Branch cut of square root I don't really understand how branch cuts work. Let's take the complex function $f(z) = \sqrt{z}$. Apparently this function is not defined for $\mathbb{R}^{-}$. But why? We defined $i$ to be $\sqrt{-1}$ and now it's multivalued?
 A: As it was already said in the comments, $\sqrt{z}$ is not undefined on the negative real line. As you mention, clearly $\sqrt{-1}=i$ and for any positive real number $r$ we have $\sqrt{-r}=i\sqrt{r}$.
Let's look first at the definition of complex $\sqrt{z}$. It is a multivalued function returning a set $\{w_1,w_2\}$ so that $w_1^2=w_2^2=z$ as its output. If we want to make it single-valued, we have to make a consistent choice between which of $w_1$ or $w_2$ should be returned. We can agree that we choose that element of $\{w_1,w_2\}$ which is on the right half-plane $(Re(z)\geq 0)$, and if $Re(w_1)=Re(w_2)=0$, we choose the one that is in the upper half-plane.
So, our function $\sqrt{z}$ has the right half-plane for its codomain. Nothing changes if we chose the left half-plane for the domain.
However, $\sqrt{z}$, as we have just defined it, is not continuous on $\mathbb C$. Suppose we apply $\sqrt{z}$ on a circle centered at $0$ of radius $1$. Let's do that by tracing out this circle by starting at $1$ and moving in the positive (counter-clockwise) direction, and applying $\sqrt{z}$ along the way. On paper, draw two separate pictures, one in which you trace out the circle, and one in which you trace out the effect of $\sqrt{z}$ as we move along the circle.
As we trace out a quarter-circle towards $i$, the function $\sqrt{z}$ continuously traces out an eight of a circle. When we hit $i$ on the circle, $\sqrt{z}$ hits $1/\sqrt{2} + i/\sqrt{2}$. Now we go past $i$ and trace the circle towards $-1$. While we are doing that, $\sqrt{z}$ is continuously tracing a quarter-circle that goes towards $i$. However, as soon as we go past $-1$ on the circle, $\sqrt{z}$ jumps from a neighbourhood of $i$ to a neighbourhood of $-i$, because this is how we defined $\sqrt{z}$. So clearly a discontinuity happened when we were traversing the negative real line at $-1$.
Nothing changes if we choose to traverse a different circle at origin intersecting the negative real line somewhere other than $-1$, so to avoid jumps in continuity, we must avoid continuously traversing the negative real line.
The easiest way to do that is to just remove the negative real line from the domain. So now our function isn't $\mathbb C\to \{z\in\mathbb C\mid Re(z)\geq 0\}$, but $\mathbb C\setminus\{z\in\mathbb R^-\}\to \{z\in\mathbb C\mid Re(z)\geq 0\}$.
And that function is continuous.
Remember that we had the other choice of defining $\sqrt{z}$, as a function $\mathbb C\to \{z\in\mathbb C\mid Re(z)\leq 0\}$. This again isn't continuous, jumps in continuity happen again at the negative real line, so now we declare this function to have domain $\mathbb C\setminus\mathbb R^-$ to make it continuous.
It turns out that we can glue these two functions together into a single function, that stays continuous even when we continuously traverse the real line. What we do is, we take two copies of $\mathbb C\setminus \mathbb R^-$, one for each choice of codomain (left or right half-plane) and we glue them along the negative real line. This is the resulting picture, and you can make it on your own by taking two pieces of paper, cutting them along the same line and twisting them along the cut so that they "glue" together.
This function has no jumps in discontinuity when continuously traversing the negative real line, because as soon as the line is crossed, in the image, a circle is being traced out continuously even past $i$, as it doesn't jump to $-i$.
Let's denote the set obtained by gluing two copies of $\mathbb C\setminus\mathbb R^-$ by $S$, and let's start traversing a "circle" (in quotes because $S$ is not flat, so the circle is distorted) with center $0$, radius $1$, starting at $1$. Declare a new function $f\colon S\to\mathbb C$ so that, before we hit the negative real line (the line of gluing), it behaves exactly the same as the $\sqrt{z}$ defined on $\mathbb C\setminus\mathbb R^-$ sending its image to the right half-plane.
But, when we hit the negative real line, our "circle" moves from the first copy of $\mathbb C\setminus\mathbb R^-$ to the second copy, where we declare $f$ to behave as the other choice of $\sqrt{z}$, i.e. the one sending its image to the left half-plane.
These two copies of $\mathbb C\setminus\mathbb R^-$ are the two branches of $\sqrt{z}$, the negative real line is the branch cut, and $S$ is an object called a Riemann surface.
A: The point is that there is no way to define $\sqrt z$ continuously on any circle about the origin.  Consider the circle $z=re^{i\theta}$.  We could define $\sqrt z=\sqrt e^{i\theta/2}$.  But if we let $z$ travel around the circle, when we get back to where we started, $\theta$ has increased by $2\pi$, and our formula gives $$\sqrt z = \sqrt re^{i(\theta+2\pi)/2}=\sqrt re^{i(\theta/2+\pi}=-re^{i\theta/2},$$ the negative of the value we had before.
Obviously, a continuous function doesn't behave this way.
What this shows is that we cannot define $\sqrt z$ continuously in any region that contains a circle about the origin.  Taking a branch cut from $0$ to $\infty$ eliminates any such circles, and it turns out to be possible to define $\sqrt z$ in such a region.
Of course, we had another choice for the definition of $\sqrt z$, and using a Riemann surface allows us to switch between the branches.
