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.

If x + i y = a + i b c + i d , prove that x 2 + y 2 2 = a 2 + b 2 c 2 + d 2 .

Given that x + i y = a + i b c + i d ⇒ ( x + i y ) 2 = a + i b c + i d ⇒ x 2 + 2 x y i + y 2 i 2 = a + i b c + i d × c - i d c - i d ⇒ x 2 + 2 x y i + y 2 i 2 = a c - a d i + b c i - b d i 2 c 2 - d 2 i 2 ⇒ x 2 - y 2 + 2 x y i = a c - a d i + b c i + b d c 2 + d 2 ∵ i 2 = - 1 ⇒ x 2 - y 2 + 2 x y i = ( a c + b d ) + ( b c - a d ) i c 2 + d 2 ∴ x 2 - y 2 + 2 x y i = a c + b d c 2 + d 2 + b c - a d c 2 + d 2 i Equating the real and imaginary parts separately, we get, x 2 - y 2 = a c + b d c 2 + d 2 and 2 x y = b c - a d c 2 + d 2 Now write x 2 + y 2 2 in terms of x 2 - y 2 2 and 2 x y ⇒ x 2 + y 2 2 = x 2 - y 2 2 + 4 x 2 y 2 ⇒ x 2 + y 2 2 = x 2 - y 2 2 + ( 2 x y ) 2 ⇒ x 2 + y 2 2 = a c + b d c 2 + d 2 2 + b c - a d c 2 + d 2 2 ⇒ x 2 + y 2 2 = ( a c + b d ) 2 + ( b c - a d ) 2 c 2 + d 2 2 ⇒ x 2 + y 2 2 = a 2 c 2 + b 2 c 2 + a 2 d 2 + b 2 d 2 c 2 + d 2 2 ⇒ x 2 + y 2 2 = a 2 + b 2 c 2 + a 2 + b 2 d 2 c 2 + d 2 2 ⇒ x 2 + y 2 2 = a 2 + b 2 ( c 2 + d 2 ) c 2 + d 2 2 ⇒ x 2 + y 2 2 = a 2 + b 2 c 2 + d 2 Hence, Proved.

If ( a + i b ) = 1 + i 1 - i , then prove that a 2 + b 2 = 1 .

Give that a + i b = 1 + i 1 - i ⇒ a + i b = 1 + i 1 - i × 1 + i 1 + i ⇒ a + i b = ( 1 + i ) 2 1 - i 2 ⇒ a + i b = 1 + 2 i + i 2 1 - i 2 ⇒ a + i b = 1 + 2 i - 1 1 + 1 ∵ i 2 = - 1 ⇒ a + i b = 2 i 2 ∴ a + i b = 0 + 1 . i Equating the real and imaginary parts separately, we get, a = 0 and b = 1 ∴ a 2 + b 2 = 0 2 + 1 2 ⇒ a 2 + b 2 = 1 Hence, proved.

Show that 7 + i 3 7 - i 3 + 7 - i 3 7 + i 3 is real.

Given expression is 7 + i 3 7 - i 3 + 7 - i 3 7 + i 3 7 + i 3 7 - i 3 + 7 - i 3 7 + i 3 = ( 7 + i 3 ) 2 + ( 7 - i 3 ) 2 ( 7 - i 3 ) ( 7 + i 3 ) = 7 + 2 21 i + 3 i 2 + 7 - 2 21 i + 3 i 2 7 - 3 i 2 = 7 + 3 ( - 1 ) + 7 + 3 ( - 1 ) 7 - 3 ( - 1 ) = 7 - 3 + 7 - 3 7 + 3 = 8 10 = 4 5 Since 4 5 is a real number. Hence, 7 + i 3 7 - i 3 + 7 - i 3 7 + i 3 is real.

If ( x + i y ) 3 = y + v i then show that y x + v y = 4 x 2 - y 2 .

Given, ( x + i y ) 3 = y + v i To Prove: y x + v y = 4 x 2 - y 2 Consider ( x + i y ) 3 = y + v i ⇒ x 3 + i y 3 + 3 i x y x + i y = y + i v ⇒ x 3 + i 3 y 3 + 3 i x 2 y + 3 i 2 x y 2 = y + i v ⇒ x 3 - i y 3 + 3 i x 2 y - 3 x y 2 = y + i v ⇒ x 3 - 3 x y 2 + i 3 x 2 y - y 3 = y + i v Equating real and imaginary parts, we get x 3 - 3 x y 2 = y and 3 x 2 y - y 3 = v ⇒ y x = x 2 - 3 y 2 and v y = 3 x 2 - y 2 Now, y x + v y = x 2 - 3 y 2 + 3 x 2 - y 2 = 4 x 2 - 4 y 2 = 4 x 2 - y 2 Hence Proved.