Let Z be a complex number satisfying |Z−4iZ−2i|=1 and |Z−8+3iZ+3i|=35 then Z can be _______

|Z−4i|=|Z−2i|⇒Z is on perpendicular bisector of (0,4) and (0,2) i.e Z is on y=3 Z=x+3i Sub in |Z−8+3iZ+3i|=35 ⇒5|x−8+6i|=3|x+6i| ⇒5|x−8+6i|=3|x+6i|⇒x2−25x+136=0(x−8)(x−17)=0x=8 or 17∴ Z=8+3i or 17+3i

Let Z be a complex number satisfying |Z−4iZ−2i|=1 and |Z−8+3iZ+3i|=35 then Z can be _______

|Z−4i|=|Z−2i|⇒Z is on perpendicular bisector of (0,4) and (0,2) i.e Z is on y=3 Z=x+3i Sub in |Z−8+3iZ+3i|=35 ⇒5|x−8+6i|=3|x+6i| ⇒5|x−8+6i|=3|x+6i|⇒x2−25x+136=0(x−8)(x−17)=0x=8 or 17∴ Z=8+3i or 17+3i

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Let Z be a complex number satisfying $| \frac{Z - 4 i}{Z - 2 i} | = 1$ and $| \frac{Z - 8 + 3 i}{Z + 3 i} | = \frac{3}{5}$ then Z can be _______

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$8 - 3 i$

$3 + 8 i$

$17 + 3 i$

$8 + 17 i$

answer is C.

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Detailed Solution

$| Z - 4 i | = | Z - 2 i | \Rightarrow$ Z is on perpendicular bisector of $(0, 4)$ and $(0, 2)$
i.e Z is on $y = 3$
$Z = x + 3 i$
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Sub in $| \frac{Z - 8 + 3 i}{Z + 3 i} | = \frac{3}{5}$
$\Rightarrow 5 | x - 8 + 6 i | = 3 | x + 6 i | \begin{array}{l} \Rightarrow 5 | x - 8 + 6 i | = 3 | x + 6 i | \\ \Rightarrow x^{2} - 25 x + 136 = 0 \\ (x - 8) (x - 17) = 0 \\ x = 8 o r 17 \\ ∴ Z = 8 + 3 i o r 17 + 3 i \end{array}$