Function's analytic continuation is its own derivative This is the question we were asked at the university by our professor for complex analysis. Not as an exam, but as a challenge. I don't think he knew the answer himself.
Find a nontrivial example of a function $f$ defined on a neighbourhood of $z\in \Bbb C$ and a path from $z$ to $z$ so that the analytic continuation of $f$ along the path is $f'$.
It's easy to find trivial example: $e^x$. What I want is $f\ne f'$.
A variation to this problem, but much easier, is to find $f$ whose analytic continuation is $f+a$, $af$, or $-f$ (for some $a\in \Bbb C$).

 They are a logarithm, a power and a square root.

But the question in title I never knew how to tackle. Can anyone shed some light on it, please?
 A: Not a positive answer, but an observation about the sort of analytic continuation which would satisfy your challenge problem.  If $f$ has the desired property, then it has that property all the way around the path $\gamma$.  That is, at any point on $\gamma$, if we continue $f$ one full time around $\gamma$ we obtain the derivative.  Following is justification (it was not immediately obvious to me, though it may be to you).
First some notation.
Let $\gamma$ denote the path in question, assumed to be parameterized with domain $[0,1]$. Since $\gamma$ is closed, for $s\in[0,1]$, we can define $\gamma_s$ to be the concatenation of the restriction $\gamma|_{[s,1]}$ followed by $\gamma|_{[0,s]}$.  Using modular arithmetic (modulo $1$) we can think of $\gamma_s$ having domain $[s,s+1]$.
Let $f_s$ denote the analytic continuation of $f$ along the path $\gamma|_{[0,s]}$ (viewed as having for its domain a neighborhood of $\gamma(s)$).  We are assuming that ${f_0}'=f_1$.
Let $A\subset[0,1]$ be the subset such that $r\in A\Leftrightarrow$ the analytic continuation of $f_r$ along $\gamma_r$ is ${f_r}'$.  We are assuming that $A$ contains $0$ (and is thus non-empty).  It is immediate from the definition of analytic functions that $A$ is open.  It is also not hard to show that $A$ is closed (since if the points where two functions are equal have an accumulation point in their common domain, they are equal in their common domain).  Thus $A=[0,1]$.
