Complex Numbers And Quadratic Equations question 358
Question: The locus of the points z which satisfy the condition arg $ ( \frac{z-1}{z+1} ) $ = $ \frac{\pi }{3} $ is
Options:
A) A straight line
B) A circle
C) A parabola
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
We have $ \frac{z-1}{z+1}=\frac{x+iy-1}{x+iy+1}=\frac{(x^{2}+y^{2}-1)+2iy}{{{(x+1)}^{2}}+y^{2}} $ Therefore $ arg\frac{z-1}{z+1}={{\tan }^{-1}}\frac{2y}{x^{2}+y^{2}-1} $ Hence $ {{\tan }^{-1}}\frac{2y}{x^{2}+y^{2}-1}=\frac{\pi }{3} $
Þ $ \frac{2y}{x^{2}+y^{2}-1}=\tan \frac{\pi }{3}=\sqrt{3} $
Þ $ x^{2}+y^{2}-1=\frac{2}{\sqrt{3}}y $
Þ $ x^{2}+y^{2}-\frac{2}{\sqrt{3}}y-1=0 $ Which is obviously a circle.
BETA
