(c) $\left| {\frac{{z – 12}}{{z – 8i}}} \right| = \frac{5}{3}$एवं $\left| {\frac{{z – 4}}{{z – 8}}} \right| = 1$

माना $z = x + iy$, तब

$\left| {\frac{{z – 12}}{{z – 8i}}} \right| = \frac{5}{3} \Rightarrow 3|z – 12| = 5|z – 8i|$

$3|(x – 12) + iy| = 5|x + (y – 8)i|$

$9{(x – 12)^2} + 9{y^2} = 25{x^2} + 25{(y – 8)^2}$     ….$(i)$

एवं $\left| {\frac{{z – 4}}{{z – 8}}} \right| = 1 \Rightarrow |z – 4| = |z – 8|$

$|x – 4 + iy| = |x – 8 + iy|$

${(x – 4)^2} + {y^2} = {(x – 8)^2} + {y^2} \Rightarrow x = 6$

समीकरण $(i) $ में $x = 6$ रखने पर ${y^2} – 25y + 136 = 0$

$y = 17,8$

अत: $z = 6 + 17i$या $z = 6 + 8i$

ट्रिक : विकल्पों द्वारा जाँच करें।