# $f(z)$ and $f(z+z^2)$ have the same singularities at 0

Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be an analytic function in a punctured neighborhood of 0 then $f(z)$ and $h(z)=f(z+z^2)$ have the same singularity at $z_0=0$.

I was able to show that every removable singularity or pole of $f$ is also the same of $h$. However, I was unable to do so for the opposite direction or for an essential singularity (Casorati-Weierstrass theorem maybe?)

This follows from the fact that the map $$H:\Bbb C\rightarrow\Bbb C,\qquad z\mapsto z+z^2$$ is a local isomorphism at $z=0$ because $H^\prime(0)=1\neq0$.

Since $H$ is a degree 2 polynomial for each point $z_0$ in the target space there are generically two points $w_1$, $w_2$ each of which has a neighborhood which is mapped by $H$ isomorphically onto a neighborhood of $z_0$. For all those points, pulling back holomorphic functions does not change their local properties (such as the order of a zero or a pole).

The exceptional points are the ramified points, namely those for which the map fails to be an analytic isomorphism locally. In the case of $H$ the only ramified point is $z_0=-1/4$ as can be recognized either observing that $H(-1/2)=-1/4$ and $H^\prime(-1/2)=0$ or, possibly more simply, observing that $k=-1/4$ is the only value such that $z^2+z-k$ has one double root.

Since again $H$ has degree 2, the ramified point $z_0=-1/4$ must have index of ramification 2, i.e. when pulling back holomorphic functions along $H$ the orders of poles or zero at $z_0=-1/4$ get multiplied by 2 at $w_0=-1/2$.

Let me see if I got this straight since I think it is pretty simple: as $\;f\;$ is analytic in a neighborhood of zero (a punctured one I presume, otherwise the function's anayltic at zero, too), then we have a Laurent series for $\;f\;$ around zero:

$$\forall\; 0<|z|<r\;,\;\;\;f(z)=\sum_{k=-\infty}^\infty a_nz^n$$

where the number of non-zero coefficients with negative index can be infinite (essential sing.), finite non-zero (pole) or zero (analicity point ore removable singularity). But then this is just the same for

$$f(z+z^2)=f\left(z(1+z)\right)=\sum_{k=-\infty}^\infty a_kz^k(1+z)^k$$

as long as $\,0<|z+z^2|=|z||1+z|<r\;\ldots$

and this means, I think, the kind of point zero is remains the same in both cases.