Proof. Suppose first that $S= S(1,\varepsilon ,\theta )$ . Set $\kappa (\theta )=\frac 18 \sin (\theta /2)$ and assume that $\varepsilon \le {\kappa (\theta )}$ . An easy computation shows that the map $z\mapsto w= 1 /z$ conjugates f to

$$ \begin{align*}g(w)= w+1+\frac {1}{w-1}\end{align*} $$

and transforms S into

$$ \begin{align*}W=\{w\in\mathbb C\colon |w|>1/\varepsilon,\ \lvert\arg w\rvert<\theta\}.\end{align*} $$

Fix $z\in S(1,\varepsilon ,\theta )$ and set $z_1=f(z)$ , $w=1/z$ and $w_1=1/z_1=g(w)$ . Consider the sectors $S_w(\theta )$ and $S_w(\theta /2)$ , where

$$ \begin{align*}S_w(\alpha)=w+\{\zeta\in\mathbb C \colon |\zeta|>0,\ \lvert\arg\zeta\rvert<\alpha\}.\end{align*} $$

Clearly, $S_w(\theta /2)\subset S_w(\theta )\subset W$ and

$$ \begin{align*}\operatorname{\mathrm{dist}} (w+1, \partial S_w(\theta/2))=\sin(\theta/ 2).\end{align*} $$

Then, since

$$ \begin{align*}\operatorname{\mathrm{dist}} (w+1, w_1)=\frac{1}{|w-1|}<\frac{1}{1/{\kappa(\theta)}-1} =\frac 1{8/\sin(\theta/2)-1}<\frac 1 4\sin(\theta/2)<\sin(\theta/2),\end{align*} $$

we see that $w_1\in S_w(\theta /2)\subset W$ , which proves that $f(S)\subset S$ . Note that

$$ \begin{align*}\operatorname{\mathrm{dist}} (w_1, \partial S_w(\theta))=|w_1-w|\sin\beta,\end{align*} $$

where $\beta $ is one of the angles determined by the ray $\overrightarrow {ww_1}$ in the sector $S_w(\theta )$ . Since ${w_1\in S_w(\theta /2)}$ , we have that $\beta \ge \theta /2$ and therefore

$$ \begin{align*} \operatorname{\mathrm{dist}} (w_1, \partial S_w(\theta))&>|w_1-w|\sin(\theta/2)\ge\bigg(1- \frac{1}{|w-1|}\bigg)\sin(\theta/2)\\ &> \big(1- \tfrac 1 4\sin(\theta/2)\big) \sin(\theta/2) >\tfrac 3 4 \sin(\theta/2). \end{align*} $$

Then, since $S_w(\theta )\subset W$ ,

$$ \begin{align*} \operatorname{\mathrm{dist}} (w_1, \partial W)\ge \operatorname{\mathrm{dist}} (w_1, \partial S_w(\theta))>\frac 34 \sin(\theta/2), \end{align*} $$

so the disk $\mathcal D=\{\zeta \in \mathbb C\colon |\zeta -w_1|<\frac 34 \sin (\theta /2)\}$ is contained in W. If we set $h(\zeta )=1/\zeta $ , we have that $h(W)=S$ and $h(\mathcal D)$ contains the disk $\{\xi \in \mathbb C\colon |\xi -h(w_1)|<r\}$ , where

$$ \begin{align*}r=\min\{|h(\zeta)-h(w_1)|\colon \zeta\in\mathcal \partial \mathcal D \}.\end{align*} $$

Notice that if $\zeta \in \mathcal \partial \mathcal D$ , then $|\zeta |\le |w_1|+3/4<2|w_1|$ . Then

$$ \begin{align*}r=\min\bigg\{\bigg|\frac{\zeta-w_1}{\zeta w_1}\bigg|\colon \zeta\in\mathcal \partial \mathcal D\bigg \}> \frac{ \frac{3}{4} \sin(\theta/2)}{2|w_1|^2}= \frac{ \frac{3}{8} \sin(\theta/2)}{|w_1|^2}. \end{align*} $$

Thus, since

$$ \begin{align*}1/{|w_1|^2}=|z_1|^2\ge |z|^2(1-|z|)^2\ge |z|^2(1-\kappa(\theta))^2\ge|z|^2 (1-1/8)^2>\tfrac 23 |z|^2,\end{align*} $$

we see that $r> \frac 14 \sin (\theta /2)|z|^2>\kappa (\theta ) |z|^2$ . Therefore we conclude that S contains the disk $\{\zeta \in \mathbb C\colon |\zeta -z_1|<\kappa (\theta ) |z|^2\}, $ which concludes the proof.

