Complex Numbers And Quadratic Equations question 704
Question: The value of $ Arg[ i\ln ( \frac{a-ib}{a+ib} ) ], $ where a and b are real numbers, is
Options:
A) $ 0 $ or $ \pi $
B) $ \frac{\pi }{2} $
C) not defined
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \ell n( \frac{a-ib}{a+ib} )=\ell n| \frac{a-ib}{a+ib} |+i[ 2n\pi +\arg ( \frac{a-ib}{a+ib} ) ] $ $ =i[ 2n\pi +\arg ( \frac{a-ib}{a+ib} ) ]Since| \frac{a-ib}{a+ib} |=1 $
$ \therefore Arg[ i\ell n( \frac{a-ib}{a+ib} ) ] $ $ =Arg[ -2n\pi -\arg ( \frac{a-ib}{a+ib} ) ]=0or\pi $ As $ 2n\pi + arg( \frac{a-ib}{a+ib} ) $ is a real number.
BETA
