Complex Numbers And Quadratic Equations question 376
Question: If $ z=x+iy $ and $ arg( \frac{z-2}{z+2} )=\frac{\pi }{6} $ , then locus of z is [RPET 2002]
Options:
A) A straight line
B) A circle
C) A parabola
D) An ellipse
Show Answer
Answer:
Correct Answer: B
Solution:
Put $ z=x+iy $ in arg $ ( \frac{z-2}{z+2} )=\frac{\pi }{6} $ $ arg( \frac{(x-2)+iy}{(x+2)+iy} )=\frac{\pi }{6} $ $ arg((x-2)+iy)-arg((x+2)+iy)=\frac{\pi }{6} $ $ {{\tan }^{-1}}\frac{y}{x-2}-{{\tan }^{-1}}\frac{y}{x+2}=\frac{\pi }{6} $ . $ {{\tan }^{-1}}( \frac{\frac{y}{x-2}-\frac{y}{x+2}}{1+\frac{y^{2}}{x^{2}-4}} )=\frac{\pi }{6} $
$ \Rightarrow \frac{xy+2y-xy+2y}{x^{2}+y^{2}-4}=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}} $ $ =x^{2}+y^{2}-4\sqrt{3y}-4=0 $ Which is equation of a circle.
BETA
