Let $D_0$ be the disc centered at $1$ of radius $1$. Let $f_0$ be the restriction of the principal branch of $\sqrt z$ to $D_0$. Let $\gamma(t)=e^{2\pi i t}$ and $\sigma(t)=e^{4\pi i t}$ for $0 \leq t \leq 1$. Find an analytic continuation $\{(f_t, D_t) : 0 \leq t \leq 1\}$ of $(f_0, D_0)$ along $\gamma$ and show that $f_1(1)=-f_0(1)$.
By definition we have $f_0 (z)=e^{\frac{1}{2} {\rm Log}(z)}$ with ${\rm Log}$ the principal branch of $\log$. $\gamma$ goes around the origin once and $\sigma$ goes around the origin twice. I don't understand the first question of finding the analytic continuation. As we go along $\gamma$ what happens to the principal branch of the square root? What is the method for solving such problems? Thx.
EDIT:thank you to all of you for these amazing answers. I never knew how to quite approach this problem. Now you gave me an idea how to do it.
Best Answer
Zhen Lin has given a complete solution of the problem in finite terms, and at minimal cost.
As an alternative I propose the following continuous version:
We are told to produce for each $t\in[0,1]$ a neighborhood $D_t$ of the point $\gamma(t)=e^{2\pi i t}$ and an analytic function $f_t:\ D_t\to{\mathbb C}$ such that (a) for all $t$ one has $$\bigl(f_t(z)\bigr)^2\ =\ z\qquad(z\in D_t)\qquad\qquad(1)$$ and that (b) whenever $D_t$ and $D_{t'}$ intersect one has $$f_{t'}(z)=f_t(z)\qquad(z\in D_t\cap D_{t'})\ .\qquad\qquad(2)$$
It is pretty clear that we should take as $D_t$ the unit disk with center $\gamma(t)$. Second, the value of $f_t$ at the center of $D_t$ should be one of $e^{\pi i t}$ and $-e^{\pi i t}$, and as $f_0(1)=1$ the only way to do this in a continuous fashion is to provide for $f_t\bigl(e^{2i\pi t}\bigr)=e^{i\pi t}$.
Note that for any $z\in D_t$ the point $w:=z\ e^{-2\pi i t}$ lies in $D_0$ where the function $f_0$ is available. This allows us to "transport" the function $f_0$ from $D_0$ to $D_t$ in the following way: We define $$f_t(z)\ :=\ e^{\pi i t}\ f_0\bigl(z\ e^{-2\pi it}\bigr)\ \qquad(z\in D_t)\ .$$ Using $\bigl(f_0(w)\bigr)^2 = w$ $\ (w\in D_0)$ it is easy to see that (1) holds. Consider now a $z\in D_t\cap D_{t'}$. The points $w:=z\ e^{-2\pi i t}$ and $w':=z\ e^{-2\pi i t'}=e^{2\pi i(t-t')} w$ are lying on an arc of radius $|z|$ in $D_0$. From the definition of $f_0$ it then follows that $f_0(w)=e^{i\pi(t'-t)}f_0(w')$, therefore we get $$f_t(z)=e^{i\pi t} f_0(w)=e^{i\pi t}e^{i\pi(t'-t)}f_0(w')=e^{i\pi t'}f_0(w')=f_{t'}(z)\ ,$$ as required by (2).
In particular we have $f_1(1)=e^{i\pi}f_0(1)=-1$.