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Question Number 35222 by abdo mathsup 649 cc last updated on 16/May/18
let ∣x∣<1 prove that arctanx =(i/2)ln(((i+x)/(i−x)))
$${let}\:\mid{x}\mid<\mathrm{1}\:{prove}\:{that}\: \\ $$$${arctanx}\:=\frac{{i}}{\mathrm{2}}{ln}\left(\frac{{i}+{x}}{{i}−{x}}\right) \\ $$
Commented by abdo 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))) .★
$${we}\:{have}\:{i}+{x}=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} \:}\:\left\{\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:+\frac{{i}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right\}\:={r}\:{e}^{{i}\theta} \\ $$$$\Rightarrow{r}=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\:{and}\:{cos}\theta=\:\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:,\:{sin}\theta\:\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$\Rightarrow\:{tan}\theta=\frac{\mathrm{1}}{{x}}\:\Rightarrow\theta\:={arctan}\left(\frac{\mathrm{1}}{{x}}\right)\:\Rightarrow \\ $$$${i}+{x}\:=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\:{e}^{{i}\:{arxtan}\left(\frac{\mathrm{1}}{{x}}\right)} \\ $$$${i}−{x}\:=\:−\left({x}−{i}\right)=−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{e}^{−{i}\:{arctan}\left(\frac{\mathrm{1}}{{x}}\right)} \:\Rightarrow \\ $$$$\frac{{i}+{x}}{{i}−{x}}\:=\:−{e}^{{i}\left({arctan}\left(\frac{\mathrm{1}}{{x}}\right)+{i}\:{arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right.} \\ $$$$=−\:{e}^{\mathrm{2}{i}\left\{\frac{\pi}{\mathrm{2}}\:−{arctanx}\right\}} =\:{e}^{−\mathrm{2}{i}\:{arctanx}} \:\:\Rightarrow \\ $$$${ln}\left(\frac{{i}+{x}}{{i}−{x}}\right)=\:−\mathrm{2}{i}\:{arctan}\left({x}\right)\:\Rightarrow \\ $$$${arctanx}=\:\frac{−\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{{i}+{x}}{{i}−{x}}\right)\:\Rightarrow \\ $$$$\bigstar\:{arctan}\left({x}\right)=\:\frac{{i}}{\mathrm{2}}{ln}\left(\frac{{i}+{x}}{{i}−{x}}\right)\:.\bigstar \\ $$
Answered by sma3l2996 last updated on 17/May/18
tany=i((e^(iy) −e^(−iy) )/(e^(iy) +e^(−iy) )) arctanx=y⇒tan(y)=i((e^(iy) −e^(−iy) )/(e^(iy) +e^(−iy) ))=x i((e^(iy) −e^(−iy) )/(e^(iy) +e^(−iy) ))=x⇔i(e^(2iy) −1)=x(e^(2iy) +1) e^(2iy) (i−x)=x+i⇔e^(2iy) =((i+x)/(i−x)) 2iy=ln(((i+x)/(i−x)))⇔y=(1/(2i))ln(((i+x)/(i−x))) y=arctanx=(i/2)ln(((i−x)/(i+x)))
$${tany}={i}\frac{{e}^{{iy}} −{e}^{−{iy}} }{{e}^{{iy}} +{e}^{−{iy}} } \\ $$$${arctanx}={y}\Rightarrow{tan}\left({y}\right)={i}\frac{{e}^{{iy}} −{e}^{−{iy}} }{{e}^{{iy}} +{e}^{−{iy}} }={x} \\ $$$${i}\frac{{e}^{{iy}} −{e}^{−{iy}} }{{e}^{{iy}} +{e}^{−{iy}} }={x}\Leftrightarrow{i}\left({e}^{\mathrm{2}{iy}} −\mathrm{1}\right)={x}\left({e}^{\mathrm{2}{iy}} +\mathrm{1}\right) \\ $$$${e}^{\mathrm{2}{iy}} \left({i}−{x}\right)={x}+{i}\Leftrightarrow{e}^{\mathrm{2}{iy}} =\frac{{i}+{x}}{{i}−{x}} \\ $$$$\mathrm{2}{iy}={ln}\left(\frac{{i}+{x}}{{i}−{x}}\right)\Leftrightarrow{y}=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{{i}+{x}}{{i}−{x}}\right) \\ $$$${y}={arctanx}=\frac{{i}}{\mathrm{2}}{ln}\left(\frac{{i}−{x}}{{i}+{x}}\right) \\ $$$$ \\ $$

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