Let z=x+iy,zˉ=x−iy ⇒⇒⇒ Now, z=1−zˉx+iy=1−(x−iy)2x=1⇒x=21 Now, ∣z∣=1⇒x2+y2=1⇒y2=1−x2y=±23 Now, tanθ=xy(θ is the argument ) =23÷21 (+ve since only principal argument) =3⇒θ=tan−13=3π Hence, z is not a real number So, statement-1 is false and 2 is true.
Let z satisfy |z|=1 and z=1- z. Statement 1: z is a real number. Statement 2 : Principal argument of z is π 3
JEE Main 2013 — Mathematics Algebra
Let z satisfy ∣z∣=1 and z=1−zˉ. Statement 1:z is a real number. Statement 2 : Principal argument of z is 3π
Held on 25 Apr 2013 · Verified 6 Jul 2026.
Options
- A
Statement 1 is true Statement 2 is true; Statement 2 is a correct explanation for Statement 1 .
- B
Statement 1 is false; Statement 2 is true
- C
Statement 1 is true, Statement 2 is false.
- D
Statement 1 is true; Statement 2 is true; Statement 2 is not a correct explanation for Statement 1 .
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