prove that arctan(x)= (i/2)ln(((i+x)/(i−x))) for ∣x∣<1


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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 byprof Abdo imad last updated on 24/Jun/18
$we have i+x=x+i=(√(x^2 +1))( (x/(√(x^2 +1))) +(i/(√(x^2 +1)))) =r e^(iθ) ⇒r=(√(x^2 +1)) and cosθ=(x/(√(x^2 +1))) sinθ=(1/(√(x^2 +1))) ⇒ tanθ=(1/x) ⇒θ=arctan((1/x)) ⇒ x+i =(√(x^2 +1)) e^(i arctan((1/x))) ⇒ ln(x+i)=(1/2)ln(x^(2 ) +1) +iarctan((1/x))and ln(−x+i)=ln(−1) +ln(x−i) =iπ +(1/2)ln(x^2 +1)−i arctan((1/x)) ⇒ (i/2)ln(((i+x)/(i−x)))=(i/2){ln(i+x)−ln(i−x)} =(i/2){ 2i arctan((1/x))−iπ} =(π/2) −arctan((1/x))=arctan(x) if 0<x<1 if −1<x<0 we use the chang.x=−t...$
$w e h a v e i + x = x + i = \sqrt{x^{2} + 1} (\frac{x}{\sqrt{x^{2} + 1}} + \frac{i}{\sqrt{x^{2} + 1}})$ $= r e^{i θ} \Rightarrow r = \sqrt{x^{2} + 1} a n d c o s θ = \frac{x}{\sqrt{x^{2} + 1}}$ $s i n θ = \frac{1}{\sqrt{x^{2} + 1}} \Rightarrow t a n θ = \frac{1}{x} \Rightarrow θ = a r c t a n (\frac{1}{x}) \Rightarrow$ $x + i = \sqrt{x^{2} + 1} e^{i a r c t a n (\frac{1}{x})} \Rightarrow$ $l n (x + i) = \frac{1}{2} l n (x^{2} + 1) + i a r c t a n (\frac{1}{x}) a n d$ $l n (- x + i) = l n (- 1) + l n (x - i)$ $= i π + \frac{1}{2} l n (x^{2} + 1) - i a r c t a n (\frac{1}{x}) \Rightarrow$ $\frac{i}{2} l n (\frac{i + x}{i - x}) = \frac{i}{2} {l n (i + x) - l n (i - x)}$ $= \frac{i}{2} {2 i a r c t a n (\frac{1}{x}) - i π}$ $= \frac{π}{2} - a r c t a n (\frac{1}{x}) = a r c t a n (x) i f 0 < x < 1$ $i f - 1 < x < 0 w e u s e t h e c h a n g . x = - t . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jun/18

	$l n (α + i β) = \frac{1}{2} l n (α^{2} + β^{2}) + i (2 n Π + {t a n}^{- 1} \frac{β}{α})$ $l n (\frac{i + x}{i - x}) = l n (i + x) - l n (i - x)$ $l n (i + x) = \frac{1}{2} l n (1 + x^{2}) + i (2 n Π + {t a n}^{- 1} \frac{1}{x})$ $l n (i - x) = \frac{1}{2} l n (1 + x^{2}) + i (2 n Π + {t a n}^{- 1} \frac{1}{- x})$ $l n (\frac{i + x}{i - x}) = i ({t a n}^{- 1} \frac{1}{x} - {t a n}^{- 1} \frac{1}{- x})$ $= i (2 {t a n}^{- 1} \frac{1}{x})$ $s o t h e v a l u e$ $\frac{i}{2} l n (\frac{i + x}{i - x})$ $\frac{i}{2} \times i (2 {t a n}^{- 1} \frac{1}{x})$ $= - {t a n}^{- 1} (\frac{1}{x}) = {t a n}^{- 1} (x) - \frac{Π}{2}$ $s i n c e$ ${t a n}^{- 1} x = k$ $t a n k = x$ $c o t k = \frac{1}{x}$ $t a n (\frac{Π}{2} - k) = \frac{1}{x}$ ${t a n}^{- 1} (\frac{1}{x}) = \frac{Π}{2} - k$ ${t a n}^{- 1} (\frac{1}{x}) + {t a n}^{- 1} (x) = \frac{Π}{2}$
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