Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
Advertisement Remove all ads

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

MCQ

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

Advertisement Remove all ads

Solution

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

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

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

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

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

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

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

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

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

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

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

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

  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 13 Complex Numbers
Q 28 | Page 65
Advertisement Remove all ads

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×