Complex Numbers And Quadratic Equations question 671
Question: If $ Re( \frac{z-1}{z+1} )=0, $ where $ 2=x+iy $ is a complex number, then which one of the following is correct?
Options:
A) $ z=1+i $
B) $ | z |=2 $
C) $ z=1-i $
D) $ | z |=1 $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{z-1}{z+1}=\frac{x+iy-1}{x+iy+1} $ $ \frac{z-1}{z+1}=\frac{x^{2}+y^{2}-1+2iy}{x^{2}+y^{2}+2x+1} $
$ \Rightarrow Re( \frac{z-1}{z+1} )=\frac{x^{2}+y^{2}-1}{x^{2}+y^{2}+2x+1}=0 $
$ \Rightarrow x^{2}+y^{2}-1=0 $
$ \Rightarrow x^{2}+y^{2}=1 $ Also, $ z\bar{z}=x^{2}+y^{2}=1 $ $ z\bar{z}={{| z |}^{2}} $
$ \Rightarrow {{| z |}^{2}}=1 $
$ \Rightarrow | z |=1 $
BETA
