Complex Numbers And Quadratic Equations question 373
Question: If $ |z^{2}-1|=|z{{|}^{2}}+1 $ , then $ z $ lies on [AIEEE 2004]
Options:
A) An ellipse
B) The imaginary axis
C) A circle
D) The real axis
Show Answer
Answer:
Correct Answer: B
Solution:
We have $ |z^{2}-1|\ =\ |z{{|}^{2}}+1 $
Þ $ |{{(x+iy)}^{2}}-1|\ =\ |x+iy{{|}^{2}}+1 $
Þ $ |(x^{2}-y^{2}-1)+2xyi|\ ={{( \sqrt{x^{2}+y^{2}} )}^{2}}+1 $
Þ $ \sqrt{{{(x^{2}-y^{2}-1)}^{2}}+{{(2xy)}^{2}}}=x^{2}+y^{2}+1 $
Þ $ x^{4}+y^{4}+1-2x^{2}y^{2}+2y^{2}-2x^{2}+4x^{2}y^{2} $ $ =x^{4}+y^{4}+1+2x^{2}y^{2}+2y^{2}+2x^{2} $
Þ $ 2x^{2}y^{2}=2x^{2}y^{2}+4x^{2} $
Þ $ x=0 $ then, $ z=x+iy=0+iy=iy $ Hence $ z $ lies on imaginary axis.
BETA
