Q202 / 270 · NDA 2016 (Phase 1) · Mathematics
Let z be a complex number satisfying
$\left|\frac{\mathrm{z}-4}{\mathrm{z}-8}\right|=1 \text { and }\left|\frac{\mathrm{z}}{\mathrm{z}-2}\right|=\frac{3}{2}$
What is $|z|$ equal to?
- Correct answerA6
- B12
- C18
- D36
Solution
$\begin{aligned}
& \left|\frac{z-4}{z-8}\right|=1 \text { and }\left|\frac{z}{z-2}\right|=\frac{3}{2} \\
& \Rightarrow|z-4|=|z-8|
\end{aligned}$
Let $z=x+i y$
$|x+i y-4|=|x+i y-8|$
Squaring both sides, we get
$\begin{aligned}
& {\left[(x-4)^2+y^2\right]=\left[(x-8)^2+y^2\right]} \\
& (x-4)^2=(x-8)^2 \\
& \Rightarrow x^2+16-8 x=x^2+64-16 x \\
& \Rightarrow 8 x=48 \Rightarrow x=6
\end{aligned}$
when $\left|\frac{z}{z-2}\right|=\frac{3}{2}$
$\Rightarrow 2|z|=3|z-2|$
Squaring both sides, we get
$\begin{aligned}
& 4\left(x^2+y^2\right)=9\left[(x-2)^2+y^2\right] \\
& \Rightarrow 4 x^2+4 y^2=9 x^2+36-36 x+9 y^2 \\
& \Rightarrow 5 x^2+5 y^2-36 x+36=0
\end{aligned}$
as we know $x=6$
$\begin{aligned}
& 5(6)^2+5 y^2-36 \times 6+36=0 \\
& \Rightarrow 5 y^2=0 \Rightarrow y=0
\end{aligned}$
Hence $x=6$ and $y=0$.
$\begin{aligned}
& \Rightarrow z=6 \\
& |z|=6
\end{aligned}$
Hence, the correct answer is option: 'A'.