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Question Number 47543 by maxmathsup by imad last updated on 11/Nov/18

solve (1+ix)^n =n with x unknown real and n integr natural .

$${solve}\:\left(\mathrm{1}+{ix}\right)^{{n}} ={n}\:\:\:{with}\:{x}\:{unknown}\:{real}\:{and}\:{n}\:{integr}\:{natural}\:. \\ $$

Answered by arcana last updated on 11/Nov/18

ln(e^((1+ix)n) )=n ln(e e^(ix) )=n ln(e)+ln(e^(ix) )=1+ln(e^(ix) )=1 ln(e^(ix) )=0=ln(1)⇒e^(ix) =cos(x)+isin(x)=1+i0 si y solo si cos(x)=1, sin(x)=0 x=kπ con k∈Z verificar deduccion

$${ln}\left({e}^{\left(\mathrm{1}+{ix}\right){n}} \right)={n}\:{ln}\left({e}\:{e}^{{ix}} \right)={n} \\ $$$${ln}\left({e}\right)+{ln}\left({e}^{{ix}} \right)=\mathrm{1}+{ln}\left({e}^{{ix}} \right)=\mathrm{1} \\ $$$${ln}\left({e}^{{ix}} \right)=\mathrm{0}={ln}\left(\mathrm{1}\right)\Rightarrow{e}^{{ix}} ={cos}\left({x}\right)+{isin}\left({x}\right)=\mathrm{1}+{i}\mathrm{0} \\ $$$$\mathrm{si}\:\mathrm{y}\:\mathrm{solo}\:\mathrm{si}\:{cos}\left({x}\right)=\mathrm{1},\:{sin}\left({x}\right)=\mathrm{0} \\ $$$${x}={k}\pi\:\mathrm{con}\:{k}\in\mathbb{Z} \\ $$$$\mathrm{verificar}\:\mathrm{deduccion} \\ $$

Commented by Smail last updated on 11/Nov/18

(1+ix)^n =ln(e^((1+ix)^n ) ) not ln(e^((1+ix)n) )

$$\left(\mathrm{1}+{ix}\right)^{{n}} ={ln}\left({e}^{\left(\mathrm{1}+{ix}\right)^{{n}} } \right)\:{not}\:{ln}\left({e}^{\left(\mathrm{1}+{ix}\right){n}} \right) \\ $$$$ \\ $$

Answered by Smail last updated on 11/Nov/18

(1+ix)^n =ne^(2ikπ) with k=(0,1,2,...,n−1) 1+ix=(n)^(1/n) e^((2ikπ)/n) 1+ix=(n)^(1/n) cos(((2kπ)/n))+i(n)^(1/n) sin(((2kπ)/n)) (n)^(1/n) cos(((2kπ)/n))=1 and (n)^(1/n) sin(((2kπ)/n))=x ((2kπ)/n)=cos^(−1) ((1/(n)^(1/n) )) +2mπ with m∈IN x=(n)^(1/n) sin(cos^(−1) ((1/(n)^(1/n) ))+2mπ) =(n)^(1/n) (√(1−(1/(n^2 )^(1/n) )))=(√((n^2 )^(1/n) −1)) x=(√((n^2 )^(1/n) −1))

$$\left(\mathrm{1}+{ix}\right)^{{n}} ={ne}^{\mathrm{2}{ik}\pi} \:\:{with}\:{k}=\left(\mathrm{0},\mathrm{1},\mathrm{2},...,{n}−\mathrm{1}\right) \\ $$$$\mathrm{1}+{ix}=\sqrt[{{n}}]{{n}}{e}^{\frac{\mathrm{2}{ik}\pi}{{n}}} \\ $$$$\mathrm{1}+{ix}=\sqrt[{{n}}]{{n}}{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)+{i}\sqrt[{{n}}]{{n}}{sin}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right) \\ $$$$\sqrt[{{n}}]{{n}}{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)=\mathrm{1}\:{and}\:\:\sqrt[{{n}}]{{n}}{sin}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)={x} \\ $$$$\frac{\mathrm{2}{k}\pi}{{n}}={cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\sqrt[{{n}}]{{n}}}\right)\:+\mathrm{2}{m}\pi\:{with}\:\:{m}\in{IN} \\ $$$${x}=\sqrt[{{n}}]{{n}}{sin}\left({cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\sqrt[{{n}}]{{n}}}\right)+\mathrm{2}{m}\pi\right) \\ $$$$=\sqrt[{{n}}]{{n}}\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\sqrt[{{n}}]{{n}^{\mathrm{2}} }}}=\sqrt{\sqrt[{{n}}]{{n}^{\mathrm{2}} }−\mathrm{1}} \\ $$$${x}=\sqrt{\sqrt[{{n}}]{{n}^{\mathrm{2}} }−\mathrm{1}} \\ $$$$ \\ $$

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