math question

z^4 = -4

z^2 = +- sqrt(-4) = +-2i

z = +- sqrt(+-2i)

=  -1+i  ,   -1-i    ,  1-i    ,  1+i 

Let's make the solution super-formal!  We can write -4 in exponential notation as 4e^(pi*i), so the equation is z^4 = 4e^(pi*i).

 

By Hamilton's Theorem, the solutions are z = 4^{1/4}*e^(pi*i/4), 4^{1/4}*e^(pi*i/4 + pi/4), 4^{1/4}*e^(pi*i/4 + 2*pi/4), and 4^{1/4}*e^(pi*i/4 + 3*pi/4).  Since 4^{1/4} = sqrt(2) and e^(pi*i/4) = (1 + i)/sqrt(2), the first solution is 1 + i.  Then the other roots work out as

 

4^{1/4}*e^(pi*i/4 + pi/4) = 1 - i,

4^{1/4}*e^(pi*i/4 + 2*pi/4) = -1 - i, and

4^{1/4}*e^(pi*i/4 + 3*pi/4) = -1 + i.