How do you find all the real and complex roots of F(x)=x^5-1?

1 Answer
Jul 20, 2016

See below

Explanation:

x^5-1=0 or

x=1e^{i phi/5} where phi=arctan0+2kpi,k=pm1,pm2,pm3,cdots

then

x_k=e^{i(2k pi)/5} for k=0,1,2,3,4 are the roots or

{(x = 1), (x = -(1+sqrt[5])/4 - i sqrt[(5- sqrt[5])/8]), (x= -(1- sqrt[5])/4 + i sqrt[(5+ sqrt[5])/8]), (x= -(1- sqrt[5])/4 - i sqrt[(5 + sqrt[5])/8]), (x = -(1+sqrt[5])/4 + i sqrt[(5 - sqrt[5])/8]):}

of course we used the Moivre's identity

e^{iphi} = cos(phi)+isin(phi)