Here is a very basic way to do it: The square roots of 16 are, of course, 4 and -4, and the root of -1 is i: the square roots of -16 are 4i and -4i. The square roots of 4 are 2 and -2 so it only remains to determine the square root of i and -i.
If a+ bi is a square root of i then \(\displaystyle (a+ bi)^2= a^2- b^2+ (2ab)i= i\) so that \(\displaystyle a^2- b^2= 0\) and 2ab= 1. (a- b)(a+ b)= 0 so either a= b or a= -b. If a= b, \(\displaystyle 2ab=2a^2= 1\) and \(\displaystyle a= \pm\frac{\sqrt{2}}{2}\) so that the square roots of i are \(\displaystyle \frac{2}{2}(1+ i)\) and \(\displaystyle -\frac{\sqrt{2}}{2}(1- i)\). Similarly, the two square roots of -i are \(\displaystyle \frac{\sqrt{2}}{2}(1- i)\) and \(\displaystyle -\frac{\sqrt{2}}{2}\).
Multiplying by 2, the fourth roots of -16 are \(\displaystyle \sqrt{2}(1+ i)\), \(\displaystyle -\sqrt{2}(1+ i)\), \(\displaystyle \sqrt{2}(1- i)\), and \(\displaystyle -\sqrt{2}(1- i)\).
