Question

Find the complex numbers z which simultaneously satisfy the equation $\left|\frac{z-12}{z-8i}\right|=\frac{5}{3}$ and $\left|\frac{z-4}{z-8}\right|=1$.

• A. 6 + 8 i or 6 + 17 i
• B. 6 + 8 i or 6 - 17 i
• C. 6 - 8 i or 6 + 17 i
• D. 6 - 8 i or 6 - 17 i

Answer 1

The correct option is A 6 + 8 i or 6 + 17 i

Put $z=x+iy$
$\left|\frac{z-4}{z-8}\right|=1\Rightarrow \left|\frac{x-4+iy}{x-8+iy}\right|=1\Rightarrow {(x-4)}^3+y^2={(x-8)}^2+y^2\Rightarrow x=6$
With $x=6,\left|\frac{z-12}{z-8i}\right|=\frac{5}{3}$
$\Rightarrow \left|\frac{-6+iy}{6+i(y-8)}\right|=\frac{5}{3}\Rightarrow 9(36+y^2)=25[36+(y-8)2]\Rightarrow y^2-25y+136=0\Rightarrow y=17,8$
Hence the required numbers are
$z=6+17i,6+8i$