Let z be a complex number such that |z|-z=2+i | vedclass

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(D) We have,$|z|-z=2+i$.Let $z=x+iy$. Then $\sqrt{x^2+y^2}-(x+iy)=2+i$.Equating the real and imaginary parts,we get:$\sqrt{x^2+y^2}-x=2$ and $-y=1$.Th

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$\frac{5}{4}$

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(D) We have,$|z|-z=2+i$.Let $z=x+iy$. Then $\sqrt{x^2+y^2}-(x+iy)=2+i$.Equating the real and imaginary parts,we get:$\sqrt{x^2+y^2}-x=2$ and $-y=1$.Thus,$y=-1$.Substituting $y=-1$ into the first equation: $\sqrt{x^2+(-1)^2}-x=2$.$\sqrt{x^2+1}=x+2$.Squaring both sides: $x^2+1=(x+2)^2 = x^2+4x+4$.Solving for $x$: $1=4x+4$ $\Rightarrow 4x=-3$ $\Rightarrow x=-\frac{3}{4}$.Therefore,$z=-\frac{3}{4}-i$.The modulus is $|z|=\sqrt{(-\frac{3}{4})^2+(-1)^2} = \sqrt{\frac{9}{16}+1} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

Let $z$ be a complex number such that $|z|-z=2+i$,where $i=\sqrt{-1}$. Then,$|z|=$

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