Simplify-i-log-x-i-x-i-

Question Number 19898 by Tinkutara last updated on 17/Aug/17

Simplify : i \log (\frac{x - i}{x + i}) .

Answered by ajfour last updated on 22/Aug/17

l e t t = i \ln (\frac{x - i}{x + i})

a n d l e t z = \frac{x - i}{x + i}

\Rightarrow ∣ z ∣= 1

s o l e t z = e^{i θ} h e r e - π < θ < 0

t = i \ln z = i [i θ] = - θ

N o w l e t \tan \emptyset = \frac{1}{x} > 0

a s x > 0, 0 < ϕ < \frac{π}{2}

\Rightarrow θ = - 2 ϕ (s e e c o m m e n t)

\Rightarrow t = i \ln (\frac{x - i}{x + i}) = - θ = 2 ϕ

= {2 \cot}^{- 1} x .

Commented by Tinkutara last updated on 18/Aug/17

Thank you very much 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 [\frac{1}{2} \log (x^{2} + 1) + i \tan^{- 1} (\frac{- 1}{x}) - \frac{1}{2} \log (x^{2} + 1) - i \tan^{- 1} (\frac{1}{x})]

= i [- 2 i \tan^{- 1} (\frac{1}{x})] = 2 \cot^{- 1} x