Question Number 29165 by abdo imad last updated on 04/Feb/18
![give the factorization inside C[x] for p(x)= x^4 −((1−i(√3))/2) .](Q29165.png)
$${give}\:{the}\:{factorization}\:{inside}\:{C}\left[{x}\right]\:{for} \\ $$$${p}\left({x}\right)=\:\:{x}^{\mathrm{4}} \:−\frac{\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:. \\ $$
Commented by abdo imad last updated on 06/Feb/18
![let find the roots lf p(x) p(z)=0 ⇔z^4 =((1−i(√3))/2) but 1−i(√3)=2( (1/2)−i((√3)/2))= 2 e^(−i(π/3)) let put z=r e^(iθ) p(z)=0⇔ r^4 =2 and 4θ=−(π/3) +2kπ k∈[[0,3]] θ_k =−(π/(12)) +((kπ)/2) so the roots are z_k =^4 (√2) e^(i(−(π/(12))+((kπ)/2))) k∈[[0,3]]its clear that the leading coefficient is 1 so p(x)= Π_(k=0) ^3 (x−z_k )=(x−z_0 )(x−z_1 )(x−z_2 )(x−z_3 )with z_0 =^4 (√2) e^(−i(π/(12))) , z_1 =^4 (√2) e^(i((5π)/(12))) , z_2 =^4 (√2) e^(i((11π)/(12))) ,z_3 =^4 (√2) e^(i((17π)/(12))) .](Q29300.png)
$${let}\:{find}\:{the}\:{roots}\:{lf}\:{p}\left({x}\right)\:{p}\left({z}\right)=\mathrm{0}\:\Leftrightarrow{z}^{\mathrm{4}} =\frac{\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:{but} \\ $$$$\mathrm{1}−{i}\sqrt{\mathrm{3}}=\mathrm{2}\left(\:\frac{\mathrm{1}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\:\mathrm{2}\:{e}^{−{i}\frac{\pi}{\mathrm{3}}} \:{let}\:{put}\:{z}={r}\:{e}^{{i}\theta} \\ $$$${p}\left({z}\right)=\mathrm{0}\Leftrightarrow\:{r}^{\mathrm{4}} =\mathrm{2}\:\:{and}\:\mathrm{4}\theta=−\frac{\pi}{\mathrm{3}}\:+\mathrm{2}{k}\pi\:\:\:\:\:{k}\in\left[\left[\mathrm{0},\mathrm{3}\right]\right] \\ $$$$\theta_{{k}} =−\frac{\pi}{\mathrm{12}}\:+\frac{{k}\pi}{\mathrm{2}}\:\:\:{so}\:{the}\:{roots}\:{are}\:{z}_{{k}} =^{\mathrm{4}} \sqrt{\mathrm{2}}\:{e}^{{i}\left(−\frac{\pi}{\mathrm{12}}+\frac{{k}\pi}{\mathrm{2}}\right)} \:\:{k}\in\left[\left[\mathrm{0},\mathrm{3}\right]\right]{its} \\ $$$${clear}\:{that}\:{the}\:{leading}\:{coefficient}\:{is}\:\mathrm{1}\:{so} \\ $$$${p}\left({x}\right)=\:\prod_{{k}=\mathrm{0}} ^{\mathrm{3}} \:\left({x}−{z}_{{k}} \right)=\left({x}−{z}_{\mathrm{0}} \right)\left({x}−{z}_{\mathrm{1}} \right)\left({x}−{z}_{\mathrm{2}} \right)\left({x}−{z}_{\mathrm{3}} \right){with} \\ $$$${z}_{\mathrm{0}} =^{\mathrm{4}} \sqrt{\mathrm{2}}\:{e}^{−{i}\frac{\pi}{\mathrm{12}}} \:\:\:,\:\:{z}_{\mathrm{1}} =^{\mathrm{4}} \sqrt{\mathrm{2}}\:{e}^{{i}\frac{\mathrm{5}\pi}{\mathrm{12}}} \:\:,\:{z}_{\mathrm{2}} =\:^{\mathrm{4}} \sqrt{\mathrm{2}}\:{e}^{{i}\frac{\mathrm{11}\pi}{\mathrm{12}}} \:,{z}_{\mathrm{3}} =^{\mathrm{4}} \sqrt{\mathrm{2}}\:{e}^{{i}\frac{\mathrm{17}\pi}{\mathrm{12}}} \:. \\ $$$$ \\ $$