(Excercise $6$ ,Conway ,page $217$ )
Let $D_0=B(1, 1) $ and $f_0$ be the restriction to $D_0$ of the principal branch of logarithm. For $n\in \Bbb{Z}$ let $\gamma(t) =e^{2\pi i nt}$ for $0\le t\le 1$ . Find a continuation $\{(f_t,D_t):t\in[0,1]\}$ along $\gamma$ of $(f_0, D_0) $ and show that $[f_1]_1=[f_0+2\pi i n]_1$.
My attempt : (This proof is motivated by the proof of analytic continuation of principal branch of $\sqrt{z}$ given here )
$D_0=B(1, 1) =B(\gamma(0), 1) $
Define $D_t:=B(\gamma(t), 1) =B(e^{2\pi in t},1)$
$f_t(\gamma(t)) $ can take value $i (2\pi n t+2 k\pi) $ for $k\in \Bbb{Z}$.
Since $f_0(\gamma(0) =0$ , we take $f(\gamma(t)) =2\pi i n t$ otherwise we loose continuity of $f_0$ at the center of $D_0$.
$z\in D_t$ implies $z \space e^{-2\pi in t}\in D_0$.
$\begin{align} f_t(z) :&=f_t(e^{2\pi i nt}) +f_0(ze^{-2 \pi i n t}) \\&= 2\pi i nt +f_0(ze^{-2 \pi i nt})\end{align} $
Hence $\gamma(t) \in D_t $ and $f_t\in \mathscr{H}{(D_t) }$.
To show
- $e^{f_t(z) }=z $.
Since $e^{f_0(z) }=z$ and $z\in D_t$ implies $ze^{-2\pi i nt}\in D_0$, $$e^{f_0(ze^{-2\pi i nt})}=ze^{-2\pi i nt}$$
$$e^{-2\pi i nt}+{f_0(ze^{-2\pi i nt})}=z$$
$$e^{f_t(z) }=z $$
- $f_t=f_t'$ on $D_t\cap D_t'$
$z\in D_t\cap D_t'$ implies $w:=ze^{-2\pi i nt}\in D_0$ and $w':=ze^{-2\pi i nt'}=e^{-2\pi i n(t'-t) } w\in D_0$
Hence $\begin{align}f_0(w) &=f_0(e^{-2\pi i n(t'-t) }+f_0(w')\\&=2 \pi i n (t'-t) +f_0(w') \end{align}$
Now $\begin{align}f_t(z) &=f_t(e^{-2\pi i nt} w) \\&=2\pi i nt+f_0(w) \\&=2 \pi i n t +2 \pi i n t(t'-t) +f_0(w') \\&= 2\pi i n t'+f_0(w') \\&=f_{t'}(z) \end{align}$
Hence $f_t=f_{t'}$ on $D_t\cap D_{t'}$.
Hence $\{(f_t,D_t):t\in [0,1]\}$ is an analytic continuation of $(f_0, D_0) $ along the path $\gamma$ where $D_t=B(e^{2\pi i nt}, 1) $ and $f_t(z) =2 \pi i nt +f_0(ze^{-2\pi i nt}) $
Now $f_1(z) = 2 \pi i n +f_0(z) $ $\space \forall z\in D_0$.
Hence $[f_1]_1=[f_0+2 \pi i n ]_1$
Is the above attempt correct?