It is well known that a holomorphic vector field $z'=f(z), z\in \mathbb{C}$ does not have any limit cycle.See the last paragraph of this post


[Orbits space of real-analytic planar foliations](https://mathoverflow.net/questions/382871/orbits-space-of-real-analytic-planar-foliations)


One can imagine several reason for this fact. For example: The flow map is a holomorphic map. So if the limit cycle is of period $T$ then $\phi\_T$ admits an infinite number of fixed point hence it is identically equal to the identity map.


Now we ask the following question



> 
> Is there a holomorphic vector field $z'=f(z)$ on an open region of $\mathbb{C}$ which admit an attractor homoclinic loop?
> 
> 
> 



Can a holomorphic vector field have an attractor homoclinic loop?
The answer is 'no' for much the same reason that the OP indicates: the existence of a homoclinic or heteroclinic connection implies that neighboring trajectories are periodic.


First, one needs to have poles in $f$ if one wishes to have heteroclinic connections between (real) saddle singularities. Indeed if $f$ is holomorphic, then the stationnary points of the underlying real vector field are either centers, foci or multiple singularities locally conformally conjugate to $z'=\frac{z^{k+1}}{1+\mu z^k}$. None of these singularities can be part of a candidate limit (poly)cycle.


Next, consider the time-coordinate $$T(z)=\int\_{z\_\*}^z \frac{dz}{f(z)}.$$ Real-time trajectories of the vector field correspond to level sets $\mathrm{Im}(T)=\textrm{cst}$. The presence of a (poly)cycle $\Gamma$ connecting poles of $f$ implies that $\int\_\Gamma \frac{dz}{f(z)}=:\tau\in \mathbb{R}$ is the period of $\Gamma$.


Let $\gamma$ be a neighboring trajectory connecting a transverse section $\Sigma$ to itself "after making one turn". Without loss of generality one may choose $\Sigma$ as a small piece of $\mathrm{Re}(T)=\textrm{cst}$. Then there exists a small curve $\sigma\subset\Sigma$ such that the concatenation $\sigma\gamma$ is a closed loop. Cauchy's formula implies that $\int\_{\sigma\gamma} \frac{dz}{f(z)}=\tau$. But only $\sigma$ contributes to the imaginary part of the integral, therefore $\sigma$ is just one point and $\gamma$ is closed.


