i ( x - i x + i ) is equal to (x R)

MCQ

$i\log \left( {\frac{{x - i}}{{x + i}}} \right)$ is equal to $(x \in R)$

A
$\pi + 2{\tan ^{ - 1}}x$
✓
$\pi - 2{\tan ^{ - 1}}x$
C
$ - \pi + 2{\tan ^{ - 1}}x$
D
$ - \pi - 2{\tan ^{ - 1}}x$

✓

Answer

Correct option: B.

$\pi - 2{\tan ^{ - 1}}x$

b
(b)Let $z = \,i\log \left( {\frac{{x - i}}{{x + i}}} \right)$$ \Rightarrow \,\frac{z}{i} = \log \left( {\frac{{x - i}}{{x + i}}} \right)$
$ \Rightarrow \frac{z}{i} = \log \,\left[ {\frac{{x - i}}{{x + i}} \times \frac{{x - i}}{{x - i}}} \right]$$ = \,\log \,\left[ {\frac{{{x^2} - 1 - 2ix}}{{{x^2} + 1}}} \right]$
$ \Rightarrow \frac{z}{i} = \log \left[ {\frac{{{x^2} - 1}}{{{x^2} + 1}} - i\frac{{2x}}{{{x^2} + 1}}} \right]$ ......$(i)$

$\,\log (a + ib) = \log (r{e^{i\theta }}) = \log r + i\theta $= $\log \sqrt {{a^2} + {b^2}} + i{\tan ^{ - 1}}(b/a)$
Hence, $\frac{z}{i} = \log \sqrt {{{\left( {\frac{{{x^2} - 1}}{{{x^2} + 1}}} \right)}^2} + {{\left( {\frac{{ - 2x}}{{{x^2} + 1}}} \right)}^2}} + i{\tan ^{ - 1}}\left( {\frac{{ - 2x}}{{{x^2} - 1}}} \right)$

[by eqn. $(i)$]

$\frac{z}{i} = \log \frac{{\sqrt {{x^4} + 1 - 2{x^2} + 4{x^2}} }}{{{{({x^2} + 1)}^2}}}$ $ + i{\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right)$
$ = \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$.

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