Complex Numbers And Quadratic Equations question 842
Question: $ i\log ( \frac{x-i}{x+i} ) $ is equal to [RPET 2000]
Options:
A) $ \pi +2{{\tan }^{-1}}x $
B) $ \pi -2{{\tan }^{-1}}x $
C) $ -\pi +2{{\tan }^{-1}}x $
D) $ -\pi -2{{\tan }^{-1}}x $
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ z=i\log ( \frac{x-i}{x+i} ) $
$ \Rightarrow \frac{z}{i}=\log ( \frac{x-i}{x+i} ) $
$ \Rightarrow \frac{z}{i}=\log [ \frac{x-i}{x+i}\times \frac{x-i}{x-i} ] $ $ =\log [ \frac{x^{2}-1-2ix}{x^{2}+1} ] $
$ \Rightarrow \frac{z}{i}=\log [ \frac{x^{2}-1}{x^{2}+1}-i\frac{2x}{x^{2}+1} ] $ ……(i) $ \because \log (a+ib)=\log (r{e^{i\theta }})=\log r+i\theta $ = $ \log \sqrt{a^{2}+b^{2}}+i{{\tan }^{-1}}(b/a) $ Hence, $ \frac{z}{i}=\log \sqrt{{{( \frac{x^{2}-1}{x^{2}+1} )}^{2}}+{{( \frac{-2x}{x^{2}+1} )}^{2}}}+i{{\tan }^{-1}}( \frac{-2x}{x^{2}-1} ) $ [by eqn. (i)] $ \frac{z}{i}=\log \frac{\sqrt{x^{4}+1-2x^{2}+4x^{2}}}{{{(x^{2}+1)}^{2}}} $ $ +i{{\tan }^{-1}}( \frac{2x}{1-x^{2}} ) $ $ =\log 1+i(2{{\tan }^{-1}}x) $ $ =0+i(2{{\tan }^{-1}}x) $
$ \therefore z=i^{2}2{{\tan }^{-1}}x=-2{{\tan }^{-1}}x $ $ =\pi -2{{\tan }^{-1}}x $ .
BETA
