A) \[0\] or \[\pi \]
B) \[\frac{\pi }{2}\]
C) not defined
D) None of these
Correct Answer: A
Solution :
\[\ell n\left( \frac{a-ib}{a+ib} \right)=\ell n\left| \frac{a-ib}{a+ib} \right|+i\left[ 2n\pi +\arg \left( \frac{a-ib}{a+ib} \right) \right]\] \[=\,\,\,i\left[ 2n\pi +\arg \left( \frac{a-ib}{a+ib} \right) \right]\,Since\,\,\left| \frac{a-ib}{a+ib} \right|=1\] \[\therefore \,\,\,\,Arg\left[ i\ell n\left( \frac{a-ib}{a+ib} \right) \right]\] \[=\,\,Arg\left[ -2n\pi -\arg \left( \frac{a-ib}{a+ib} \right) \right]=0\,\,or\,\,\pi \] As \[2n\pi + arg\left( \frac{a-ib}{a+ib} \right)\] is a real number.
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