$\sqrt{-1}=\pm i$ (What does $\sqrt{-1}\cdot \sqrt{-1}$ equal ?) 
$\textbf{Question:}$ Am I using the definition of $\sqrt{-1}$ (see below) correctly here or not by giving four cases for $\sqrt{-1}\cdot \sqrt{-1}$?

Also, if I am using this definition correctly, is it true that $(ab)^c=a^cb^c$  for any $a, b, c \in \mathbb{C}$?



Below is the following definition I use to define $\sqrt{-1}$ over $\mathbb{C}$. I was told to use this definition previously on this site.
$\textbf{Definition:}$ $\sqrt{-1}=\pm i$.
So, the other day someone wrote $\sqrt{-1}\cdot \sqrt{-1}=i\cdot i =-1$ and said all other solutions were not possible.
However, I feel that this can be broken down into $4$ cases all of which are acceptable:

Case 1: $\sqrt{-1}\sqrt{-1}=i\cdot i=-1$.
Case 2: $\sqrt{-1}\sqrt{-1}=(-i)(-i)=-1$.
Case 3: $\sqrt{-1}\sqrt{-1}=i(-i)=-i^2=1$.
Case 4: $\sqrt{-1}\sqrt{-1}=(-i)i=-i^2=1$.

So, are there four cases (2 possibilities) or not here?
 A: Every number $z\in\mathbb{C}\setminus\{0\}$ has two square roots. So, unless you specify which square root you have in mind, you should not use the expression $\sqrt z$. On the other hand, if you specify which square root you have in mind for each $z\in\mathbb{C}\setminus\{0\}$, then either $\sqrt{-1}=i$ or $\sqrt{-1}=-i$, but not both.
And, in $\mathbb C$, the equality $(ab)^c=a^cb^c$ is meaningless, unless you define $a^b$ for each $a,b\in\mathbb C$. But that will not be easy.
A: Writing $\sqrt{-1}$ is ambiguous because $-1$ has two square roots (namely, $i$ and $-i$). 
When we take the square root of a positive real number (e.g., $\sqrt{16}$) we assume by convention that we take the positive square root, i.e., $\sqrt{16}=4$. If we wanted to refer to the other root, we would have to write $-\sqrt{16}=-4$.
Your four cases are akin to saying $\sqrt{16}=\pm 4$, and so:


*

*$\sqrt{16}\sqrt{16}=4\cdot 4 = 16$

*$\sqrt{16}\sqrt{16}=(-4)\cdot (-4) = 16$

*$\sqrt{16}\sqrt{16}=4\cdot (-4) = -16$

*$\sqrt{16}\sqrt{16} =( -4 )\cdot 4 = -16$.


The only line that is correct is line 1 because by convention $\sqrt{16}=4$.
Similarly, when you write $\sqrt{-1}$ you need to specify which root you are referring to. By convention mathematicians write $i=\sqrt{-1}$, even though it is also true that $(-i)^2=-1$. So again, only line 1 is correct.
This post explains why $(ab)^c = a^cb^c$ does not always hold for complex numbers, but it assumes some knowledge of complex analysis (which I'm afraid is necessary to truly understand what's going on).
