Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# 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

#### 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?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 13 Complex Numbers
Q 28 | Page 65