Complex Numbers And Quadratic Equations question 644

Question: If $ |z|=\max {|z-1|,|z+1|} $ then

Options:

A) $ |z+\bar{z}|=\frac{1}{2} $

B) $ z+\bar{z}=1 $

C) $ |z+\bar{z}|=1 $

D) None of these

Show Answer

Answer:

Correct Answer: D

Solution:

If $ | z-1 |>| z+1 |, $ then max $ {|z-1|,|z+1|}=|z-1| $
$ \Rightarrow $ If $ |z{{|}^{2}}+1-z-\bar{z}>|z{{|}^{2}}+1+z+\bar{z} $ then $ |z|=|z-1| $
$ \Rightarrow $ If $ z+\bar{z}<0 $ then $ |z{{|}^{2}}=|z{{|}^{2}}+1-z-\bar{z} $
$ \Rightarrow $ If $ z+\bar{z}<0 $ then $ z+\bar{z}=1, $ which is not possible. Again If $ |z+1|>|z-1| $ then max $ {|z-1|,|z+1|}=|z+1| $
$ \Rightarrow $ If $ |z{{|}^{2}}+1+z+\bar{z}>|z{{|}^{2}}+1-z-\bar{z} $ then $ |z|=|z+1| $
$ \Rightarrow $ If $ z+\bar{z}>0 $ then $ |z{{|}^{2}}=|z{{|}^{2}}+1+z+\bar{z} $
$ \Rightarrow $ If $ z+\bar{z}>0 $ then $ z+\bar{z}=-1 $ Not possible again. Therefore the given result cannot hold.



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