Consistent branch choice I found in my class notes the following comment regarding branch choice:

It is important to choose a branch consistently, otherwise one can get absurd results, for example: $-1 = i^2 = \sqrt{-1}\times\sqrt{-1} = \sqrt{(-1)\times(-1)} = \sqrt{1} = 1$

I don't understand this example. Can you please explain to me which branch we started with and when did we changed branches?
Thank you!
 A: I think what you're missing is that the issue isn't about choosing a branch to consistently define a function, but choosing branches to make an identity valid.
It might help to visualize things dynamically. At the start, you have three points: $a = 1$, $b = 1$, and $c = 1$, and we choose $\sqrt{a} = \sqrt{b} = \sqrt{c} = 1$, and we have the identities $c = ab$ and $\sqrt{c} = \sqrt{a} \sqrt{b}$.
If we slide $a,b,c$ around, and we slide the chosen square root along consistently, then the identity remains true. e.g. if we move $a$ counterclockwise to $2i$, and correspondingly we drag $\sqrt{a}$ along to $(1+i)$. If we move $b$ counterclockwise to $-1$, then correspondingly we move $\sqrt{b}$ to $i$. And $c = ab$ has also moved counterclockwise three quarters of the way around the plane to $-2i$ and our choice of $\sqrt{c}$ has been dragged over to $-1+i$.
And we indeed have $\sqrt{c} = \sqrt{a} \sqrt{b}$ for these choices of square roots.
If we had chosen the principal square root for everything, then when $c$ crossed the branch cut, our choice of $\sqrt{c}$ would have jumped to the other branch, making the identity inconsistent.
A: In the heart of the problem is the following problem:

Given a region in $\mathbb{C}$, can we define a "logarithm"? Meaning, can we have a function $g(z)$ such that $e^{g(z)}=z?$

This is because, if we can do that, we can define a "square root" unambiguously:
$\displaystyle \sqrt{x} := e^{\frac{1}{2}\log(x)}$
and it is obvious why this is called THE "square root".
Well, we can do this if it is a simply connected region where $0$ is not in and $1$ is in. (We then define $\displaystyle \log(z):=\int_{\gamma}\frac{1}{\zeta}d\zeta$, where $\gamma$ is a path connecting $1$ and $z$, and everything goes okay.)
So, what you are writing is the following:
$-1=i^2=i.i=...$
Now enters the problem: Why is $\sqrt{x.y}=\sqrt{x}.\sqrt{y}$? It isn't.
Let the region we are be $\mathbb{C}$ minus the "lower" complex line. We can define a logarithm here. Let $x=-1$ and $y=-1$. $\sqrt{xy}=\sqrt{1}=1$ obviously, from the definition.
But $\displaystyle \sqrt{-1}=e^{\frac{1}{2}\log(-1)}$
$\displaystyle \log(-1)=\int_{\gamma}\frac{1}{\zeta}d\zeta=\pi i$, where $\gamma$ is a path connecting $1$ and $-1$, so:
$\displaystyle \sqrt{-1}=e^{\frac{1}{2}\log(-1)}=e^{\pi i /2}=i$, meaning $\sqrt{-1} . \sqrt{-1}=-1$.
Thus, the problem is that you used a equality that doesn't hold in general. And, to explain this, you need to understand properly what is a "square root".
OBS: You can define a logarithm even if $1$ is not in the region.
