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

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(B) Given $|z|+z=2+i$. Let $z=x+iy$,where $x, y \in \mathbb{R}$. Then $|z|=\sqrt{x^2+y^2}$. Substituting these into the equation: $\sqrt{x^2+y^2}+x+iy

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$\frac{5}{4}$

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(B) Given $|z|+z=2+i$. Let $z=x+iy$,where $x, y \in \mathbb{R}$. Then $|z|=\sqrt{x^2+y^2}$. Substituting these into the equation: $\sqrt{x^2+y^2}+x+iy=2+i$. Equating real and imaginary parts,we get: $y=1$ and $\sqrt{x^2+y^2}+x=2$. Substituting $y=1$ into the second equation: $\sqrt{x^2+1}=2-x$. Squaring both sides: $x^2+1=(2-x)^2 = 4-4x+x^2$. $1=4-4x$ $\Rightarrow 4x=3$ $\Rightarrow x=\frac{3}{4}$. Thus,$z=\frac{3}{4}+i$. Therefore,$|z|=\sqrt{(\frac{3}{4})^2+1^2} = \sqrt{\frac{9}{16}+1} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

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

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