let ∣x∣<1 prove that arctanx =(i/2)ln(((i+x)/(i−x)))


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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 byabdo mathsup 649 cc last updated on 17/May/18
$we have i+x=(√(1+x^2 )) {(x/(√(1+x^2 ))) +(i/(√(1+x^2 )))} =r e^(iθ) ⇒r=(√(1+x^2 )) and cosθ= (x/(√(1+x^2 ))) , sinθ =(1/(√(1+x^2 ))) ⇒ tanθ=(1/x) ⇒θ =arctan((1/x)) ⇒ i+x =(√(1+x^2 )) e^(i arxtan((1/x))) i−x = −(x−i)=−(√(1+x^2 )) e^(−i arctan((1/x))) ⇒ ((i+x)/(i−x)) = −e^(i(arctan((1/x))+i arctan((1/x))) =− e^(2i{(π/2) −arctanx}) = e^(−2i arctanx) ⇒ ln(((i+x)/(i−x)))= −2i arctan(x) ⇒ arctanx= ((−1)/(2i))ln(((i+x)/(i−x))) ⇒ ★ arctan(x)= (i/2)ln(((i+x)/(i−x))) .★$
$w e h a v e i + x = \sqrt{1 + x^{2}} {\frac{x}{\sqrt{1 + x^{2}}} + \frac{i}{\sqrt{1 + x^{2}}}} = r e^{i θ}$ $\Rightarrow r = \sqrt{1 + x^{2}} a n d c o s θ = \frac{x}{\sqrt{1 + x^{2}}}, s i n θ = \frac{1}{\sqrt{1 + x^{2}}}$ $\Rightarrow t a n θ = \frac{1}{x} \Rightarrow θ = a r c t a n (\frac{1}{x}) \Rightarrow$ $i + x = \sqrt{1 + x^{2}} e^{i a r x t a n (\frac{1}{x})}$ $i - x = - (x - i) = - \sqrt{1 + x^{2}} e^{- i a r c t a n (\frac{1}{x})} \Rightarrow$ $\frac{i + x}{i - x} = - e^{i (a r c t a n (\frac{1}{x}) + i a r c t a n (\frac{1}{x})}$ $= - e^{2 i {\frac{π}{2} - a r c t a n x}} = e^{- 2 i a r c t a n x} \Rightarrow$ $l n (\frac{i + x}{i - x}) = - 2 i a r c t a n (x) \Rightarrow$ $a r c t a n x = \frac{- 1}{2 i} l n (\frac{i + x}{i - x}) \Rightarrow$ $★ a r c t a n (x) = \frac{i}{2} l n (\frac{i + x}{i - x}) . ★$
	Answered by sma3l2996 last updated on 17/May/18

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