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

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(D) Let $Z = a + ib$. Then $|Z| = \sqrt{a^2 + b^2}$. Given $|Z| + Z = 2 + i$. Substituting the values,we get $\sqrt{a^2 + b^2} + a + ib = 2 + i$. Comp

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$\frac{5}{4}$

Vedclass · Answer

(D) Let $Z = a + ib$. Then $|Z| = \sqrt{a^2 + b^2}$. Given $|Z| + Z = 2 + i$. Substituting the values,we get $\sqrt{a^2 + b^2} + a + ib = 2 + i$. Comparing the real and imaginary parts,we get $b = 1$ and $\sqrt{a^2 + b^2} + a = 2$. Since $b = 1$,$\sqrt{a^2 + 1} = 2 - a$. Squaring both sides,$a^2 + 1 = (2 - a)^2$. $a^2 + 1 = 4 - 4a + a^2$. $4a = 3$,so $a = \frac{3}{4}$. Now,$|Z| = \sqrt{a^2 + b^2} = \sqrt{(\frac{3}{4})^2 + 1^2} = \sqrt{\frac{9}{16} + 1} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

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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