Complex Numbers And Quadratic Equations question 129
Question: If $ \sqrt{3}+i=(a+ib)(c+id) $ , then $ {{\tan }^{-1}}( \frac{b}{a} )+ $ $ {{\tan }^{-1}}( \frac{d}{c} ) $ has the value
Options:
A) $ \frac{\pi }{3}+2n\pi ,n\in I $
B) $ n\pi +\frac{\pi }{6},n\in I $
C) $ n\pi -\frac{\pi }{3},n\in I $
D) $ 2n\pi -\frac{\pi }{3},n\in I $
Show Answer
Answer:
Correct Answer: B
Solution:
We have $ \sqrt{3}+i=(a+ib)(c+id) $
$ \therefore ac-bd=\sqrt{3} $ and $ ad+bc=1 $ Now tan-1 $ ( \frac{b}{a} )+{{\tan }^{-1}}( \frac{d}{c} ) $ $ \begin{aligned} & ={{\tan }^{-1}}( \frac{\frac{b}{a}+\frac{d}{c}}{1-\frac{b}{a}.\frac{d}{c}} )={{\tan }^{-1}}( \frac{bc+ad}{ac-bd} )={{\tan }^{-1}}( \frac{1}{\sqrt{3}} ) \\ & \\ \end{aligned} $ $ =n\pi +\frac{\pi }{6},n\in I $
BETA
