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Let $z$ be a complex number such that $|z|-z=2+i$, where $i=\sqrt{-1}$. Then, $|z|=$
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Verified Answer
The correct answer is:
$\frac{5}{4}$
We have, $|z|-z=2+i$
Let $z=x+i y \quad \therefore \sqrt{x^2+y^2}-x-i y=2+i$
Equating real part and imaginary part we get,
$\begin{array}{lll}
& \sqrt{x^2+y^2}-x=2 \text { and } y=-1 \\
\therefore & \sqrt{x^2+1}=x+2 \Rightarrow x^2+1=x^2+4 x+4 \\
\Rightarrow & x=-\frac{3}{4} \Rightarrow z=-\frac{3}{4}-i \\
\therefore & |z|=\sqrt{\frac{9}{16}+1}=\sqrt{\frac{25}{16}}=\frac{5}{4}
\end{array}$
Let $z=x+i y \quad \therefore \sqrt{x^2+y^2}-x-i y=2+i$
Equating real part and imaginary part we get,
$\begin{array}{lll}
& \sqrt{x^2+y^2}-x=2 \text { and } y=-1 \\
\therefore & \sqrt{x^2+1}=x+2 \Rightarrow x^2+1=x^2+4 x+4 \\
\Rightarrow & x=-\frac{3}{4} \Rightarrow z=-\frac{3}{4}-i \\
\therefore & |z|=\sqrt{\frac{9}{16}+1}=\sqrt{\frac{25}{16}}=\frac{5}{4}
\end{array}$
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