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Question Number 72877 by mhmd last updated on 04/Nov/19

by using theorem demwover find x^4 =1? pleas sir help me

byusingtheoremdemwoverfindx4=1?pleassirhelpme

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

rootsatCx=z=reiθsox4=1⇔z4=1⇔r4ei4θ=ei2kπ⇒r=1and4θ=2kπ⇒θk=kπ2withk∈[[0,3]]therootsareZk=eikπ2and0⩽k⩽3z0=1,Z1=eiπ2=i,z2=eiπ=−1,Z3=ei3π2=−ialsowehaveZ4−1=(Z−1)(Z+1)(Z−i)(Z+i)rootsatRx4−1=0⇔(x2−1)(x2+1)=0⇔x2−1=0⇔x=+−1

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