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

Given, x + i y = a + i b a - i b Taking mod on both the sides: | x + i y | = a + i b a - i b ⇒ x 2 + y 2 = a 2 + b 2 a 2 + b 2 ⇒ x 2 + y 2 = 1 Hence, proved.

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

Given, x + i y = a + i b a - i b Taking mod on both the sides: | x + i y | = a + i b a - i b ⇒ x 2 + y 2 = a 2 + b 2 a 2 + b 2 ⇒ x 2 + y 2 = 1 Hence, proved.

Find the real values of θ for which the complex number 1 + i cos θ 1 - 2 i cos θ is purely real.

Given, 1 + i cos θ 1 - 2 i cos θ = 1 + i cos θ 1 - 2 i cos θ × 1 + 2 i cos θ 1 + 2 i cos θ = 1 + 2 i cos θ + i cos θ - 2 cos θ 1 + 4 cos 2 θ = 1 - 2 cos θ + i 3 cos θ 1 + 4 cos 2 θ For it to be purely real, the imaginary part must be zero. 3 cos θ = 0 This is true for odd multiples of π 2 . ∴ θ = ( 2 n + 1 ) π 2 , n ∈ Z

If 1 + i 1 - i 3 - 1 - i 1 + i 3 = x + i y , find ( x , y )

Given, 1 + i 1 - i = 1 + i 1 - i × 1 + i 1 + i = ( 1 + i ) 2 1 2 - i 2 = 1 2 + i 2 + 2 i 1 + 1 ∵ i 2 = - 1 = 1 - 1 + 2 i 2 = i ...(i) Also, 1 - i 1 + i = 1 - i 1 + i × 1 - i 1 - i = ( 1 - i ) 2 1 2 - i 2 = 1 2 + i 2 - 2 i 1 + 1 ∵ i 2 = - 1 = 1 - 1 - 2 i 2 = - i ...(ii) It is given that, 1 + i 1 - i 3 - 1 - i 1 + i 3 = x + i y ⇒ ( i ) 3 - ( - i ) 3 = x + i y [From (i) and (ii)] ⇒ i 3 + i 3 = x + i y ⇒ 2 i 3 = x + i y ⇒ 0 - 2 i = x + i y ∵ i 3 = - i ⇒ x = 0 and y = - 2 Thus, ( x , y ) = ( 0 , - 2 )

If 1 - i 1 + i 100 = a + i b , find ( a , b ) .

Given, 1 − i 1 + i 100 = a + i b Now, 1 - i 1 + i = 1 - i 1 + i × 1 - i 1 - i = ( 1 - i ) 2 1 2 - i 2 = 1 2 + i 2 - 2 i 1 + 1 ∵ i 2 = - 1 = 1 - 1 - 2 i 2 = - 2 i 2 = - i ...(i) It is given that, 1 - i 1 + i 100 = a + i b ⇒ ( - i ) 100 = a + i b [from (i)] ⇒ i 4 × 25 = a + i b ⇒ 1 + 0 i = a + i b ∵ i 4 = 1 ⇒ a = 1 and b = 0 Thus, ( a , b ) = ( 1 , 0 )

If a = cos θ + i sin θ , find the value of 1 + a 1 - a .

Given, a = cos θ + i sin θ Now, 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 sin θ + cos θ - cos 2 θ + i cos θ sin θ + i sin θ - i sin θ cos θ + i 2 sin 2 θ ( 1 - cos θ ) 2 - i 2 sin 2 θ = 1 - cos 2 θ - sin 2 θ + 2 i sin θ 1 + cos 2 θ - 2 i cos θ + sin 2 θ ∵ i 2 = - 1 = sin 2 θ - sin 2 θ + 2 i sin θ 2 - 2 i cos θ ∵ cos 2 θ + sin 2 θ = 1 = i sin θ 1 - cos θ = 2 i sin θ 2 cos θ 2 2 sin 2 θ 2 = 2 i cot θ 2 Thus, 1 + a 1 - a = 2 i cot θ 2 .