If z 1 = 2 - i , z 2 = 1 + i , find z 1 + z 2 + 1 z 1 - z 2 + i .

Given : z 1 = 2 - i , z 2 = 1 + i ∴ z 1 + z 2 + 1 z 1 - z 2 + i = 2 - i + 1 + i + 1 2 - i - 1 - i + i = 4 1 - i Also, | 1 - i | = 1 2 + i 2 ( ∵ | a + b i | = a 2 + b 2 ) = 2 ∴ z 1 + z 2 + 1 z 1 - z 2 + i = 4 2 = 2 2

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