Suppose now that $S= \widetilde S(1,\varepsilon ,\theta )$ , take $\kappa (\theta )=\frac 18 \sin (\pi /4-\theta /2)$ and assume that ${\varepsilon \le {\kappa (\theta )}}$ . Then the map $z\mapsto w=1/z$ transforms S into

$$ \begin{align*}\widetilde W=W\cup\{w\in\mathbb C\colon\operatorname{\mathrm{Re}}(we^{-i\theta})>1/\varepsilon\}\cup\{w\in\mathbb C\colon \operatorname{\mathrm{Re}}(we^{i\theta})>1/\varepsilon\}.\end{align*} $$

Note that $S_w(\pi /4-\theta /2)\subset S_w(\pi /2-\theta )\subset W$ . Then the proof follows exactly as in previous case, with $\pi /2-\theta $ instead of $\theta $ . Finally, to unify the selection of $\kappa $ we can take $\kappa (\theta )= \frac 18\min \{\sin (\theta /2), \sin (\pi /4-\theta /2)\}$ , which works for both cases.

Proof of Proposition 2.2

From the expression for $x_1^my_1^n$ we obtain that

(1) $$ \begin{align} (x_1^my_1^n)^d=(x^my^n)^d-(x^my^n)^{2d}+(x^my^n)^{2d}\sigma(x,y) \end{align} $$

and

(2) $$ \begin{align} \frac 1{(x_1^my_1^n)^d}=\frac 1{(x^my^n)^d}+ 1+\sigma_1(x,y), \end{align} $$

where $\sigma =O(x,y)$ and $\sigma _1=O(x,y)$ . Moreover, notice that $x_1^my_1^n/(x^my^n)=1+ O(x,y)$ is arbitrarily close to 1 provided x and y are small enough. Then, given $\theta \in (0,\pi /2)$ , we can find $\delta _G(\theta )>0$ such that for all $(x,y)$ with $|x|<\delta _G(\theta )$ and $|y|<\delta _G(\theta )$ we have that

$$ \begin{align*} |\sigma(x,y)|< \kappa (\theta), |\sigma_1(x,y)|<\frac 14 \quad \text{and} \quad \bigg\lvert\!\arg\frac{x_1^my_1^n}{x^my^n}\bigg\rvert<\frac{\pi-2\theta}d, \end{align*} $$

where $\kappa (\theta )$ is given by Lemma 2.3. Set $\varepsilon _G(\theta )= {\kappa (\theta )}.$

Let us prove assertion 1 of the proposition. Take $\varepsilon \le \varepsilon _G(\theta )$ , let $\mathcal S$ be one of the components of $S(d,\varepsilon ,\theta )$ or $\widetilde S(d,\varepsilon ,\theta )$ and take $(x,y)$ such that $x^my^n\in \mathcal S$ , $|x|<\delta _G(\theta )$ and $|y|<\delta _G(\theta )$ . Set $f(z)=z-z^2$ . Note that $S=\{z^d\colon z\in \mathcal S\}$ is one of the sets $S(1,\varepsilon ,\theta )$ or $\widetilde S(1,\varepsilon ,\theta )$ , so by equation (1) and Lemma 2.3 we have that

$$ \begin{align*} |(x_1^my_1^n)^d-f((x^my^n)^d)|<\kappa (\theta) |x^my^n|^{2d}\le \operatorname{\mathrm{dist}} (f((x^my^n)^d),\partial S), \end{align*} $$

which means that $(x_1^my_1^n)^d\in S$ . Since the components of $S(d,\varepsilon ,\theta )$ or $\widetilde S(d,\varepsilon ,\theta )$ are separated by a sector of opening at least $(\pi -2\theta )/d$ and $\lvert \arg (x_1^my_1^n/(x^my^n))\rvert < (\pi -2\theta )/d$ , we conclude that $(x_1^my_1^n)^d$ belongs to the same component $\mathcal {S}$ .

