Let $z$ satisfy $|z|=1$ and $z=1-\bar{z}$. Statement $1: z$ is a real number. Statement 2 : Principal…
Let $z$ satisfy $|z|=1$ and $z=1-\bar{z}$.
Statement $1: z$ is a real number.
Statement 2 : Principal argument of z is $\frac{\pi}{3}$
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Statement 1 is true Statement 2 is true; Statement 2 is a correct explanation for Statement 1 .
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Statement 1 is false; Statement 2 is true
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Statement 1 is true, Statement 2 is false.
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Statement 1 is true; Statement 2 is true; Statement 2 is not a correct explanation for Statement 1 .
Solution
Let $z=x+i y, \bar{z}=x-i y$
$
\begin{aligned}
& \text { Now, } z=1-\bar{z} \\
\Rightarrow & x+i y=1-(x-i y) \\
\Rightarrow & 2 x=1 \Rightarrow x=\frac{1}{2} \\
& \text { Now, }|z|=1 \Rightarrow x^2+y^2=1 \Rightarrow y^2=1-x^2 \\
\Rightarrow & y=\pm \frac{\sqrt{3}}{2}
\end{aligned}
$
Now, $\tan \theta=\frac{y}{x} \quad(\theta$ is the argument $)$
$
=\frac{\sqrt{3}}{2} \div \frac{1}{2}
$
(+ve since only principal argument)
$
\begin{gathered}
\quad=\sqrt{3} \\
\Rightarrow \quad \theta=\tan ^{-1} \sqrt{3}=\frac{\pi}{3}
\end{gathered}
$
Hence, $z$ is not a real number
So, statement-1 is false and 2 is true
Asked in: JEE Main 2013 (25 Apr Online)
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