prove-that-arctan-x-i-2-ln-i-x-i-x-for-x-lt-1-

Question Number 38112 by maxmathsup by imad last updated on 21/Jun/18

p r o v e t h a t a r c t a n (x) = \frac{i}{2} l n (\frac{i + x}{i - x}) f o r ∣ x ∣< 1

Commented by prof Abdo imad last updated on 24/Jun/18

w e h a v e i + x = x + i = \sqrt{x^{2} + 1} (\frac{x}{\sqrt{x^{2} + 1}} + \frac{i}{\sqrt{x^{2} + 1}})

= r e^{i θ} \Rightarrow r = \sqrt{x^{2} + 1} a n d c o s θ = \frac{x}{\sqrt{x^{2} + 1}}

s i n θ = \frac{1}{\sqrt{x^{2} + 1}} \Rightarrow t a n θ = \frac{1}{x} \Rightarrow θ = a r c t a n (\frac{1}{x}) \Rightarrow

x + i = \sqrt{x^{2} + 1} e^{i a r c t a n (\frac{1}{x})} \Rightarrow

l n (x + i) = \frac{1}{2} l n (x^{2} + 1) + i a r c t a n (\frac{1}{x}) a n d

l n (- x + i) = l n (- 1) + l n (x - i)

= i π + \frac{1}{2} l n (x^{2} + 1) - i a r c t a n (\frac{1}{x}) \Rightarrow

\frac{i}{2} l n (\frac{i + x}{i - x}) = \frac{i}{2} {l n (i + x) - l n (i - x)}

= \frac{i}{2} {2 i a r c t a n (\frac{1}{x}) - i π}

= \frac{π}{2} - a r c t a n (\frac{1}{x}) = a r c t a n (x) i f 0 < x < 1

i f - 1 < x < 0 w e u s e t h e c h a n g . x = - t \dots

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jun/18

l n (α + i β) = \frac{1}{2} l n (α^{2} + β^{2}) + i (2 n Π + {t a n}^{- 1} \frac{β}{α})

l n (\frac{i + x}{i - x}) = l n (i + x) - l n (i - x)

l n (i + x) = \frac{1}{2} l n (1 + x^{2}) + i (2 n Π + {t a n}^{- 1} \frac{1}{x})

l n (i - x) = \frac{1}{2} l n (1 + x^{2}) + i (2 n Π + {t a n}^{- 1} \frac{1}{- x})

l n (\frac{i + x}{i - x}) = i ({t a n}^{- 1} \frac{1}{x} - {t a n}^{- 1} \frac{1}{- x})

= i (2 {t a n}^{- 1} \frac{1}{x})

s o t h e v a l u e

\frac{i}{2} l n (\frac{i + x}{i - x})

\frac{i}{2} \times i (2 {t a n}^{- 1} \frac{1}{x})

= - {t a n}^{- 1} (\frac{1}{x}) = {t a n}^{- 1} (x) - \frac{Π}{2}

s i n c e

{t a n}^{- 1} x = k

t a n k = x

c o t k = \frac{1}{x}

t a n (\frac{Π}{2} - k) = \frac{1}{x}

{t a n}^{- 1} (\frac{1}{x}) = \frac{Π}{2} - k

{t a n}^{- 1} (\frac{1}{x}) + {t a n}^{- 1} (x) = \frac{Π}{2}