Let z be a complex number such that left|frac | vedclass

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(C) Given $\left|\frac{z-i}{z+2i}\right|=1$,which implies $|z-i|=|z+2i|$.This means $z$ lies on the perpendicular bisector of the points $(0, 1)$ and

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$\frac{7}{2}$

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(C) Given $\left|\frac{z-i}{z+2i}ight|=1$,which implies $|z-i|=|z+2i|$.This means $z$ lies on the perpendicular bisector of the points $(0, 1)$ and $(0, -2)$.The perpendicular bisector is the line $	ext{Im}(z) = -\frac{1}{2}$.Let $z = x - \frac{i}{2}$.Given $|z| = \frac{5}{2}$,we have $x^2 + (-\frac{1}{2})^2 = (\frac{5}{2})^2$.$x^2 + \frac{1}{4} = \frac{25}{4} \Rightarrow x^2 = 6$.Now,$|z+3i| = |x - \frac{i}{2} + 3i| = |x + \frac{5i}{2}|$.$|z+3i| = \sqrt{x^2 + (\frac{5}{2})^2} = \sqrt{6 + \frac{25}{4}} = \sqrt{\frac{24+25}{4}} = \sqrt{\frac{49}{4}} = \frac{7}{2}$.

Let $z$ be a complex number such that $\left|\frac{z-i}{z+2i}\right|=1$ and $|z|=\frac{5}{2}$. Then the value of $|z+3i|$ is:

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