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If θ is the Amplitude of a + I B a − I B , than Tan θ = - Mathematics

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प्रश्न

If θ is the amplitude of \[\frac{a + ib}{a - ib}\] , than tan θ =

पर्याय

  • \[\frac{2a}{a^2 + b^2}\]

  • \[\frac{2ab}{a^2 - b^2}\]

  • \[\frac{a^2 - b^2}{a^2 + b^2}\]

  • none of these

MCQ
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उत्तर

\[\frac{2ab}{a^2 - b^2}\]

\[z = \frac{a + ib}{a - ib} \times \frac{a + ib}{a + ib}\]

\[ \Rightarrow z = \frac{a^2 + i^2 b^2 + 2abi}{a^2 - i^2 b^2}\]

\[ \Rightarrow z = \frac{a^2 - b^2 + 2abi}{a^2 + b^2}\]

\[ \Rightarrow z = \frac{a^2 - b^2}{a^2 + b^2} + i\frac{2ab}{a^2 + b^2}\]

\[ \Rightarrow \text { Re }\left( z \right) = \frac{a^2 - b^2}{a^2 + b^2}, \text { Im }\left( z \right) = \frac{2ab}{a^2 + b^2}\]

\[\tan \alpha = \left| \frac{Im\left( z \right)}{Re\left( z \right)} \right|\]

\[ = \frac{2ab}{a^2 - b^2}\]

\[\alpha = \tan^{- 1} \left( \frac{2ab}{a^2 - b^2} \right)\]

\[\text { Since, z lies in the first quadrant . Therefore, } \]

\[\arg (z) = \alpha = \tan^{- 1} \left( \frac{2ab}{a^2 - b^2} \right)\]

\[\tan \theta = \frac{2ab}{a^2 - b^2}\]

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पाठ 13: Complex Numbers - Exercise 13.6 [पृष्ठ ६५]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 13 Complex Numbers
Exercise 13.6 | Q 28 | पृष्ठ ६५

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