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(B) Let $z = x + iy$,where $x, y \in \mathbb{R}$.Given $|z| + z = 3 + i$.Substituting $z = x + iy$,we get $\sqrt{x^2 + y^2} + x + iy = 3 + i$.Equating

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$\frac{5}{3}$

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(B) Let $z = x + iy$,where $x, y \in \mathbb{R}$.Given $|z| + z = 3 + i$.Substituting $z = x + iy$,we get $\sqrt{x^2 + y^2} + x + iy = 3 + i$.Equating the real and imaginary parts:$1) \sqrt{x^2 + y^2} + x = 3$$2) y = 1$Substitute $y = 1$ into the first equation:$\sqrt{x^2 + 1} = 3 - x$Squaring both sides:$x^2 + 1 = (3 - x)^2$$x^2 + 1 = 9 - 6x + x^2$$6x = 8 \implies x = \frac{8}{6} = \frac{4}{3}$.Now,$|z| = \sqrt{x^2 + y^2} = 3 - x = 3 - \frac{4}{3} = \frac{9 - 4}{3} = \frac{5}{3}$.Thus,$|z| = \frac{5}{3}$.

Let $z$ be a complex number such that $|z| + z = 3 + i$,where $i = \sqrt{-1}$. Then $|z| = $

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