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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 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))) .★

$${we}\:{have}\:{i}+{x}=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} \:}\:\left\{\frac{{x}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:+\frac{{i}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right\}\:={r}\:{e}^{{i}\theta} \\ $$ $$\Rightarrow{r}=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\:{and}\:{cos}\theta=\:\frac{{x}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:,\:{sin}\theta\:\:=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$ $$\Rightarrow\:{tan}\theta=\frac{\mathrm{1}}{{x}}\:\Rightarrow\theta\:={arctan}\left(\frac{\mathrm{1}}{{x}}\right)\:\Rightarrow \\ $$ $${i}+{x}\:=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\:{e}^{{i}\:{arxtan}\left(\frac{\mathrm{1}}{{x}}\right)} \\ $$ $${i}−{x}\:=\:−\left({x}−{i}\right)=−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{e}^{−{i}\:{arctan}\left(\frac{\mathrm{1}}{{x}}\right)} \:\Rightarrow \\ $$ $$\frac{{i}+{x}}{{i}−{x}}\:=\:−{e}^{{i}\left({arctan}\left(\frac{\mathrm{1}}{{x}}\right)+{i}\:{arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right.} \\ $$ $$=−\:{e}^{\mathrm{2}{i}\left\{\frac{\pi}{\mathrm{2}}\:−{arctanx}\right\}} =\:{e}^{−\mathrm{2}{i}\:{arctanx}} \:\:\Rightarrow \\ $$ $${ln}\left(\frac{{i}+{x}}{{i}−{x}}\right)=\:−\mathrm{2}{i}\:{arctan}\left({x}\right)\:\Rightarrow \\ $$ $${arctanx}=\:\frac{−\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{{i}+{x}}{{i}−{x}}\right)\:\Rightarrow \\ $$ $$\bigstar\:{arctan}\left({x}\right)=\:\frac{{i}}{\mathrm{2}}{ln}\left(\frac{{i}+{x}}{{i}−{x}}\right)\:.\bigstar \\ $$

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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