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4-1.Complex numbers
hard
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
A
$\sqrt{10}$
B
$2 \sqrt{3}$
C
$\frac{7}{2}$
D
$\frac{15}{4}$
(JEE MAIN-2020)
Solution
$\left|\frac{z-i}{z+2 i}\right|=1$
$\Rightarrow|z-i|=|z+2 i|$
$\Rightarrow \quad z$ lies on perpendicular bisector of $(0,1)$ and $(0,-2)$
$\Rightarrow \quad \operatorname{Im} z=-\frac{1}{2}$
Let $z=x-\frac{i}{2}$
$\because \quad|z|=\frac{5}{2} \quad \Rightarrow \quad x^{2}=6$
$\therefore \quad|z+3 i|=\left|x+\frac{5 i}{2}\right|=\sqrt{x^{2}+\frac{25}{4}}$
$=\sqrt{6+\frac{25}{4}}=\frac{7}{2}$
Standard 11
Mathematics