Now, let $(x,y)$ be such that $x^my^n\in \mathcal S$ with $|x_j|<\delta _G(\theta )$ and $|y_j|<\delta _G(\theta )$ for all $j\ge 0$ . From (2) we obtain that

$$ \begin{align*} \frac 1{(x_j^my_j^n)^d}=\frac 1{(x^my^n)^d}+ j+\sum_{l=0}^{j-1}\sigma_1(x_l,y_l), \end{align*} $$

so

$$ \begin{align*} \frac 1{|x_j^my_j^n|^d}\le \frac 1{|x^my^n|^d}+j+ \frac 14 j \le 2\bigg(\frac 1{|x^my^n|^d}+j\bigg) \end{align*} $$

for all $j\ge 0$ , which gives the lower bounds in the inequalities of assertion 1. On the other hand, from the equation above we have that

$$ \begin{align*} \frac 1{|x_j^my_j^n|^d} &\ge \operatorname{\mathrm{Re}} \frac 1{(x_j^my_j^n)^d} \ge \operatorname{\mathrm{Re}} \frac 1{(x^my^n)^d}+j- \sum_{l=0}^{j-1}|\sigma_1(x_l,y_l)|\\ &\ge \operatorname{\mathrm{Re}} \frac 1{(x^my^n)^d}+j - \frac 14j = \operatorname{\mathrm{Re}} \frac 1{(x^my^n)^d}+ \frac 34 j, \end{align*} $$

for all $j\ge 0$ . If $\mathcal S$ is a component of $S(d,\varepsilon ,\theta )$ , since $\lvert \arg ( 1/ {(x^my^n)^d}\rvert <\theta $ , we have that $\operatorname {\mathrm {Re}} ( 1/{(x^my^n)^d})\ge {\cos \theta }/{|x^my^n|^d}$ and therefore

$$ \begin{align*} \frac 1{|x_j^my_j^n|^d}\ge \frac {\cos \theta}{|x^my^n|^d}+ \frac 34 j \ge \frac{\cos\theta}2\bigg(\frac 1{|x^my^n|^d}+j \bigg) \end{align*} $$

for all $j\ge 0$ , which gives the upper bound in the first inequality of assertion 1. If $\mathcal S$ is a component of $\widetilde S(d,\varepsilon ,\theta )$ we obtain that for all $j\ge 6/|x^my^n|^d$ ,

$$ \begin{align*} \frac 1{|x_j^my_j^n|^d} \ge \operatorname{\mathrm{Re}}\frac 1{(x^my^n)^d}+\frac 34 j \ge -\frac 1{|x^my^n|^d}+ \frac 34 j \ge \frac 12 \bigg(\frac 1{|x^my^n|^d} +j\bigg), \end{align*} $$

which concludes the proof of assertion 1.

Let us prove assertion 2. From the expression for $x_1^my_1^n$ we can write $x_1^my_1^n=x^my^n(1-\zeta ),$ where $\zeta =(1/d+\tau ) (x^my^n)^d$ and $\tau =O(x,y)$ . Then, up to reducing $\delta _\nu $ , we have that if $ |x|<\delta _\nu $ and $ |y|<\delta _\nu $ then

$$ \begin{align*} |\zeta|<1,\quad \lvert\arg(1-\zeta)\rvert<\theta/d,\quad |1/d+\tau|>\delta_\nu, \quad \lvert\arg(1/d+\tau)\rvert<\theta^*, \end{align*} $$

where $\theta ^*= \min \{\theta /2, \pi /4-\theta /2\}$ . Set

$$ \begin{align*}\tilde\mu=e^{-2\pi\nu/(d\delta_\nu\sin\theta^*)}\min\{\mu,\delta_\nu, \delta_G\}.\end{align*} $$

