Transformations of diffeomorphism $f(z)=e^{i\alpha}z+\overline{z}^3+z^2\overline{z}$ that eliminates $\bar z^3$

Question

Find a transformation of the form $z=w+a\overline{w}^3$ such that $$f(z)=e^{i\alpha}z+\overline{z}^3+z^2\overline{z}$$ where $\alpha\neq2\pi p/q,\ q=1,2,3,4,$ becomes $$\tilde{f}(w)=e^{i\alpha}w+w^2\overline{w}+O\left(|w|^4\right).$$ The solution should be $$a=\frac{e^{3i\alpha}}{1-e^{4i\alpha}},$$ but I get $$a=-e^{-i\alpha}.$$ What am I doing wrong?

Answer 1

I think you composed $f$ with the transformation $z=z(w)$, obtaining $$e^{i\alpha}(w+a\bar w^3)+\bar w^3+w^2\bar w \tag{1}$$ But you should have conjugated $f$ with that transformation. That also includes composition with the inverse $w=w(z)$ on the left.

The inverse transformation is $w=z-a\bar z^3+O(|z^4|)$. Applying it to (1) yields $$e^{i\alpha}(w+a\bar w^3)+\bar w^3+w^2\bar w -ae^{-3i \alpha} \bar w^3 +(|w|^4)\tag{2}$$ Equate the coefficient of $\bar w^3$ to zero to get the result.