Why is $\sqrt{-2} \sqrt{-3} \neq \sqrt{6}$? Why is $\sqrt{-2} \cdot \sqrt{-3} \neq \sqrt{6}$? Are there other examples where regular arithmetic goes wrong for complex numbers?
 A: You have to be careful with square roots, because for any $y$ we have $2$ values of $x$ satisfying $x^2=y$, while the square root function gives you only one of these. In your case, even though $(\sqrt{-2} \cdot \sqrt{-3})^2 = \sqrt{6}^2$, we have that $\sqrt{-2} \cdot \sqrt{-3} = -\sqrt{6}$, the other possible solution to $x^2=6$. So you must keep in mind that $\sqrt{a^2}=a$ is not necessarily true. A related erroneous assumption often made is $\sqrt{ab}=\sqrt{a}\sqrt{b}$. While $\sqrt{ab}=\sqrt{a}\sqrt{b}$ is true for some, it does not hold for all $a,b$. What if it did? This would make the following "proof" valid:
$$1=\sqrt{1^2}=\sqrt{(-1)^2}=\sqrt{-1}\sqrt{-1}=-1$$
thus it clearly cannot be.
Edit: Note that many people do not even define $\sqrt{y}$ for $y<0$, for the reasons described in Didier's comment. In this post I let $\sqrt{y}$ be defined as the solution to $x^2=y$ with positive imaginary component, as seems to be done in the original post.
A: The argument of a complex number is only determined upto  addition/subtraction of multiples of $2\pi$.
Now when multiplying complex numbers $z_1, z_2$ we can visualise the answer as having modulus $|z_1||z_2|$ and argument in the set $\{\text{arg}(z_1) + \text{arg}(z_2) + 2k\pi\mid k\in\mathbb{Z}\}$.
So taking the square root should essentially (positive) square root the modulus and half all possible arguments.
Let's see how this works with $\sqrt{-4}$. As a complex number $-4$ has modulus $4$ and possible arguments $\{\ldots, -\pi, \pi,\ldots\}$.
If we consider the argument to be $\pi$ then the square root should be $2i$ (since this is of modulus $2$ and principal argument $\frac{\pi}{2}$).
However, if we consider the argument to be $-\pi$ then the square root should be $-2i$ (since this is of modulus $2$ and principal argument $-\frac{\pi}{2}$).
All other possible arguments give one of these two complex numbers.
You see how we get a choice of square root, and it is not really favourable to choose one over the other. The complex numbers don't have a nice ordering like the real numbers do.
Compare this with square roots of positive real numbers. Considering the number as having argument $0$ we get the positive square root but considering the number as having argument $2\pi$ we get the negative square root. We usually choose the positive square root over the negative one but both are equally important.
So really something like $\sqrt{-2}\sqrt{-3}$ has $4$ different values depending on which square root you take for each term.
A: Interestingly, the great Euler, in his Algebra, writes explicitly that $\sqrt{-2}\sqrt{-3}=\sqrt{6}$. (This is in Article $148$, p. $43$.) Earlier, he writes that $\sqrt{9}=3$, so he is not treating square root as a multiple-valued function. Your question reminded me that I saw this many years ago, and wondered why he did it.  Very nice book, better than current stuff that does the basics of algebra.    