Take $x^my^n\in \widetilde S_k(d,\varepsilon ,\theta )$ such that $|x|<\tilde \mu $ and $|y|<\tilde \mu $ . If $\lvert \arg (x^my^n)^d\rvert <\theta $ , we have that $x^my^n\in S_k(d,\varepsilon ,\theta )$ and, since $\tilde \mu \le \mu $ , we can take $j_0=0$ and we are done. Thus we can assume that $\lvert \arg (x^my^n)^d\rvert \in [\theta , \theta +\pi /2)$ . Moreover, we assume that $\arg (x^my^n)^d\in [\theta , \theta +\pi /2)$ ; the other case is analogous. Then

$$ \begin{align*}\arg\zeta=\arg(1/d+\tau)+\arg (x^my^n)^d\in (\theta-\theta^*, \theta+\pi/2+\theta^*)\subset (\theta/2, \theta/2+3\pi/4)\end{align*} $$

and hence

$$ \begin{align*}\operatorname{\mathrm{Im}}\zeta>|\zeta|\min\{\sin(\theta/2),\sin(\theta/2+3\pi/4)\}=|\zeta|\sin\theta^*.\end{align*} $$

Then, since $|\zeta |<1$ , $\sin (\arg (1-\zeta ))=- {\operatorname {\mathrm {Im}}\zeta }/{|1-\zeta |} < -\tfrac 1 2 \sin \theta ^*|\zeta |$ and therefore

$$ \begin{align*}-\theta <\arg (1-\zeta)^d<d\sin(\arg (1-\zeta))< -\frac d2\sin\theta^*|\zeta|.\end{align*} $$

It follows that

$$ \begin{align*}0\le\arg(x^my^n)^d- \theta &< \arg(x^my^n)^d+\arg(1-\zeta)^d < \arg(x^my^n)^d-\frac d2 \sin\theta^*|\zeta|\\ &<\arg(x^my^n)^d-\frac d2 \delta_\nu{ \sin\theta^*} |x^my^n|^d; \end{align*} $$

that is, $\arg (x_1^my_1^n)^d=\arg (x^my^n)^d+\arg (1-\zeta )^d$ satisfies

(3) $$ \begin{align} 0< \arg(x_1^my_1^n)^d< \arg(x^my^n)^d-\frac d2 \delta_\nu \sin\theta^* |x^my^n|^d. \end{align} $$

Then $ |x^my^n|^d<(2/(d\delta_{\nu}\sin\theta^*))\arg (x^my^n)^d< 2\pi/(d\delta_{\nu}\sin\theta^*)$ so

$$ \begin{align*}|x_1|\le |x|(1+\nu |x^my^n|^d)<|x| e^{\nu |x^my^n|^d}<\tilde\mu e^{ 2\pi\nu/(d\delta_\nu\sin\theta^*)}=\min\{\mu,\delta_\nu,\delta_G\}\end{align*} $$

and, analogously, $|y_1|<\min \{\mu , \delta _\nu ,\delta _G\}$ . Thus, if $\arg (x_1^my_1^n)^d<\theta $ , since $x_1^my_1^n\in \widetilde S_k(d,\varepsilon , \theta )$ because of assertion 1, we have that $x_1^my_1^n\in S_k(d,\varepsilon ,\theta )$ , we can choose $j_0=1$ and we are done. We assume then that $\arg (x_1^my_1^n)^d\in [\theta , \theta +\pi /2)$ . Proceeding exactly as above, we obtain that

$$ \begin{align*}0 <\arg(x_2^my_2^n)^d < \arg(x_1^my_1^n)^d-\frac d2 \delta_\nu{ \sin\theta^*} |x_1^my_1^n|^d \end{align*} $$

and, in view of (3),

$$ \begin{align*} 0< \arg(x_2^my_2^n)^d <\arg(x^my^n)^d-\frac d2 \delta_\nu{ \sin\theta^*} |x^my^n|^d-\frac d2 \delta_\nu{ \sin\theta^*} |x_1^my_1^n|^d. \end{align*} $$

