Let z satisfy ∣ z ∣ = 1 and z = 1 − z ˉ . Statement 1 : z is a real number. Statement 2 : Principal argument of z is 3 π

Question

Let z satisfy ∣ z ∣ = 1 and z = 1 − z ˉ . Statement 1 : z is a real number. Statement 2 : Principal argument of z is 3 π ​

TestHub · Accepted Answer

Let z = x + i y , z ˉ = x − i y ⇒ ⇒ ⇒ ​ Now, z = 1 − z ˉ x + i y = 1 − ( x − i y ) 2 x = 1 ⇒ x = 2 1 ​ Now, ∣ z ∣ = 1 ⇒ x 2 + y 2 = 1 ⇒ y 2 = 1 − x 2 y = ± 2 3 ​ ​ ​ Now, tan θ = x y ​ ( θ is the argument ) = 2 3 ​ ​ ÷ 2 1 ​ (+ve since only principal argument) = 3 ​ ⇒ θ = tan − 1 3 ​ = 3 π ​ ​ Hence, z is not a real number So, statement-1 is false and 2 is true.

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