• question_answer Find the complex number z satisfying the equations $\left| \frac{z-12}{z-8i} \right|=\frac{5}{3},\left| \frac{z-4}{z-8} \right|=1$ [Roorkee 1993] A) 6 B) $6\pm 8i$ C) $6+8i,\,6+17i$ D) None of these

We have $\left| \frac{z-12}{z-8i} \right|=\frac{5}{3}$and $\left| \frac{z-4}{z-8} \right|=1$ Let$z=x+iy$, then  $\left| \frac{z-12}{z-8i} \right|=\frac{5}{3}\Rightarrow 3|z-12|=5|z-8i|$ Þ $3|(x-12)+iy|=5|x+(y-8)i|$ Þ $9{{(x-12)}^{2}}+9{{y}^{2}}=25{{x}^{2}}+25{{(y-8)}^{2}}$     ....(i) and $\left| \frac{z-4}{z-8} \right|=1\Rightarrow |z-4|=|z-8|$ Þ $|x-4+iy|=|x-8+iy|$ Þ ${{(x-4)}^{2}}+{{y}^{2}}={{(x-8)}^{2}}+{{y}^{2}}\Rightarrow x=6$ Putting $x=6$in (i), we get ${{y}^{2}}-25y+136=0$ \ $y=17,8$ Hence $z=6+17i$or $z=6+8i$ Trick: Check it with options.