Let z satisfy left| z right| = 1 an | vedclass

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Question

Let $z=x+i y$,  $\bar{z}=x-i y$

Now, $z=1-\bar{z}$

$\Rightarrow \,\, x+i y=1-(x-i y)$

$\Rightarrow \,\, 2 x=1 \Rightarrow x=\frac{1}{2}$

Vedclass · Accepted Answer

Statement $1$ is false; Statement $2$ is true

Vedclass · Answer

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

Let $z$ satisfy $\left| z \right| = 1$ and $z = 1 - \vec z$.

Statement $1$ : $z$ is a real number

Statement $2$ : Principal argument of $z$ is $\frac{\pi }{3}$

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