Complex functions with certain properties. I'm trying to prove that on $U=\mathbb{C}- \{0\}$, there exists no non-constant holomorphic function such that $f(z)=f(2z)\ \forall z\in U$, but on the other hand, that there does exist a nonzero holomorphic function such that $g(2z)=\sqrt{2}zg(z)$.
For the first part, it came out very simply from considering the Laurent series and its uniqueness, but only if I can assume that the Laurent series can be considered convergent for the whole annulus $U$ when centred at $0$; is this justifiable? 
The second part I literally have no clue. Considering the Laurent series of both sides quickly gives series that simply don't look like they'll be the same for non-trivial coefficients!
 A: 
but only if I can assume that the Laurent series can be considered convergent for the whole annulus $U$ when centred at $0$; is this justifiable?

Yes. The Laurent series always converges (locally uniformly) in the largest annulus where the function is holomorphic. If $f$ is holomorphic on the annulus $K(0;r,R) = \{ z : r < \lvert z\rvert < R\}$, and  $z \in K(0;r,R)$, then we can pick two radii $r'$ and $R'$ with $r < r' < \lvert z\rvert < R' < R$, and Cauchy's integral formula yields
$$f(z) = \frac{1}{2\pi i} \int_{\lvert \zeta\rvert = R'} \frac{f(\zeta)}{\zeta-z}\,d\zeta - \frac{1}{2\pi i}\int_{\lvert \zeta\rvert = r'} \frac{f(\zeta)}{\zeta-z}\,d\zeta,$$
and we can expand $\frac{1}{\zeta-z}$ into geometric series,
$$\frac{1}{\zeta-z} = \frac{1}{\zeta}\sum_{k=0}^\infty \left(\frac{z}{\zeta}\right)^k$$
on $\lvert \zeta\rvert = R'$ and
$$\frac{1}{\zeta-z} = -\frac{1}{z}\sum_{k=0}^\infty \left(\frac{\zeta}{z}\right)^k$$
on $\lvert\zeta\rvert = r'$. Both series converge uniformly on the respective circles, allowing to interchange summation and integration since $f(\zeta)$ is bounded on the (compact) circles. That yields the Laurent series of $f$, which therefore is convergent in $z$.
For the second part, the identity $g(2z) = \sqrt{2}\,zg(z)$ yields
$$\sum_{n=-\infty}^\infty a_n z^n = g(z) = \frac{g(2z)}{\sqrt{2}\,z} = \sum_{m=-\infty}^\infty \frac{2^ma_m z^m}{\sqrt{2}\,z} = \sum_{n=-\infty}^\infty \left(2^{n+\frac12}a_{n+1}\right)z^n$$
and hence the recurrence
$$a_n = 2^{n+\frac12}a_{n+1}.$$
If there is such a $g\not\equiv 0$, it will have one non-zero coefficient in its Laurent expansion, and the recurrence shows all coefficients will be nonzero. All coefficients can then be explicitly written in terms of $a_0$, and it remains to show that the thus-determined formal series does indeed converge on all of $\mathbb{C}\setminus\{0\}$.
