Let z = 1 + i and z 1 = 1 + i z ¯ z ¯ ( 1 - z ) + 1 z · Then 12 π arg z 1 is equal to

Step-by-step solution

Given: z = 1 + i z ¯ = 1 - i And, 1 z = 1 1 + i × 1 - i 1 - i = 1 - i 2 So, z 1 = 1 + i z ¯ z ¯ ( 1 - z ) + 1 z = 1 + i - i 2 ( 1 - i ) ( - i ) + 1 - i 2 = 1 + i - i 2 - i + i 2 + 1 2 - i 2 = 2 + i - 3 i - 1 2 = 4 + 2 i - 3 i - 1 = - ( 4 + 2 i ) ( 3 i - 1 ) ( 3 i ) 2 - ( 1 ) 2 = 12 i - 4 + 6 i 2 - 2 i 10 = 10 i - 10 10 = i - 1 Arg z 1 = 3 π 4 ∴ 12 π arg z 1 = 12 π × 3 π 4 = 9