How do you find all solutions to #x^4+(1+i)=0#?

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Answer 1 · 2016-09-28T16:52:22

#x = root(8)(2)e^(i(5/16pi+k pi/2))# for # k = 0,pm1,pm2,cdots#

#x^4 = -(1+i)#

Using de Moivre's identity

#e^(iphi)=cosphi+isinphi# we have

#-(1+i) = -sqrt(2)(1/sqrt(2)+i/sqrt(2))=-sqrt(2)(cos phi_0+i sin phi_0)#

with #phi_0 = pi/4+2kpi, k = 0,pm1,pm2,cdots#

but

#-sqrt(2)(cos phi_0+i sin phi_0) = sqrt(2)e^(ipi)e^(i(pi/4+2kpi)) = sqrt(2)e^(i(5/4pi+2kpi)) #

#x^4 = sqrt(2)e^(i(5/4pi+2kpi)) #

finally

#x = root(8)(2)e^(i(5/16pi+k pi/2))#

for # k = 0,pm1,pm2,cdots#