Question

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

Answer 1

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

Explanation:

`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))`

(we used `e^(ipi)+1=0` after Euler
https://en.wikipedia.org/wiki/Leonhard_Euler)

`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`