Complex Numbers And Quadratic Equations question 138
Question: Find the complex number z satisfying the equations $ | \frac{z-12}{z-8i} |=\frac{5}{3},| \frac{z-4}{z-8} |=1 $ [Roorkee 1993]
Options:
A) 6
B) $ 6\pm 8i $
C) $ 6+8i,6+17i $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
We have $ | \frac{z-12}{z-8i} |=\frac{5}{3} $ and $ | \frac{z-4}{z-8} |=1 $ Let $ z=x+iy $ , then $ | \frac{z-12}{z-8i} |=\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}}+9y^{2}=25x^{2}+25{{(y-8)}^{2}} $ ….(i) and $ | \frac{z-4}{z-8} |=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.
BETA
