Question Number 72877 by mhmd last updated on 04/Nov/19
$${by}\:{using}\:{theorem}\:{demwover}\:{find}\:\:{x}^{\mathrm{4}} =\mathrm{1}? \\ $$$${pleas}\:{sir}\:{help}\:{me} \\ $$
Commented by mathmax by abdo last updated on 04/Nov/19
![roots at C x=z=r e^(iθ) so x^4 =1 ⇔z^4 =1 ⇔r^4 e^(i4θ) =e^(i2kπ) ⇒ r=1 and 4θ =2kπ ⇒θ_k =((kπ)/2) with k∈[[0,3]] the roots are Z_k =e^((ikπ)/2) and 0≤k≤3 z_0 =1 ,Z_1 =e^((iπ)/2) =i ,z_2 =e^(iπ) =−1 ,Z_3 =e^(i((3π)/2)) =−i also we have Z^4 −1 =(Z−1)(Z+1)(Z−i)(Z+i) roots at R x^4 −1 =0 ⇔(x^2 −1)(x^2 +1)=0 ⇔x^2 −1=0 ⇔x=+^− 1](Q72879.png)
$${roots}\:{at}\:{C}\:\:\:\:{x}={z}={r}\:{e}^{{i}\theta} \:\:{so}\:{x}^{\mathrm{4}} =\mathrm{1}\:\Leftrightarrow{z}^{\mathrm{4}} =\mathrm{1}\:\Leftrightarrow{r}^{\mathrm{4}} {e}^{{i}\mathrm{4}\theta} ={e}^{{i}\mathrm{2}{k}\pi} \:\Rightarrow \\ $$$${r}=\mathrm{1}\:{and}\:\mathrm{4}\theta\:=\mathrm{2}{k}\pi\:\Rightarrow\theta_{{k}} =\frac{{k}\pi}{\mathrm{2}}\:{with}\:{k}\in\left[\left[\mathrm{0},\mathrm{3}\right]\right] \\ $$$${the}\:{roots}\:{are}\:{Z}_{{k}} ={e}^{\frac{{ik}\pi}{\mathrm{2}}} \:\:\:{and}\:\mathrm{0}\leqslant{k}\leqslant\mathrm{3} \\ $$$${z}_{\mathrm{0}} =\mathrm{1}\:\:,{Z}_{\mathrm{1}} ={e}^{\frac{{i}\pi}{\mathrm{2}}} ={i}\:\:,{z}_{\mathrm{2}} ={e}^{{i}\pi} =−\mathrm{1}\:\:,{Z}_{\mathrm{3}} ={e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{2}}} \:=−{i}\:{also}\:{we}\:{have} \\ $$$${Z}^{\mathrm{4}} −\mathrm{1}\:=\left({Z}−\mathrm{1}\right)\left({Z}+\mathrm{1}\right)\left({Z}−{i}\right)\left({Z}+{i}\right) \\ $$$${roots}\:{at}\:{R}\:\:{x}^{\mathrm{4}} −\mathrm{1}\:=\mathrm{0}\:\Leftrightarrow\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0}\:\Leftrightarrow{x}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\:\Leftrightarrow{x}=\overset{−} {+}\mathrm{1} \\ $$