Question Number 19898 by Tinkutara last updated on 17/Aug/17
$$\mathrm{Simplify}\::\:{i}\:\mathrm{log}\:\left(\frac{{x}\:−\:{i}}{{x}\:+\:{i}}\right). \\ $$
Answered by ajfour last updated on 22/Aug/17
![let t=iln (((x−i)/(x+i))) and let z=((x−i)/(x+i)) ⇒ ∣z∣=1 so let z=e^(iθ) here − π < θ < 0 t=iln z=i[ iθ ]=−θ Now let tan ∅ =(1/x)> 0 as x>0 , 0 < φ < (π/2) ⇒ θ=−2φ (see comment) ⇒ t=iln (((x−i)/(x+i)))= −θ =2φ =2cot^(−1) x .](Q19910.png)
$${let}\:{t}={i}\mathrm{ln}\:\left(\frac{{x}−{i}}{{x}+{i}}\right) \\ $$$${and}\:{let}\:{z}=\frac{{x}−{i}}{{x}+{i}}\: \\ $$$$\Rightarrow\:\mid{z}\mid=\mathrm{1} \\ $$$${so}\:{let}\:\:{z}={e}^{{i}\theta} \:\:\:{here}\:\:−\:\pi\:<\:\theta\:<\:\mathrm{0} \\ $$$$\:\:\:\:\:\:{t}={i}\mathrm{ln}\:{z}={i}\left[\:{i}\theta\:\right]=−\theta \\ $$$$\:\:\:\:\:{Now}\:\:{let}\:\:\mathrm{tan}\:\emptyset\:=\frac{\mathrm{1}}{{x}}>\:\mathrm{0}\:\: \\ $$$${as}\:{x}>\mathrm{0}\:,\:\:\:\mathrm{0}\:<\:\phi\:<\:\frac{\pi}{\mathrm{2}} \\ $$$$\:\Rightarrow\:\:\:\theta=−\mathrm{2}\phi\:\:\left({see}\:{comment}\right) \\ $$$$\Rightarrow\:\:\:\:\:\:\:{t}={i}\mathrm{ln}\:\left(\frac{{x}−{i}}{{x}+{i}}\right)=\:−\theta\:=\mathrm{2}\phi \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{2cot}^{−\mathrm{1}} {x}\:. \\ $$
Commented by Tinkutara last updated on 18/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Commented by ajfour last updated on 18/Aug/17
Answered by Tinkutara last updated on 22/Aug/17
![i[log (x − i) − log (x + i)] = i[(1/2)log (x^2 + 1) + itan^(−1) (((−1)/x)) − (1/2)log (x^2 + 1) − itan^(−1) ((1/x))] = i[−2itan^(−1) ((1/x))] = 2 cot^(−1) x](Q20112.png)
$${i}\left[\mathrm{log}\:\left({x}\:−\:{i}\right)\:−\:\mathrm{log}\:\left({x}\:+\:{i}\right)\right] \\ $$$$=\:{i}\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:+\:{i}\mathrm{tan}^{−\mathrm{1}} \left(\frac{−\mathrm{1}}{{x}}\right)\:−\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:−\:{i}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right)\right] \\ $$$$=\:{i}\left[−\mathrm{2}{i}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right)\right]\:=\:\mathrm{2}\:\mathrm{cot}^{−\mathrm{1}} {x} \\ $$