let-x-lt-1-prove-that-arctanx-i-2-ln-i-x-i-x-

Question Number 35222 by abdo mathsup 649 cc last updated on 16/May/18

l e t ∣ x ∣< 1 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})

Commented by abdo mathsup 649 cc last updated on 17/May/18

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

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

\Rightarrow t a n θ = \frac{1}{x} \Rightarrow θ = a r c t a n (\frac{1}{x}) \Rightarrow

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

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

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

= - e^{2 i {\frac{π}{2} - a r c t a n x}} = e^{- 2 i a r c t a n x} \Rightarrow

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

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

★ a r c t a n (x) = \frac{i}{2} l n (\frac{i + x}{i - x}) . ★

Answered by sma3l2996 last updated on 17/May/18

t a n y = i \frac{e^{i y} - e^{- i y}}{e^{i y} + e^{- i y}}

a r c t a n x = y \Rightarrow t a n (y) = i \frac{e^{i y} - e^{- i y}}{e^{i y} + e^{- i y}} = x

i \frac{e^{i y} - e^{- i y}}{e^{i y} + e^{- i y}} = x \Leftrightarrow i (e^{2 i y} - 1) = x (e^{2 i y} + 1)

e^{2 i y} (i - x) = x + i \Leftrightarrow e^{2 i y} = \frac{i + x}{i - x}

2 i y = l n (\frac{i + x}{i - x}) \Leftrightarrow y = \frac{1}{2 i} l n (\frac{i + x}{i - x})

y = a r c t a n x = \frac{i}{2} l n (\frac{i - x}{i + x})