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

1 Answer
Jul 20, 2016

Answer:

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)#