Then

$$ \begin{align*}|x^my^n|^d+ |x_1^my_1^n|^d<\frac 2 {d\delta_\nu{ \sin\theta^*} }\arg(x^my^n)^d<\frac {2\pi} {d\delta_\nu{ \sin\theta^*} }\end{align*} $$

so

$$ \begin{align*} |x_2|&\le |x|(1+\nu |x^my^n|^d)(1+\nu |x_1^my_1^n|^d)<|x| e^{\nu |x^my^n|^d+\nu |x_1^my_1^n|^d}\\ &<\tilde\mu e^{2\pi\nu/(d\delta_\nu\sin\theta^*)}=\min\{\mu,\delta_\nu, \delta_G\} \end{align*} $$

and, analogously, $|y_2|<\min \{\mu ,\delta _\nu ,\delta _G\}$ . Thus, if $\arg (x_2^my_2^n)^d<\theta $ , we have that $x_2^my_2^n\in S_k(d,\varepsilon ,\theta )$ , we can take $j_0=2$ and we are done. We assume then that $\arg (x_2^my_2^n)^d\in [\theta , \theta +\pi /2)$ and repeat the argument. If assertion 2 were false, this process would continue indefinitely and we would obtain, for every $j\ge 0$ , that $|x_j|<\delta _G$ , $|y_j|<\delta _G$ and

$$ \begin{align*} |x^my^n|^d+ |x_1^my_1^n|^d+\cdots + |x_j^my_j^n|^d<\frac {2\pi} {d\delta_\nu{ \sin\theta^*} }.\end{align*} $$

But the inequalities in assertion 1 show that $\sum _{j\ge 0}|x_j^my_j^n|^d$ diverges, which is a contradiction.

Let us prove assertion 3. Let $\theta \in (0,\pi /2)$ . Choose $\delta _1>0$ such that for all $(x,y)$ with $|x|<\delta _1$ and $|y|<\delta _1$ we have $|\sigma _1(x,y)|<\tfrac 12 \tan \theta $ and set

$$ \begin{align*}\tilde\delta_G(\theta)=\min\{\delta_1, \delta_G,\varepsilon_G ^{1/((m+n)d)}\}.\end{align*} $$

Take $(x,y)$ such that $|x_j|<\tilde \delta _G(\theta )$ and $|y_j|<\tilde \delta _G(\theta )$ for all $j\ge 0$ . From (2) we have

$$ \begin{align*} \bigg\vert\!\operatorname{\mathrm{Im}} \frac 1{(x_j^my_j^n)^d}\bigg\vert\le \bigg\vert\!\operatorname{\mathrm{Im}}\frac 1{(x^my^n)^d}\bigg|+ \sum_{l=0}^{j-1}|\sigma_1(x_l,y_l)| \le \bigg\vert\!\operatorname{\mathrm{Im}}\frac 1{(x^my^n)^d}\bigg|+\frac 12 j\tan\theta. \end{align*} $$

On the other hand, as we showed in the proof of assertion 1,

$$ \begin{align*} \operatorname{\mathrm{Re}} \frac 1{(x_j^my_j^n)^d} \ge \operatorname{\mathrm{Re}}\frac 1{(x^my^n)^d}+\frac 34 j. \end{align*} $$

Therefore, for j big enough,

$$ \begin{align*} 0< \frac {\lvert\operatorname{\mathrm{Im}} ( 1/{(x_j^my_j^n)^d})\rvert}{\operatorname{\mathrm{Re}} {( 1/{(x_j^my_j^n)^d})}}\le \frac {\lvert\operatorname{\mathrm{Im}}( 1/{(x^my^n)^d})\rvert+ (1/2) j\tan \theta} { \operatorname{\mathrm{Re}}( 1/{(x^my^n)^d})+(3/4) j}< \tan\theta. \end{align*} $$

This means that $\lvert \arg ( 1/{(x_j^my_j^n)^d})\rvert <\theta $ for j big enough, so we can fix $j_0\ge 0$ such that $\lvert \arg (x_{j_0}^my_{j_0}^n)^d\rvert <\theta $ . Moreover, since $|(x_{j_0}^my_{j_0}^n)^d|<\tilde \delta _G(\theta ) ^{(m+n)d}\le \varepsilon _G (\theta )$ , we have that $x_{j_0}^my_{j_0}^n\in S(d,\varepsilon _G (\theta ),\theta )$ . Thus $x_{j_0}^my_{j_0}^n\in S_k(d,\varepsilon _G (\theta ),\theta )$ for some $k\in \{0,\ldots , d-1\}$ and it follows from assertion 1 that $x_{j}^my_{j}^n\in S_k(d,\varepsilon _G (\theta ),\theta )$ for all $j\ge j_0$ and $x_{j}^my_{j}^n\to 0$ . Clearly, given $\varepsilon>0$ , up to increasing $j_0$ , we have that $x_{j}^my_{j}^n\in S_k(d,\varepsilon ,\theta )$ for all $j\ge j_0$ . Moreover, arguing exactly as in the proof of assertion 1 we have that the two last inequalities of that assertion hold. This concludes the proof of Proposition 2.2.

