An exercise in Conway about an analytic continuation of the square root 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.
 A: 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 $f_1(1)=e^{i\pi}f_0(1)=-1$.
A: In fact, by compactness it suffices to find only finitely many analytic continuations along the path. In the case of the square root, we have an intuitive idea of what the complete analytic function looks like, so this is easy enough to carry out explicitly. I claim it suffices to find four, at $\frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1$. Indeed, let $f_{\frac{1}{4}}$ be defined on the open disc of radius $1$ about $i$ by
$$f_{\frac{1}{4}}(r \exp i \theta) = \sqrt{r} \exp \frac{1}{2} \theta i, \text{ where } \frac{1}{2} \pi < \theta < \frac{3}{2} \pi$$
[Check that this is well-defined and holomorphic.] A sketch shows the domain of $f_{\frac{1}{4}}$ overlaps with the domain of $f_0$ and direct calculation shows that they agree on the overlap. We repeat this process, defining $f_{\frac{1}{2}}$ on the open disc of radius $1$ about $-1$, $f_{\frac{3}{4}}$ on the open disc of radius $1$ about $-i$, and then finally $f_1$ on the open disc of radius $1$ about $1$. Then, another straightforward calculation shows that $f_0 = -f_1$, as required.
For more general functions, more sophisticated methods may be necessary, but I hope the above example illustrates the general principle.
A: I assume that $D_t$ is the disc of radius $1$ centered at $\gamma(t)$. Define
$$
f_t(z)=\sqrt{\frac{z}{\gamma(t)}}\;\gamma(\frac{t}{2})
$$
Since $\gamma(\frac{1}{2})=-1$ and $\gamma(0)=\gamma(1)=1$ we get that $f_1(z)=-f_0(z)$ for this choice of $f_t$.
Now, we need to verify that the $f_t$ match on intersecting $D_t$. For any $z\in D_t\cap D_s$ where $|t-s|<\frac{1}{4}$, we have that $\frac{z}{\gamma(t)}$ and $\frac{z}{\gamma(s)}$ are both in $D_0$ and
$$
\begin{align}
f_t(z) &= \sqrt{\frac{z}{\gamma(t)}}\;\gamma(\frac{t}{2})\\
&=\sqrt{\frac{z}{\gamma(s)}\gamma(s-t)}\;\gamma(\frac{t}{2})\\
&=\sqrt{\frac{z}{\gamma(s)}}\;\gamma(\frac{s-t}{2})\;\gamma(\frac{t}{2})\\
&=\sqrt{\frac{z}{\gamma(s)}}\;\gamma(\frac{s}{2})\\
&=f_s(z)
\end{align}
$$
We can split the square root in the third equation since both factors are in the right half plane. Thus, the extensions match on $D_t\cap D_s$.
Thus, $\{(f_t,D_t):t\in[0,1]\}$ is a chart for the extension of the square root.
