I saw a problem on Brilliant.org recently that sparked quite a heated discussion. The question was: Compute $\sqrt{-4} * \sqrt{-9}$. Now, If $\sqrt{-4}$ = 2i, the answer can only be -6. However, if $\sqrt{-4}$ = $\pm$2i, the answer can be $\pm$6. What would be the correct answer to this question?
Let me also say this: If we can split $\sqrt{-4}$ to $\sqrt{-1}*\sqrt{4}$, can we write the value of $\sqrt{-4}$ as $\pm$2i?
It's a silly thing to argue about because there's no standard definition of the square root for negative numbers. For nonnegative numbers there's no problem: we define $\sqrt{4}$ to be the positive number $x$ that solves $x^2=4$. This definition breaks down more generally because there's no natural extension of "positive" and "negative" to non-real numbers. If you like, you can create your own definition:
This looks reasonable to me; the problem is it's not a standard definition that everyone has accepted like in the $a\geq0$ case.
Edit: Another answer mentions the "principal branch" of the square-root function. The problem with this, and the reason there's confusion about the square root of negative numbers, is that the principal branch of $\sqrt{z}$ avoids the ray $(-\infty,0]$ in the complex plane. For any other complex number $z$, there's a standard choice of square root: $$z^{1/2}=|z|^{1/2}e^{i\theta/2}$$ where $\theta\in(-\pi,\pi)$ is the angle $z$ makes with the positive real axis. Of course the negative real numbers correspond to values of $z$ with $\theta=\pm\pi$, which is outside of its domain. Depending on which sign of $\pi$ you use, you get different signs for $z^{1/2}$. The reason mathematicians decline to make such a choice is because the square root function would necessarily cease to be continuous (and in particular, complex analytic) at this branch cut.