Find the solution set for x³ - i = 2 and express answer in the form a + bi

x³ - i = 2
x³ = 2 + i ==> convert this to polar form
e.g a + ib in polar form
let a = rcosΘ and b = rsinΘ
then r = sqrt(a^2 + b^2)
and tanΘ = b/a

In ur question
r = sqrt(2^2 + 1^2)
r = sqrt5
and rcosΘ = 2 , rsinΘ = 1
tanΘ = 1/2

x³ = r(cosΘ + isinΘ) where cosΘ = 2/sqrt5 and sinΘ = 1/sqrt5
x = r^1/3(cosΘ + isinΘ)^1/3
x = r^1/3(cos(Θ/3) + isin(Θ/3))
Since by de moivre's formula (cosx + isinx)^n = cos(nx) + isin(nx)

x = r^(1/3)cis [(2kπ+Θ)/3]
[2kπ+Θ)/3 is the generalised value of Θ