We already know that z=i solves z²+1=0. But if z=-i, then

z² =(-i)*(-i) = (-1)² i² = -1

Thus, z=-i is a second solution. These are the only two solutions, because

z²+1 = (z+i)(z-i)

To find all solutions to x⁴+1=0 we first note that

z⁴+1 = (z²+i)(z²-i)

Thus, we need to find the roots of i and -i next. In other words, we need to find a complex number z = x+iy such that

z² = (x+iy)² = x² - y² + 2ixy = i

Lokking at the real and imaginary parts separately, this results in two equations:

(1)       x²-y² = 0
(2)       2xy = 1

From there, and using a similar system to solve z² = -i we get the following four solutions to z⁴=-1:

z₁ = 1/(1+i)
z₂ = 1/(-1+i)
z₃ = 1/(-1-i)
z₄ = 1/(1-i)

Later it will turn out that any n-th degree polynomial has n roots, and they sometimes have some nice geometric properties.