Complex Numbers And Quadratic Equations question 657
Question: If $ |z-2|=\min {|z-1|,|z-5|}, $ where z is a complex number, then
Options:
A) $ Re(z)=\frac{3}{2} $
B) $ Re(z)=\frac{7}{2} $
C) $ Re(z)\in { \frac{3}{2},\frac{7}{2} } $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ |z-2|=min{ | z-1 |<| z-5 | } $ i.e., $ | z-2 | = | z-1 |,where| z-1 |<| z- 5 | $
$ \Rightarrow Re(z) = \frac{3}{2}which satisfy | z-5 |<| z-1 | $ Also, $ | z-2 |=| z-5 |, where | z-5 |<| z-1 | $
$ \Rightarrow Re(z)= \frac{7}{2} which satisfy| z-5 || z-1 | $
BETA
