If $\sqrt{a}\sqrt{b}=\sqrt{ab}$ only holds for positive real $a$ & $b$, then why can we say $\sqrt{-a}=\sqrt{-1\cdot a}=\sqrt{-1}\sqrt{a}=i\sqrt{a}$? I am a little bit bummed that I have this question as I'm sure it has been asked before (I couldn't find the answer) but...

If $\sqrt{a}\sqrt{b} = \sqrt{ab}$ is only true for positive reals $a$ and $b$. Then what allows us to say the following? $$\sqrt{-a} = \sqrt{-1\cdot a} = \sqrt{-1}\sqrt{a} = i\sqrt{a}$$

I don't know what allows the second equal sign. Is this just convention?
 A: The square root function is usually understood by convention to return the "principal" square root of a complex number. For real numbers, this just means that the square root of a positive real is a positive real and the square root of a negative real is imaginary with positive imaginary part. Under this convention, it happens to be valid that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ if $a$ and $b$ are real with at least one of them nonnegative. But the equality fails if both are negative. For example
$$
1=\sqrt{(-1)(-1)}\neq\sqrt{-1}\sqrt{-1}=i\cdot i=-1.
$$
A: Be aware of the setting $i=\sqrt{-1}.$ The law $\sqrt{a}\sqrt{b}=\sqrt{ab}$ does only apply to real numbers. It is wrong for complex numbers because
$$
-1= i^2=i\cdot i =\sqrt{-1} \cdot \sqrt{-1}\neq \sqrt{(-1)^2}=1
$$
The best way to see it is in my opinion to operate with the equation $i^2=-1$ only; not roots. They are an abbreviation for real numbers.
Another possibility is to use Euler's formula $-1=e^{i \pi}.$ Thus
$$
-1=e^{i\pi}=\left(e^{i\pi/2}\right)^2=i^2\text{ and }\sqrt{(-1)^2}=\sqrt{\left(e^{i\pi}\right)^2}=\sqrt{e^{2i\pi}}=\sqrt{1}=1
$$
This uses the power $1/2$ instead of roots and you avoid formulas which simply do not hold for complex numbers.
A: $\DeclareMathOperator{\Log}{Log} $
Let $a \in \mathbb R$ and define a square root via $\sqrt{z} := \exp(\tfrac 12\Log z)$ with $\Log z$ having a branch cut along the negative imaginary axis, i.e. $\Log(z) := \log|z| + i \theta$ with $\log$ the real logarithm and $\theta \in [-\pi/4, 3\pi/4)$. Then your formula above says:
$$\exp(\tfrac 12\Log(-a)) = \exp(\tfrac 12\Log(-1\cdot a))= \exp(\tfrac 12\Log(-1))\exp(\tfrac{1}{2}\Log a)=i\exp(\tfrac{1}{2}\Log a)$$
Thus the two things to check here are that $\Log(-1\cdot a) = \Log(-1) + \Log a$ and that $\exp(\tfrac 12\Log(-1)) = i$. For the latter, note $\Log(-1) = \log(1) + i\pi= i\pi$ so that $\exp(\tfrac 12\Log(-1))=e^{i \pi/2}=i$. For the former, note that if $x,y \in \mathbb C$ with  $-\pi/4 \le \arg(x)+\arg(y)<3 \pi/4$ then $\arg(x)+\arg(y) = \arg(xy)$ (note this does not hold if we cross a branch... Can you see why?
) so we have:
$$\Log(x)+\Log(y)=\log|x|+\log|y|+i(\arg(x) + \arg(y)) \stackrel{!}{=}\log(|xy|)+i\arg(xy) := \Log(xy)$$
where equality $!$ follows from the multiplicativity of the real logarithm, the fact $|x||y| = |xy|$, and the fact $\arg(x) + \arg(y) = \arg(xy)$.
Applying this result with $x = -1$ and $y = a$ we see $\arg(x) = \pi$, $\arg(y) = 0$, and $\arg(xy) = \arg(-a) = \pi$ so the above lemma holds and we are done. Note the choice of the branch along the negative imaginary axis was arbitrary; the argument above holds without any problem for any branch cut from $0$ to $\infty$ lying strictly in the open lower half-plane. If you want to choose a branch intersecting the closed upper half-plane (e.g. the standard branch) you'll have to keep track of the extra $2 \pi i$ factor added or removed every time you cross the branch, though the argument will go through in essentially the same way.

Note: As @Subrosar says, we always have $\Log(x) + \Log(y) = \Log(xy)$ when one of $x$ or $y$ is nonnegative and the choice of branch cut is a straight line from $0$ to $\infty$, and this happens simply because one of $\arg(x)$ or $\arg(y)$ is zero; assuming WLOG $\arg(x) = 0$, we have $\arg(x) = 0 \le \arg(xy) = \arg(y)$, so $xy$ does not cross the branch cut since it is simply a scaled copy of $y$. Note this fails if the branch cut is not a straight line, as it could be that $y$ and $xy$ lie on separate sides of the branch cut.
