This is problem 9-1 from Milnor, dynamics of one complex variable (arxiv).
Let $f_{\alpha}(z) = z + \alpha z^2 + z^3$. Show that $f_{\alpha}$ can be perturbed so that the double fixed point at the origin splits up into two fixed points which are both attractive if and only if $\alpha^2$ lies inside the disk of radius 1/2 centered at 1/2.
Question: it doesn't seem possible that origin can split up into two attractive fixed points ever. Where am I wrong in my argument?
Assume the perturbation $f_{\alpha, t}$ has fixed points analytic in $t$, i.e. $$f_{\alpha, t}(z) - z = (z - \beta_1(t))(z - \beta_2(t))(z - \gamma(t))$$ where $\beta_1(0) = \beta_2(0) = 0$, and $\gamma(0) = -\alpha$, where $\beta_1, \beta_2, \gamma$ are analytic functions defined locally near 0.
Now $\beta_1(t), \beta_2(t)$ corresponds to the deformation of origin. The multipliers at these points are $$f_{\alpha,t}'(\beta_1(t)) = 1 + (\beta_1(t) - \beta_2(t)) (\beta_1(t) - \gamma(t)) = 1 + \alpha (\beta_1(t) - \beta_2(t)) + \text{(higher order terms)}$$ $$f_{\alpha,t}'(\beta_2(t)) = 1 + (\beta_2(t) - \beta_1(t)) (\beta_2(t) - \gamma(t)) = 1 + \alpha (\beta_2(t) - \beta_1(t)) + \text{(higher order terms)}$$
This then doesn't make sense, since $(\beta_1(t) - \beta_2(t))$ and $(\beta_2(t) - \beta_1(t))$ are negation of each other, so there is no way $$|1 + \alpha(\beta_1(t) - \beta_2(t))| < 1 \text{ and } |1 + \alpha(\beta_2(t) - \beta_1(t))| < 1$$ hold simultaneously, i.e. there is no way both $\beta_1(t)$ and $\beta_2(t)$ are attractive fixed points of $f_{\alpha, t}(z)$.
Questions:
- Is there anything wrong with this argument?
- If not, I am assuming that analyticity of roots is too strong a sense of perturbation for the statement to hold. How should I make sense of Milnor's statement then?
Thank you!