Originally Posted by

**Deveno** defining i = √(-1) is somewhat problematic. the reason being, we don't know how to distinguish between i and -i (algebraically a+ib and a-ib act identitically: the map

z-->z* is a field automorphism).

the best we can say is i is "a" square root of -1 (in other words, orientation in the complex plane is chosen arbitrarily). behold the following "proof":

1 = √(1) = √((-1)(-1)) = √(-1)√(-1) = (i)(i) = -1,

which highlights the risky business of dealing with complex square roots. what happens is: multiplication in the complex plane can "rotate by more than a whole circle", the exponential function is no longer 1-1. so things like principal values and branches become very important (if we regard the complex number 1 = 1+0i as having angle 2π, it's "principal square root" is -1, if we regard it as having angle 0, it's principal square root is 1, the "standard convention").

perhaps the least confusing way to answer the OP's question is this:

for any real number a:

if a > 0, there are two real numbers that satisfy x^{2} = a. we call the positive one √a, and the negative one -√a.

if a = 0, there is only one number satisfying x^{2} = 0, namely: 0.

if a < 0, there are two complex numbers satisfying x^{2} = a, either one of which can be called the/a square root (there's no sense of "positiveness" in the complex plane).