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}}$

$=\sqrt{6+\frac{25}{4}}=\frac{7}{2}$

List $I$	List $II$
$P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j=1$	$1.$ True
$Q.$ There exists a $k \in\{1,2, \ldots ., 9\}$ such that $z_{1 .} . z=z_k$ has no solution $z$ in the set of complex numbers.	$2.$ False
$R.$ $\frac{\left\|1-z_1\right\|\left\|1-z_2\right\| \ldots . .\left\|1-z_9\right\|}{10}$ equals	$3.$ $1$
$S.$ $1-\sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}\right)$ equals	$4.$ $2$

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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