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Question
If θ is the amplitude of \[\frac{a + ib}{a - ib}\] , than tan θ =
Options
\[\frac{2a}{a^2 + b^2}\]
\[\frac{2ab}{a^2 - b^2}\]
\[\frac{a^2 - b^2}{a^2 + b^2}\]
none of these
Solution
\[\frac{2ab}{a^2 - b^2}\]
\[z = \frac{a + ib}{a - ib} \times \frac{a + ib}{a + ib}\]
\[ \Rightarrow z = \frac{a^2 + i^2 b^2 + 2abi}{a^2 - i^2 b^2}\]
\[ \Rightarrow z = \frac{a^2 - b^2 + 2abi}{a^2 + b^2}\]
\[ \Rightarrow z = \frac{a^2 - b^2}{a^2 + b^2} + i\frac{2ab}{a^2 + b^2}\]
\[ \Rightarrow \text { Re }\left( z \right) = \frac{a^2 - b^2}{a^2 + b^2}, \text { Im }\left( z \right) = \frac{2ab}{a^2 + b^2}\]
\[\tan \alpha = \left| \frac{Im\left( z \right)}{Re\left( z \right)} \right|\]
\[ = \frac{2ab}{a^2 - b^2}\]
\[\alpha = \tan^{- 1} \left( \frac{2ab}{a^2 - b^2} \right)\]
\[\text { Since, z lies in the first quadrant . Therefore, } \]
\[\arg (z) = \alpha = \tan^{- 1} \left( \frac{2ab}{a^2 - b^2} \right)\]
\[\tan \theta = \frac{2ab}{a^2 - b^2}\]
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