Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Let $z$ satisfy $\left|z\right|=1 and z=1-\overline{z}.$
Statement 1 : $z$ is a real number.
Statement 2 : Principal argument of $z$ is $\frac{\pi}{3}$

JEE MainJEE Main 2013Complex Numbers and Quadratic Equations

Solution:

Let $z= x+ iy, \overline{z} =x-iy $
Now, $z = 1- \overline{z}$
$\Rightarrow x + iy=1 -(x-iy)$
$\Rightarrow 2x=1 $
$\Rightarrow x= \frac{1}{2}$
Now, $\left|z\right|=1$
$\Rightarrow x^{2}+y^{2}=1 $
$\Rightarrow y^{2}=1-x^{2}$
$\Rightarrow y=\pm \frac{\sqrt{3}}{2}$
Now, tan $\theta=\frac{y}{x}$ $(\theta$ is the argument
$=\frac{\sqrt{3}}{2} \div\frac{1}{2}$
(+ve since only principal argument)
$=\sqrt{3}$
$\Rightarrow \theta=tan^{-1} \sqrt{3}=\frac{\pi}{3}$
Hence, $z$ is not a real number
So, statement-1 is false and 2 is true.