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Question Number 49248 by maxmathsup by imad last updated on 04/Dec/18

1) solve z^4 =1+i(√3) 2) factorize p(x)=x^4 −1−i(√3)inside C[x] 3)factorze inside R[x] the polynom p(x).

$$\left.\mathrm{1}\right)\:{solve}\:{z}^{\mathrm{4}} =\mathrm{1}+{i}\sqrt{\mathrm{3}} \\ $$$$\left.\mathrm{2}\right)\:{factorize}\:{p}\left({x}\right)={x}^{\mathrm{4}} −\mathrm{1}−{i}\sqrt{\mathrm{3}}{inside}\:{C}\left[{x}\right] \\ $$$$\left.\mathrm{3}\right){factorze}\:{inside}\:{R}\left[{x}\right]\:{the}\:{polynom}\:{p}\left({x}\right). \\ $$

Answered by Smail last updated on 05/Dec/18

z^4 =2e^(i((π/3)+2kπ)) z=(2)^(1/4) e^(i((π/(12))+((kπ)/2))) z_0 =(2)^(1/4) e^(i(π/(12))) , z_1 =i(2)^(1/4) e^(i(π/(12))) =iz_0 z_2 =−(2)^(1/4) e^(i(π/(12))) =−z_0 , z_3 =−iz_0 2) p(x)=x^4 −z^4 =(x−z_0 )(x+z_0 )(x−iz_0 )(x+iz_0 ) 3)

$${z}^{\mathrm{4}} =\mathrm{2}{e}^{{i}\left(\frac{\pi}{\mathrm{3}}+\mathrm{2}{k}\pi\right)} \\ $$$${z}=\sqrt[{\mathrm{4}}]{\mathrm{2}}{e}^{{i}\left(\frac{\pi}{\mathrm{12}}+\frac{{k}\pi}{\mathrm{2}}\right)} \\ $$$${z}_{\mathrm{0}} =\sqrt[{\mathrm{4}}]{\mathrm{2}}{e}^{{i}\frac{\pi}{\mathrm{12}}} \:\:,\:\:{z}_{\mathrm{1}} ={i}\sqrt[{\mathrm{4}}]{\mathrm{2}}{e}^{{i}\frac{\pi}{\mathrm{12}}} ={iz}_{\mathrm{0}} \\ $$$${z}_{\mathrm{2}} =−\sqrt[{\mathrm{4}}]{\mathrm{2}}{e}^{{i}\frac{\pi}{\mathrm{12}}} =−{z}_{\mathrm{0}} \:\:,\:\:{z}_{\mathrm{3}} =−{iz}_{\mathrm{0}} \\ $$$$\left.\mathrm{2}\right) \\ $$$${p}\left({x}\right)={x}^{\mathrm{4}} −{z}^{\mathrm{4}} =\left({x}−{z}_{\mathrm{0}} \right)\left({x}+{z}_{\mathrm{0}} \right)\left({x}−{iz}_{\mathrm{0}} \right)\left({x}+{iz}_{\mathrm{0}} \right) \\ $$$$\left.\mathrm{3}\right) \\ $$$$ \\ $$

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