Can we add fractional powers of negative numbers? This question might be silly and very basic. But my friend and me happened to argue on this for long. My argument was, if $-2 \sqrt3=\sqrt{12}$ which came from $\sqrt{(-2)(-2)} \sqrt{3} $ . If this is true, let's consider $-1\sqrt {-1}= \sqrt{(-1)(-1)}\sqrt{-1}=\sqrt{(-1)^3}=\sqrt{-1}$ which should also be true. But his argument is that, it is not applicable in second case.  
 A: Unfortunately, exponentiation tends not to behave so nicely on the complex numbers. When we work with non-negative reals $a$ raised to real exponents $b$, we get nice identities like:
$$a_1^{b}a_2^b=(a_1a_2)^b$$
but this doesn't work so well on the complex numbers. In particular, we need to figure out what we even mean by exponentiating on complex numbers. Raising a complex number to an integer power is clearly well-defined, since we just multiply it by itself (or divide by it) a proper number of times.
In the positive reals $a$, it's easy to define $a^{\frac{1}n}$ for integer $n$, since there is a unique positive solution to $x^n=a$, and we can say that's the $n^{th}$ root of $a$. From there it's easy to define exponentiation for any rational exponent and extend by continuity to any real exponent.
This doesn't work in the complex numbers since there are $n$ solutions to $x^n=a$ and it's not clear which one we're supposed to choose; the best we can do in general is to say something like:

Suppose $x$ satisfies $x^2=-1$ and $y$ satisfies $y^2=-4$. Then $xy$ is a root of $(xy)^2=4$.

which is kind of like our typical rules, where $x$ is trying to be $\sqrt{-1}$ and $y$ is trying to be $\sqrt{-4}$, but it's quite possible that, when we choose a root for $x$ (say $i$) and a root for $y$ (say $2i$) that we multiply them together and get $-2$ - which we want to say is $\sqrt{4}$ - which is not the anticipated result. In general, if we multiply a root of $x^n=c_1$ by a root of $y^n=c_2$, we will get a root of $(xy)^n=c_1c_2$, which is similar to the normal exponent laws, but since there are two roots to each of these equations, we can't strengthen this statement much.
A: A number whose square is a negative number is clearly not real. For this purpose, the concept of imaginary numbers was developed. The imaginary unit $i$ is defined as a number with the property that $i^2 = -1$. The product of $i$ and any real number is called an imaginary number, and the sum of an imaginary and real number is said to be complex. In short, a complex number is of the form $a + ib$, where both $a$ and $b$ are real numbers.
Many of the properties that you are familiar with for real numbers do not apply for the complex numbers. One of these is that $\sqrt{a}{b} = \sqrt a * \sqrt b$, which is true for all positive real numbers $a$ and $b$.
This is, in general, false for complex numbers.
It's important to note that $\sqrt x$ is the principal square root of a real number, that is, the non-negative real numbers. That's why $-2\sqrt 3$ is not the $\sqrt 12$ : it's $- \sqrt 12$. Now, complex numbers are neither positive or negative. as the square roots of negative numbers are complex, we can say that negative numbers don't have principal square roots: so it makes no sense to speak of $\sqrt -1$.
A: For nonzero complex numbers $z$, there are two numbers $r$ such that $r^2=z$.  If you want the square root symbol to denote a function, you have to specify a convention for picking which square root you mean.  The usual convention has it that $\sqrt a$ is positive when $a$ is positive, in which case $-2\sqrt3\not=\sqrt{12}$.  You are free, of course, to choose some other convention, but if you do you need to explain it.  
Moreover, and what's most important, there is no convention under which the formula $\sqrt{ab}=\sqrt a\sqrt b$ holds for all complex numbers $a$ and $b$.  That's not an obvious fact, but it's nonetheless, if unfortunately, true.
A: For a real number $x \geq 0$, $\sqrt{x}$ denotes not any square root of $x$, but the unique nonnegative real square root of $x$. So, $\sqrt{(-2) (-2)} = \sqrt{4}$ is $2$, not $-2$, even though $(-2)^2 = 4$.
Every nonzero complex number $z$ has exactly two square roots, i.e., complex numbers $w$ such that $w^2 = z$; since $w^2 = (-w)^2$, the two square roots are always additive inverses of each other. In general, there's no canonical choice of which square root to pick — it should really be thought of as a multivalued function, defined on this branched surface. But for a positive real number $x$, one of the square roots is a positive real number and the other is a negative real number, so the positive root is a natural choice, and we denote it $\sqrt{x}$.
The property $\sqrt{xy} = \sqrt{x} \sqrt{y}$ of nonnegative real square roots comes from the fact that we've made the nonnegative choice of square root in all cases. For arbitrary complex numbers, all we can say in general is that, if $v^2 = w^2$, then $v = \pm w$, so $(ab)^{1/2} = a^{1/2} b^{1/2}$ is true for some choices of square roots.
