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

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Question

$\left|\frac{z-i}{z+2 i}\right|=1$

$\Rightarrow|z-i|=|z+2 i|$

$\Rightarrow \quad z$ lies on perpendicular bisector of $(0,1)$ and $(0,-2)$

$\

Vedclass · Accepted Answer

$\frac{7}{2}$

Vedclass · Answer

$\left|\frac{z-i}{z+2 i}ight|=1$

$\Rightarrow|z-i|=|z+2 i|$

$\Rightarrow \quad z$ lies on perpendicular bisector of $(0,1)$ and $(0,-2)$

$\Rightarrow \quad \operatorname{Im} z=-\frac{1}{2}$

Let $z=x-\frac{i}{2}$

$\because \quad|z|=\frac{5}{2} \quad \Rightarrow \quad x^{2}=6$

$	herefore \quad|z+3 i|=\left|x+\frac{5 i}{2}ight|=\sqrt{x^{2}+\frac{25}{4}}$

$=\sqrt{6+\frac{25}{4}}=\frac{7}{2}$

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

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