Possible solutions for an analytic continuation given a condition on the germs Let $\gamma$ be a path in $\mathbb{C}$ such that $\gamma(0)=\gamma(1)=a$ and consider an analytic continuation {$(f_t, D_t): 0\leq t \leq 1$} along the path $\gamma$ such that $[f_1]_a = [f'_0]_a$. We also require that $f'_0$ is non zero. What are the possible solutions for $f_0$? 
For any $K \in \mathbb{C}$ then $Ke^z$ is a solution to the problem but are there other functions satisfying this problem? By going along $-\gamma$ are we somehow integrating and getting a primitive? Thx.
 A: This is not an answer; but perhaps someone has further ideas on this problem.
By assumption the function $f_0$ maps a neighborhood $U$ of $a$ bijectively onto a neighborhood $V$ of $w_0:=f(a)$, and there is an analytic inverse function $\phi:\ V\to U$, $\ w\mapsto z:=\phi(w)$. Let $h(w):=f'\bigl(\phi(w)\bigr)$; then we have
$$f_0'(z)\ =\ h\bigl(f_0(z)\bigr)\qquad(z\in U)\ ,$$
i.e., the function $f_0$ satisfies the differential equation $$y'=h(y)\ .\qquad\qquad(1)$$ It follows from general principles that all function elements $f_t$ arising in the analytic continuation satisfy $(1)$ (as long as it makes sense, depending on $h$). This means that also the function $f_1=f_0'$ defined on $U$ satisfies $(1)$. Now the solutions of $(1)$ differ by a translation in the independent variable $z$. This means that
$$f_0'(z) = f_1(z)=f_0(z-c)\qquad(z\in U)$$
for some $c\in{\mathbb C}$.
That's how far I got, but I couldn't exclude the possibility $f_0=\sin$, $c={3\pi\over2}$ without using that $\sin$ is defined uniquely for all $z\in{\mathbb C}$.
