Question
The value of $${\text{Arg}}\left[ {i\ln \left( {\frac{{a - ib}}{{a + ib}}} \right)} \right],$$ where $$a$$ and $$b$$ are real numbers, is
A.
$$0\,\,{\text{or}}\,\,\pi $$
B.
$$\frac{\pi }{2}$$
C.
not defined
D.
None of these
Answer :
$$0\,\,{\text{or}}\,\,\pi $$
Solution :
$$\eqalign{
& \ln \left( {\frac{{a - ib}}{{a + ib}}} \right) = \ln \left| {\frac{{a - ib}}{{a + ib}}} \right| + i\left[ {2n\pi + \arg \left( {\frac{{a - ib}}{{a + ib}}} \right)} \right] \cr
& = i\left[ {2n\pi + \arg \left( {\frac{{a - ib}}{{a + ib}}} \right)} \right]{\text{ Since, }}\left| {\frac{{a - ib}}{{a + ib}}} \right| = 1 \cr
& \therefore {\text{Arg}}\left[ {i\ln \left( {\frac{{a - ib}}{{a + ib}}} \right)} \right] \cr
& = {\text{Arg}}\left[ { - 2n\pi - \arg \left( {\frac{{a - ib}}{{a + ib}}} \right)} \right] = 0\,\,{\text{or }}\pi \cr
& {\text{As }}2n\pi + \arg \left( {\frac{{a - ib}}{{a + ib}}} \right)\,\,{\text{is a real number}}{\text{.}} \cr} $$