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)}#
