If #a in RR# then #a^2 >= 0#. So there is no Real square root of #-16#.

If #i# is the imaginary unit, then #i^2 = -1# and we find that:

#(4i)^2 = 4^2*i^2 = 16 * -1 = -16#

So #4i# is a square root of #-16#.

Also:

#(-4i)^2 = (-4)^2*i^2 = 16 * -1 = -16#

So #-4i# is a square root of #-16#.

If #x in RR# and #x < 0# then #sqrt(x)# stands for the principal square root of #x# defined as:

#sqrt(x) = i sqrt(-x)#

In our case:

#sqrt(-16) = i sqrt(16) = 4i#

Note that you do need to be slightly cautious when dealing with square roots of negative numbers. In particular, the property #sqrt(ab) = sqrt(a)sqrt(b)# fails if #a, b < 0#:

#1 = sqrt(1) = sqrt(-1 * -1) != sqrt(-1)sqrt(-1) = (sqrt(-1))^2 = -1#