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Question Number 34770 by abdo mathsup 649 cc last updated on 10/May/18

$${let}\:{f}\left({x}\right)=\:{ln}\left(\mathrm{1}+{ix}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{1}\right)\:{extrsct}\:{Re}\left({f}\left({x}\right)\right)\:{and}\:{Im}\left({f}\left({x}\right)\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:{f}^{'} \left({x}\right)\:{by}\:{two}\:{methods} \\ $$

Commented by abdo mathsup 649 cc last updated on 13/May/18

1) we have proved that f(x)=(1/2)ln(1+x^4 ) +i arctan(x^2 ) ⇒ Re(f(x))= (1/2)ln(1+x^4 ) and Im(f(x))= arctan(x^2 ) 2) for ∣u∣<1 ln(1+u) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) x^n ⇒ (1/2)ln(1+x^4 ) = (1/2) Σ_(n=1) ^∞ (((−1)^(n−1) )/n) x^(4n) (d/dx)(arctanu)= (1/(1+u^2 )) = Σ_(n=0) ^∞ (−1)^n u^(2n) ⇒ arctanu =Σ_(n=0) ^∞ (((−1)^n )/(2n+1))u^(2n+1) ⇒ arctan(x^2 ) = Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) x^(4n+2) so f(x)= (1/2) Σ_(n=1) ^∞ (((−1)^(n−1) )/n) x^(4n) +i Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) x^(4n+2) the radius of convergence is R =1

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{proved}\:{that} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)\:+{i}\:{arctan}\left({x}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${Re}\left({f}\left({x}\right)\right)=\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)\:{and}\:{Im}\left({f}\left({x}\right)\right)=\:{arctan}\left({x}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{2}\right)\:{for}\:\mid{u}\mid<\mathrm{1}\:{ln}\left(\mathrm{1}+{u}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:{x}^{{n}} \:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:{x}^{\mathrm{4}{n}} \\ $$$$\frac{{d}}{{dx}}\left({arctanu}\right)=\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\left(−\mathrm{1}\right)^{{n}} \:{u}^{\mathrm{2}{n}} \Rightarrow \\ $$$${arctanu}\:\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}{u}^{\mathrm{2}{n}+\mathrm{1}} \:\Rightarrow \\ $$$${arctan}\left({x}^{\mathrm{2}} \right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:{x}^{\mathrm{4}{n}+\mathrm{2}} \\ $$$$\:{so}\: \\ $$$${f}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:{x}^{\mathrm{4}{n}} \:\:\:+{i}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:{x}^{\mathrm{4}{n}+\mathrm{2}} \\ $$$${the}\:{radius}\:{of}\:{convergence}\:{is}\:{R}\:=\mathrm{1} \\ $$

Commented by abdo mathsup 649 cc last updated on 13/May/18

3) we have f(x)=(1/2)ln(1+x^4 ) +i arctan(x^2 ) ⇒ f^′ (x) = ((2x^3 )/(1+x^4 )) + i((2x)/(1+x^4 )) =((2x)/(1+x^4 ))( x^2 +i) another method f(x) =ln(1+ix^2 ) ⇒f^′ (x) = ((2ix)/(1+ix^2 )) = ((2ix( 1−ix^2 ))/(1+x^4 )) = ((2ix +2x^3 )/(1+x^4 )) = ((2x^3 )/(1+x^4 )) +i((2x)/(1+x^4 )) .

$$\left.\mathrm{3}\right)\:{we}\:{have}\:{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)\:+{i}\:{arctan}\left({x}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:=\:\:\frac{\mathrm{2}{x}^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{4}} }\:\:+\:{i}\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{4}} }\:=\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{4}} }\left(\:{x}^{\mathrm{2}} \:+{i}\right) \\ $$$${another}\:{method} \\ $$$${f}\left({x}\right)\:={ln}\left(\mathrm{1}+{ix}^{\mathrm{2}} \right)\:\Rightarrow{f}^{'} \left({x}\right)\:=\:\frac{\mathrm{2}{ix}}{\mathrm{1}+{ix}^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{2}{ix}\left(\:\mathrm{1}−{ix}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{4}} }\:=\:\frac{\mathrm{2}{ix}\:\:+\mathrm{2}{x}^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{4}} }\:=\:\frac{\mathrm{2}{x}^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{4}} }\:+{i}\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{4}} }\:\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/May/18

ln(1+ix^2 )=(1/2)ln{1^2 +(x^2 )^2 }+itan^(−1) (x^2 /1) =(1/2)(x^4 −(((x^4 )^2 )/2)+(((x^4 )^3 )/3)....)+i(x^2 −(((x^2 )^3 )/3)+(((x^2 )^5 )/5)...) =(1/2)(x^4 −(x^8 /2)+(x^(12) /3)....)+i(x^2 −(x^6 /3)+(x^(10) /5)−....)

$${ln}\left(\mathrm{1}+{ix}^{\mathrm{2}} \right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left\{\mathrm{1}^{\mathrm{2}} +\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} \right\}+{itan}^{−\mathrm{1}} \left({x}^{\mathrm{2}} /\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{4}} −\frac{\left({x}^{\mathrm{4}} \right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\left({x}^{\mathrm{4}} \right)^{\mathrm{3}} }{\mathrm{3}}….\right)+{i}\left({x}^{\mathrm{2}} −\frac{\left({x}^{\mathrm{2}} \right)^{\mathrm{3}} }{\mathrm{3}}+\frac{\left({x}^{\mathrm{2}} \right)^{\mathrm{5}} }{\mathrm{5}}…\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{4}} −\frac{{x}^{\mathrm{8}} }{\mathrm{2}}+\frac{{x}^{\mathrm{12}} }{\mathrm{3}}….\right)+{i}\left({x}^{\mathrm{2}} −\frac{{x}^{\mathrm{6}} }{\mathrm{3}}+\frac{{x}^{\mathrm{10}} }{\mathrm{5}}−….\right) \\ $$$$ \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 11/May/18