Complex Numbers And Quadratic Equations question 78
Question: If $ |z|=1,(z\ne -1) $ and $ z=x+iy, $ then $ ( \frac{z-1}{z+1} ) $ is [RPET 1997]
Options:
A) Purely real
B) Purely imaginary
C) Zero
D) Undefined
Show Answer
Answer:
Correct Answer: B
Solution:
$ z=x+iy\Rightarrow |z{{|}^{2}}=x^{2}+y^{2}=1 $ …..(i) Now, $ ( \frac{z-1}{z+1} )=\frac{(x-1)+iy}{(x+1)+iy}\times \frac{(x+1)-iy}{(x+1)-iy} $ $ =\frac{(x^{2}+y^{2}-1)+2iy}{{{(x+1)}^{2}}+y^{2}} $ $ =\frac{2iy}{{{(x+1)}^{2}}+y^{2}} $ [by equation (i)] Hence, $ ( \frac{z-1}{z+1} ) $ is purely imaginary.
BETA
