For this, we will use Euler's Formula e^(itheta) = cos(theta)+isin(theta)
Using this, we can express any complex number as Re^(itheta) for some R,theta in RR
First, note that -1 = cos(pi) +isin(pi) = e^(ipi)
However, we can add or subtract 2pi from an angle without changing the value of the complex number, so as a more general form, we have
-1 = e^(i(pi+2pik)) for some integer k
As R=1 in this case, we know that any roots of -1 will also have R=1. Then, we can write the general form of the roots as follows:
Let w=e^(itheta)
=>e^(i(pi+2pik))=(e^(itheta))^4=e^(i(4theta))
=>4theta = pi+2pik
=>theta = pi/4+pi/2k
Restricting theta to [0,2pi) this gives us
theta in{pi/4, (3pi)/4, (5pi)/4, (7pi)/4}
=>w in {e^(ipi/4), e^(i(3pi)/4),e^(i(5pi)/4),e^(i(7pi)/4)}
Thus, converting back to to a+bi form using Euler's formula gain, we end with the solution set
{sqrt(2)/2(1+i),sqrt(2)/2(-1+i),sqrt(2)/2(-1-i),sqrt(2)/2(1-i)}
