A question about analytic continuation 
Let $\Omega = \{z: \frac{1}{2} < |z| < 2\}$. For $n = 1,2,3, ...$ let $X_n$ be the set of all $f \in H(\Omega)$ that are nth derivatives of some $g \in H(\Omega)$. [In other words, $X_n$ is the range of the differential operator $D^n$ with domain $H(\Omega)$.]


(a) Show that $f \in X_1$ if and only if $\int_\gamma f(z) dz = 0$, where $\gamma$ is the positively oriented unit circle.


(b) Show that $f \in X_n$ for every n if and only if $f$ extends to a holomorphic function in $D(0; 2)$.

For (a) $f \in X_1$ that $\int_\gamma f(z) dz = 0$ is easy get from Cauchy Theorem, but I cannot figure out the converse part. Maybe I can define a $F(z) = \int_{[a,z]}f$ Show that F is analytic in $\Omega$ and $F'=f$
But I totally have no idea about the (b), any relation between the analytic continuous and existence of f.
 A: A holomorphic function in the annulus $\Omega$ can be developed into a Laurent series:
$$
 f(z) = \sum_{n=-\infty}^\infty a_n z^n
$$
and the coefficients are uniquely determined and can be computed with
$$
a_n = \frac{1}{2 \pi i} \oint_\gamma \frac{f(z)}{z^{n+1}} \, dz
$$
where $\gamma$ is any closed path surrounding the origin once in positive direction.
Show first that $f \in X_n$ if and only if
$$ 
a_{-1} = a_{-2} = \cdots = a_{-n} = 0 \, .
$$
Then $f \in X_1$ if and only if $a_{-1} = \frac{1}{2 \pi i} \oint_\gamma f(z) \, dz$ is zero.
For the second part show that the following statements are equivalent:

*

*$f \in X_n$ for every positive integer $n$.

*$a_n = 0$ for every negative integer $n$.

*$f$ can be developed into a Taylor series which converges for $|z| < 2$.

*$f$ extends to a holomorphic function in $D(0; 2)$.

A: While @Martin's comment solves the problem directly, a more conceptual approach uses the fact that the hypothesis that $f$ is a derivative of arbitrary order implies $\int_{|z|=1}f(z)z^ndz=0, n \ge 1$ which is easily proved by induction.
But now if a continuous function $f$ on the unit circle satisfies the above, then its Cauchy transform, $f_C(z)=\frac{1}{2\pi i}\int_{z}\frac{f(w)dw}{w-z}$, is equal to its Poisson transform, $f_P(re^{i\alpha})=\frac{1}{2\pi}\int_0^{2\pi}f(e^{i\theta})P_r(\theta-\alpha)d\theta$, just by doing the integrals in question explicitly in terms of $z, \bar z$, so in other words, its Poisson transform is analytic inside the unit disc; in our case $f$ being analytic is harmonic on the annulus so it is equal to its Poisson transform in the part of the annulus contained within the unit disc, and the above implies that $f$ is equal there an analytic function defined inside the unit disc, hence it extends analytically to the unit disc, so to $B(0,2)$
