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let-f-x-ln-1-ix-with-x-lt-1-1-extract-Re-f-x-and-Im-f-x-2-developp-f-x-at-integr-serie-

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Question Number 34634 by abdo mathsup 649 cc last updated on 09/May/18
let f(x)=ln(1+ix) with ∣x∣<1 1) extract Re(f(x)) and Im(f(x)) 2) developp f(x) at integr serie.
$${let}\:{f}\left({x}\right)={ln}\left(\mathrm{1}+{ix}\right)\:{with}\:\mid{x}\mid<\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{extract}\:{Re}\left({f}\left({x}\right)\right)\:{and}\:{Im}\left({f}\left({x}\right)\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\left({x}\right)\:{at}\:{integr}\:{serie}. \\ $$
Commented by math khazana by abdo last updated on 10/May/18
1)we have 1+ix=(√(1+x^2 ))( (1/( (√(1+x^2 )))) +i (x/( (√(1+x^2 ))))) = r e^(iθ) ⇒r=(√(1+x^2 )) and cosθ =(1/( (√(1+x^2 )))) , sinθ =(x/( (√(1+x^2 )))) ⇒tanθ =x ⇒θ=arctanx ⇒ ln(1+ix) =ln(√(1+x^2 )) +i arctan x =(1/2)ln(1+x^2 ) +i arc tanx ⇒ Re(f(x))=(1/2)ln(1+x^2 ) and Im(f(x))= artanx 2) we have ln^′ (1+u) = (1/(1+u)) =Σ_(n=0) ^∞ (−1)^n u^n ⇒ ln(1+u) = Σ_(n=0) ^∞ (((−1)^n )/(n+1))u^(n+1) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n)u^n ⇒ (1/2)ln(1+x^2 ) = Σ_(n=1) ^∞ (((−1)^(n−1) )/(2n)) x^(2n) also arctan^′ x = (1/(1+x^2 )) = Σ_(n=0) ^∞ (−1)^n x^(2n) ⇒ arctanx = Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) x^(2n+1) finally f(x) = Σ_(n=1) ^∞ (((−1)^(n−1) )/(2n)) x^(2n) +i Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) x^(2n+1) .
$$\left.\mathrm{1}\right){we}\:{have}\:\mathrm{1}+{ix}=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\left(\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:+{i}\:\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right) \\ $$$$=\:{r}\:{e}^{{i}\theta} \:\:\Rightarrow{r}=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} \:}\:\:{and}\:\:{cos}\theta\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:, \\ $$$${sin}\theta\:=\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:\Rightarrow{tan}\theta\:={x}\:\Rightarrow\theta={arctanx}\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{ix}\right)\:={ln}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} \:}\:+{i}\:{arctan}\:{x} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:+{i}\:{arc}\:{tanx}\:\Rightarrow \\ $$$${Re}\left({f}\left({x}\right)\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:\:{and}\:{Im}\left({f}\left({x}\right)\right)=\:{artanx} \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{ln}^{'} \left(\mathrm{1}+{u}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}+{u}}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:{u}^{{n}} \\ $$$$\Rightarrow\:{ln}\left(\mathrm{1}+{u}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}{u}^{{n}+\mathrm{1}} \:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}{u}^{{n}} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\mathrm{2}{n}}\:{x}^{\mathrm{2}{n}} \:\:\:\:\:{also} \\ $$$${arctan}^{'} {x}\:\:=\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{2}{n}} \:\Rightarrow \\ $$$${arctanx}\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:{x}^{\mathrm{2}{n}+\mathrm{1}} \:\:\:{finally} \\ $$$${f}\left({x}\right)\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\mathrm{2}{n}}\:{x}^{\mathrm{2}{n}} \:\:\:+{i}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:{x}^{\mathrm{2}{n}+\mathrm{1}} \:\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 09/May/18
ln(1+ix)=(1/2)ln(1^2 +x^2 )+itan^(−1) ((x/1)) =(1/2)ln(1+x^2 )+itan^(−1) (x) ln(1+t)=t−(t^2 /2)+(t^3 /3)−(t^4 /4)+.... tan^(−1) (x)=x−(x^3 /3)+(x^5 /5)−.... so required ans is =(1/2)(x^2 −(x^4 /2)+(x^6 /3)−(x^8 /4)...)+i(x−(x^3 /3)+(x^5 /5)−....)
$${ln}\left(\mathrm{1}+{ix}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)+{itan}^{−\mathrm{1}} \left(\frac{{x}}{\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)+{itan}^{−\mathrm{1}} \left({x}\right) \\ $$$${ln}\left(\mathrm{1}+{t}\right)={t}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}+\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−\frac{{t}^{\mathrm{4}} }{\mathrm{4}}+…. \\ $$$${tan}^{−\mathrm{1}} \left({x}\right)={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}}−…. \\ $$$${so}\:{required}\:{ans}\:{is} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} −\frac{{x}^{\mathrm{4}} }{\mathrm{2}}+\frac{{x}^{\mathrm{6}} }{\mathrm{3}}−\frac{{x}^{\mathrm{8}} }{\mathrm{4}}…\right)+{i}\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}}−….\right) \\ $$

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