Here is the definitions I am working with
Define the map $$ (z-1)^{\frac{1}{2}}$$ defined as the inverse of $$\begin{align} f: \mathbb{C} & \rightarrow \mathbb{C} \\ z &\mapsto z^{2}+1 \end{align}$$.
Establishing multivalues is immediate from the fact that $1$ and $-1$ get mapped to the same value by $f$.
To establish that $1$ is a branch point I think I should go about contradiction.
Suppose not. Then, by definition, there exists $\epsilon> 0$ such that there is a branch of the inverse map $g$ defined on the disk $D(1,\epsilon)$ with the point $1$ omitted. Now consider the contour \begin{align*} \gamma(t) = 1 + \frac{\epsilon}{2} e^{it}, \quad 0\leq t \leq 2\pi. \end{align*} Then I had the idea to show \begin{align*} \int_{\gamma }^{} g(z) \, \mathrm{d}z \neq 0. \end{align*} This proved hard because I can't find an expression for \begin{align*} g(r e^{i\theta }) \end{align*} to integrate. I also thought about defining $g$ in terms of its derivative but this was to no avail as \begin{align*} f' (g(z)) g'(z) = 1 \implies g'(z) = \frac{1}{f' (g(z))}= \frac{1}{2 g(z)}. \end{align*} This removes any hopes to define the function as an integral.
Question: Formally how do I show that the square root function has a branch point at $0$?
A branch point is defined by the fact there is no continuous instance of the function on a (small) circle around it - as continuity implies analyticity here (assuming of course the function is multivalued analytic to start with except at singularities); in this case assume there is a continuous hence analytic $f(z)^2=z$ on the circle of radius $1$ say around zero; one can use the argument principle and integrate $2f'/f=1/z$ on the circle giving that an even integer is equal to an odd integer which is a contradiction, but one can also argue from continuity only as follows:
$f(1)=\pm 1$ so wlog we can assume $f(1)=1$ by changing $f \to -f$ and then $f(-1)=\pm i$ so $\Re f(-1)=0$ and $-1$ is the only point on the circle with that property; hence $f$ continuous means $\Re f(z)>0, z \ne -1$ so in particular, we have $f(e^{i(\pi-\epsilon)})=e^{i(\pi-\epsilon)/2}$ since that is the positive real part square root, but then $f(e^{i(\pi+\epsilon)})=-e^{i(\pi+\epsilon)/2}$ since now that is the positive real part square root and taking $\epsilon \to 0$ gives a contradiction to continuity at $-1$