Complex Numbers And Quadratic Equations question 668
Question: If n is a positive integer grater than unity and z is a complex satisfying the equation $ z^{n}={{(z+1)}^{n}}, $ then
Options:
A) $ Re(z)<2 $
B) $ Re(z)>0 $
C) $ Re(z)=0 $
D) z lies on $ x=-\frac{1}{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ {z^{n}}={{(z+1)}^{n}}\Rightarrow {{| z |}^{n}}={{| z+1 |}^{n}} $ or $ | z |=| z+1 | $ So the distance of point z remain same from (0, 0) and(-1, 0). So, z lies on perpendicular bisector of line joining (0, 0) and (-1, 0) that is on $ x = -\frac{1}{2} $
BETA
