Question 420170
The fourth roots of -16 are definitely possible, using complex numbers. Let {{{z}}} be a complex number such that z^4 = -16 (we know the magnitude of z must be 2 since 2^4 = 16). We can express z in the form {{{2e^(i*theta)}}} and obtain


{{{z = 2e^(i*theta) = 2(cos(theta) + i*sin(theta))}}} (by Euler's formula)


Hence,


{{{z^4 = 2^4e^(4i*theta) = 16e^(4i*theta) = -16}}}


Therefore, {{{e^(4i*theta) = -1}}}. From DeMoivre's theorem, {{{e^(4i*theta) = cos(4*theta) + i*sin(4*theta)}}}. This number must equal -1, so {{{cos(4*theta) = -1}}} and {{{sin(4*theta) = 0}}}. This implies {{{4*theta = pi + k*2pi}}} --> {{{theta = (pi/4) + k*(pi/2)}}}, where k is any integer. For now, we can say that {{{theta = pi/4}}} and the fourth root of -16 is {{{e^(i*pi/4)}}}. In a+bi form, this is equal to {{{cos(pi/4) + i*sin(pi/4) = (sqrt(2)/2) + (sqrt(2)/2)i}}}.