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Question Number 38112 by maxmathsup by imad last updated on 21/Jun/18
prove that arctan(x)= (i/2)ln(((i+x)/(i−x))) for ∣x∣<1
$${prove}\:{that}\:\:{arctan}\left({x}\right)=\:\frac{{i}}{\mathrm{2}}{ln}\left(\frac{{i}+{x}}{{i}−{x}}\right)\:{for}\:\mid{x}\mid<\mathrm{1} \\ $$
Commented by prof 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...
$${we}\:{have}\:{i}+{x}={x}+{i}=\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\left(\:\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}}\:+\frac{{i}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}}\right) \\ $$$$={r}\:{e}^{{i}\theta} \:\Rightarrow{r}=\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:{and}\:{cos}\theta=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}} \\ $$$${sin}\theta=\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\:\Rightarrow\:{tan}\theta=\frac{\mathrm{1}}{{x}}\:\Rightarrow\theta={arctan}\left(\frac{\mathrm{1}}{{x}}\right)\:\Rightarrow \\ $$$${x}+{i}\:=\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:{e}^{{i}\:{arctan}\left(\frac{\mathrm{1}}{{x}}\right)} \:\Rightarrow \\ $$$${ln}\left({x}+{i}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}\:} +\mathrm{1}\right)\:+{iarctan}\left(\frac{\mathrm{1}}{{x}}\right){and} \\ $$$${ln}\left(−{x}+{i}\right)={ln}\left(−\mathrm{1}\right)\:+{ln}\left({x}−{i}\right) \\ $$$$={i}\pi\:\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)−{i}\:{arctan}\left(\frac{\mathrm{1}}{{x}}\right)\:\Rightarrow \\ $$$$\frac{{i}}{\mathrm{2}}{ln}\left(\frac{{i}+{x}}{{i}−{x}}\right)=\frac{{i}}{\mathrm{2}}\left\{{ln}\left({i}+{x}\right)−{ln}\left({i}−{x}\right)\right\} \\ $$$$=\frac{{i}}{\mathrm{2}}\left\{\:\mathrm{2}{i}\:{arctan}\left(\frac{\mathrm{1}}{{x}}\right)−{i}\pi\right\} \\ $$$$=\frac{\pi}{\mathrm{2}}\:−{arctan}\left(\frac{\mathrm{1}}{{x}}\right)={arctan}\left({x}\right)\:{if}\:\mathrm{0}<{x}<\mathrm{1} \\ $$$${if}\:−\mathrm{1}<{x}<\mathrm{0}\:{we}\:{use}\:{the}\:{chang}.{x}=−{t}… \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jun/18
ln(α+iβ)=(1/2)ln(α^2 +β^2 )+i(2nΠ+tan^(−1) (β/α)) ln(((i+x)/(i−x)))=ln(i+x)−ln(i−x) ln(i+x)=(1/2)ln(1+x^2 )+i(2nΠ+tan^(−1) (1/x)) ln(i−x)=(1/2)ln(1+x^2 )+i(2nΠ+tan^(−1) (1/(−x))) ln(((i+x)/(i−x)))=i(tan^(−1) (1/x)−tan^(−1) (1/(−x))) =i(2tan^(−1) (1/x)) so the value (i/2)ln(((i+x)/(i−x))) (i/2)×i(2tan^(−1) (1/x)) =−tan^(−1) ((1/x))=tan^(−1) (x)−(Π/2) since tan^(−1) x=k tank=x cotk=(1/x) tan((Π/2)−k)=(1/x) tan^(−1) ((1/x))=(Π/2)−k tan^(−1) ((1/x))+tan^(−1) (x)=(Π/2)
$${ln}\left(\alpha+{i}\beta\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \right)+{i}\left(\mathrm{2}{n}\Pi+{tan}^{−\mathrm{1}} \frac{\beta}{\alpha}\right) \\ $$$${ln}\left(\frac{{i}+{x}}{{i}−{x}}\right)={ln}\left({i}+{x}\right)−{ln}\left({i}−{x}\right) \\ $$$${ln}\left({i}+{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)+{i}\left(\mathrm{2}{n}\Pi+{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}}\right) \\ $$$${ln}\left({i}−{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)+{i}\left(\mathrm{2}{n}\Pi+{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{−{x}}\right) \\ $$$${ln}\left(\frac{{i}+{x}}{{i}−{x}}\right)={i}\left({tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}}−{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{−{x}}\right) \\ $$$$={i}\left(\mathrm{2}{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}}\right) \\ $$$${so}\:{the}\:{value} \\ $$$$\frac{{i}}{\mathrm{2}}{ln}\left(\frac{{i}+{x}}{{i}−{x}}\right) \\ $$$$\frac{{i}}{\mathrm{2}}×{i}\left(\mathrm{2}{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}}\right) \\ $$$$=−{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right)={tan}^{−\mathrm{1}} \left({x}\right)−\frac{\Pi}{\mathrm{2}} \\ $$$${since} \\ $$$${tan}^{−\mathrm{1}} {x}={k} \\ $$$${tank}={x} \\ $$$${cotk}=\frac{\mathrm{1}}{{x}} \\ $$$${tan}\left(\frac{\Pi}{\mathrm{2}}−{k}\right)=\frac{\mathrm{1}}{{x}} \\ $$$${tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right)=\frac{\Pi}{\mathrm{2}}−{k} \\ $$$${tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right)+{tan}^{−\mathrm{1}} \left({x}\right)=\frac{\Pi}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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