Proof. From the expression for F, we can write $x_1=x(1-\zeta )$ , where $\zeta =(-a+\sigma ) (x^my^n)^d$ and $\sigma =O(x,y)$ . By the choice of $\theta _0$ , we can take $\delta _0>0$ such that for all $(x,y)$ with $|x|<\delta _0$ and $|y|<\delta _0$ we have

$$ \begin{align*}\delta_0<|\!-a+\sigma|< 1/\delta_0, \quad \lvert\arg(-a+\sigma)\rvert+2\theta_0<\pi/2.\end{align*} $$

Set $\eta =\frac 1{2}\delta _0 {\cos (\pi /2-\theta _0)}>0$ and $\varepsilon _0=\delta _0\cos (\pi /2-\theta _0)$ . Let us show that if $\varepsilon \le \varepsilon _0$ and $k\in \{0,\ldots , d-1\}$ then for all $(x,y)\in D_k(\varepsilon ,\theta ,\delta _0)$ we have

(6) $$ \begin{align} |x_1|\le |x|(1-\eta|x^my^n|^d)\le |x|. \end{align} $$

Consider $(x,y)\in D_k(\varepsilon ,\theta ,\delta _0)$ with $\varepsilon \le \varepsilon _0$ . We have that

$$ \begin{align*}|\zeta|<|x^my^n|^d/\delta_0<\varepsilon/\delta_0< 1\end{align*} $$

and

$$ \begin{align*}\lvert\arg\zeta\rvert\le\lvert\arg (-a+\sigma)\rvert+\lvert\arg(x^my^n)^d\rvert<(\pi/2 -2\theta_0 ) +\theta< \pi/2-\theta_0.\end{align*} $$

Thus $|\zeta |<\varepsilon /\delta _0\le \cos (\pi /2 -\theta _0)< \cos (\arg \zeta )$ and so $|\zeta |^2<\operatorname {\mathrm {Re}}\zeta $ . Then

$$ \begin{align*}|1-\zeta|^2=1-2\operatorname{\mathrm{Re}}\zeta+|\zeta|^2<1-\operatorname{\mathrm{Re}}\zeta<1-\operatorname{\mathrm{Re}}\zeta +\tfrac 14 (\operatorname{\mathrm{Re}}\zeta)^2=(1-\tfrac 12 \operatorname{\mathrm{Re}}\zeta)^2\end{align*} $$

and therefore $|1-\zeta |<|1-\tfrac 12 \operatorname {\mathrm {Re}}\zeta |=1-\frac 12 \operatorname {\mathrm {Re}}\zeta $ , so

$$ \begin{align*}|1-\zeta|<1-\tfrac 12 \operatorname{\mathrm{Re}} \zeta= 1-\tfrac 12\cos (\arg\zeta) |\zeta|<1-\tfrac 12\cos (\pi/2 -\theta_0)|\zeta|\end{align*} $$

and hence $|1-\zeta |<1-\frac {1} {2 }\delta _0\cos (\pi /2 -\theta _0) |x^my^n|^d,$ which proves (6). Proceeding in the same way, up to reducing $\delta _0$ if necessary, we get that for all $(x,y)\in D_k(\varepsilon ,\theta ,\delta _0)$ with $\varepsilon \le \varepsilon _0$ ,

