Complex Numbers And Quadratic Equations question 371
Question: The locus of the point z satisfying $ arg( \frac{z-1}{z+1} )=k, $ (where k is non zero) is [Orissa JEE 2002]
Options:
A) Circle with centre on y-axis
B) Circle with centre on x-axis
C) A straight line parallel to x-axis
D) A straight line making an angle $ 60^{o} $ with the x-axis
Show Answer
Answer:
Correct Answer: A
Solution:
$ arg( \frac{z-1}{z+1} )=k $
$ \Rightarrow $ $ arg( \frac{(x-1)+iy}{(x+1)+iy} )=k $
$ \Rightarrow $ $ arg[ (x-1)+iy ]-arg[ (x+1)+iy ]=k $
$ \Rightarrow $ $ {{\tan }^{-1}}( \frac{y}{x-1} )-{{\tan }^{-1}}( \frac{y}{x+1} )=k $
$ \Rightarrow $ $ {{\tan }^{-1}}[ \frac{\frac{y}{(x-1)}-\frac{y}{(x+1)}}{1+\frac{y^{2}}{x^{2}-1}} ]=k $
$ \Rightarrow $ $ \tan k=\frac{y(x+1)-y(x-1)}{x^{2}+y^{2}-1}=\frac{2y}{x^{2}+y^{2}-1} $
$ \Rightarrow $ $ \frac{2y}{\tan k}=x^{2}+y^{2}-1 $
$ \Rightarrow $ $ x^{2}+y^{2}-\frac{2y}{\tan k}-1=0 $ It is an equation of circle whose centre is $ (-g,-f)=(0,\cot k) $ on y-axis.
BETA
