i log left( frac{x - i}{x + i} right) is equa | vedclass

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(B) Let $z = i \log \left( \frac{x - i}{x + i} \right)$.Then $\frac{z}{i} = \log \left( \frac{x - i}{x + i} \right)$.Multiplying the numerator and den

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $z = i \log \left( \frac{x - i}{x + i} ight)$.Then $\frac{z}{i} = \log \left( \frac{x - i}{x + i} ight)$.Multiplying the numerator and denominator by $(x - i)$,we get:$\frac{z}{i} = \log \left( \frac{(x - i)^2}{x^2 + 1} ight) = \log \left( \frac{x^2 - 1 - 2ix}{x^2 + 1} ight) = \log \left( \frac{x^2 - 1}{x^2 + 1} - i \frac{2x}{x^2 + 1} ight)$.Using the polar form $\log(a + ib) = \log \sqrt{a^2 + b^2} + i 	an^{-1}(\frac{b}{a})$,where $a = \frac{x^2 - 1}{x^2 + 1}$ and $b = \frac{-2x}{x^2 + 1}$:$\sqrt{a^2 + b^2} = \sqrt{\frac{(x^2 - 1)^2 + 4x^2}{(x^2 + 1)^2}} = \sqrt{\frac{x^4 + 2x^2 + 1}{(x^2 + 1)^2}} = 1$.So,$\frac{z}{i} = \log(1) + i 	an^{-1} \left( \frac{-2x / (x^2 + 1)}{(x^2 - 1) / (x^2 + 1)} ight) = 0 + i 	an^{-1} \left( \frac{-2x}{x^2 - 1} ight) = i 	an^{-1} \left( \frac{2x}{1 - x^2} ight)$.Using the identity $2 	an^{-1} x = 	an^{-1} \left( \frac{2x}{1 - x^2} ight)$ (for $|x| Thus,$z = i^2(2 	an^{-1} x) = -2 	an^{-1} x$. Considering the branch of the logarithm,the expression evaluates to $\pi - 2 	an^{-1} x$.

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

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