Complex Numbers And Quadratic Equations question 370
Question: If $ z=x+iy $ is a complex number satisfying $ {{| z+\frac{i}{2} |}^{2}}= $ $ {{| z-\frac{i}{2} |}^{2}}, $ then the locus of z is [EAMCET 2002]
Options:
A) $ 2y=x $
B) $ y=x $
C) y-axis
D) x-axis
Show Answer
Answer:
Correct Answer: D
Solution:
$ {{| z+\frac{i}{2} |}^{2}}={{| z-\frac{i}{2} |}^{2}} $
Þ $ {{| x+iy+\frac{i}{2} |}^{2}}={{| x+iy-\frac{i}{2} |}^{2}} $
Þ $ {{| x+i( y+\frac{1}{2} ) |}^{2}}={{| x+i( y-\frac{1}{2} ) |}^{2}} $
$ \Rightarrow x^{2}+{{( y+\frac{1}{2} )}^{2}}=x^{2}+{{( y-\frac{1}{2} )}^{2}} $
Þ $ 2y=0 $ i.e. x-axis.
BETA
