sin θ 1 - cos θ - i sin θ × 1 - cos θ + i s i n θ 1 - cos θ + i s i n θ = ( 1 + i sin θ ) 2 - ( cos θ ) 2 ( 1 - cos θ ) 2 - ( i sin θ ) 2 . . . [ ( a - b ) ( a + b ) = a 2 - b 2 ] = 1 + i 2 sin 2 θ + 2 i sin θ - cos 2 θ 1 - 2 cos θ + cos 2 θ - i 2 sin 2 θ = 1 + 2 i sin θ + ( - 1 ) sin 2 θ - cos 2 θ 1 - 2 cos θ + cos 2 θ - ( - 1 ) sin 2 θ . . . [ i 2 = - 1 ] = 1 + 2 i sin θ - 1 1 - 2 cos θ + 1 . . . [ sin 2 x + cos 2 x = 1 ] = 2 i sin θ 2 ( 1 - cos θ ) = i 2 sin ( θ 2 ) cos ( θ 2 ) 2 sin 2 ( θ 2 ) . . . [ sin 2 x = 2 sin x cos x and 1 - 2 sin 2 x = cos 2 x ] = ( cot θ 2 ) i . . . [ cos x sin x = cot x ]

If z 1 = ( 2 - i ) and z 2 = ( 1 + i ) , find z 1 + z 2 + 1 z 1 - z 2 + i .

Given: z 1 = ( 2 - i ) and z 2 = ( 1 + i ) . To find: z 1 + z 2 + 1 z 1 - z 2 + t Consider z 1 + z 2 + 1 z 1 - z 2 + i = 2 - i + 1 + i + 1 2 - i - 1 - i + i = 4 1 - i = 4 1 - i × 1 + i 1 + i = 4 ( 1 + i ) ( 1 ) 2 - ( i ) 2 . . . . . . ( a - b ) ( a + b ) = a 2 - b 2 = 4 ( 1 + i ) 1 - ( - 1 ) . . . . . . [ ∵ i 2 = - 1 ] = 2 ( 1 + i ) = 2 1 + i = 2 × 1 2 + 1 2 = 2 2 .