(7) $$ \begin{align} |y_1|\le |y|(1-\eta|x^my^n|^d)\le |y|. \end{align} $$

From the expression for F, if $x,y\in \mathbb C^*$ are small enough, we have that

$$ \begin{align*} \dfrac{|y_1|}{|x_1^my_1^n|^\gamma}&=\dfrac{|y||1+b(x^my^n)^d(1+O(x,y))|}{|x^my^n|^\gamma|1-(\gamma/d) (x^my^n)^d(1+O(x,y))|}\\ &=\dfrac{|y|}{|x^my^n|^\gamma}\bigg|1+(x^my^n)^d\bigg[b+\frac{\gamma}{d}+O(x,y)\bigg]\bigg|, \end{align*} $$

so we can write

$$ \begin{align*} \dfrac{|y_1|}{|x_1^my_1^n|^\gamma} =\dfrac{|y|}{|x^my^n|^\gamma}|1-\zeta_1|, \end{align*} $$

where $\zeta _1=(-b-\gamma /d+\tau )(x^my^n)^d$ with $\tau =O(x,y)$ . By the choice of $\theta _0$ , reducing $\delta _0$ if necessary, we get that for all $(x,y)$ with $|x|<\delta _0$ and $|y|<\delta _0$ we have

$$ \begin{align*} \delta_0<|-b-\gamma/d+\tau|<1/\delta_0,\quad \lvert\arg (-b-\gamma/d+\tau)\rvert+2\theta_0<\pi/2. \end{align*} $$

Let us show that if $\varepsilon \le \varepsilon _0$ and $k\in \{0,\ldots , d-1\}$ , then for all $(x,y)\in D_k(\varepsilon ,\theta ,\delta _0)$ we have

(8) $$ \begin{align} \dfrac{|y_1|}{|x_1^my_1^n|^\gamma}\le \dfrac{|y|}{|x^my^n|^\gamma}(1-\eta|x^my^n|^d)\le \dfrac{|y|}{|x^my^n|^\gamma}. \end{align} $$

Consider $(x,y)\in D_k(\varepsilon ,\theta ,\delta _0)$ with $\varepsilon \le \varepsilon _0$ . We have that

$$ \begin{align*}|\zeta_1|<|x^my^n|^d/\delta_0<\varepsilon/\delta_0\le 1 \end{align*} $$

and

$$ \begin{align*}\lvert\arg\zeta_1\rvert\le\lvert\arg (-b-\gamma/d+\tau)\rvert+\lvert\arg(x^my^n)^d\rvert<(\pi/2-2\theta_0 ) +\theta< \pi/2-\theta_0,\end{align*} $$

so proceeding exactly as above, we obtain that $|1-\zeta _1|<1-\eta |x^my^n|^d$ which proves (8). If $n\ge 1$ , analogously we get, up to reducing $\delta _0$ if necessary, that for $(x,y)\in D_k(\varepsilon ,\theta ,\delta _0)$ with $\varepsilon \le \varepsilon _0$ we have

(9) $$ \begin{align} \dfrac{|x_1|}{|x_1^my_1^n|^\gamma}\le \dfrac{|x|}{|x^my^n|^\gamma}(1-\eta|x^my^n|^d)\le \dfrac{|x|}{|x^my^n|^\gamma}. \end{align} $$

Notice that, because of equation (4), F satisfies the hypotheses of Proposition 2.2. Set

$$ \begin{align*}\delta_\theta=\min\{1, \delta_0,\delta_F(\theta)\} \quad \text{and}\quad \varepsilon_\theta=\min\{1, \varepsilon_0,\varepsilon_F(\theta), \delta_0^{d/\gamma}, \delta_F(\theta)^{d/\gamma}\},\end{align*} $$

where $\varepsilon _F(\theta )$ and $\delta _F(\theta )$ are the constants given by Proposition 2.2. Now, consider ${\varepsilon \le \varepsilon _\theta }$ , $\delta \le \delta _\theta $ and $k\in \{0,\ldots , d-1\}$ and let us show that $D_k(\varepsilon , \theta ,\delta )$ and $U_k(\varepsilon ,\theta )$ are invariant by F and that estimates (5) hold. First, take $(x,y)\in D_k(\varepsilon ,\theta ,\delta )$ . Then, it follows from (6) and (7) that $|x_1|\le |x|<\delta $ and $|y_1|\le |y|<\delta $ . Moreover, since $\varepsilon \le \varepsilon _F(\theta )$ , $\delta \le \delta _F(\theta )$ and $x^my^n\in S_k(d,\varepsilon ,\theta )$ , we have by assertion 1 of Proposition 2.2 that $x^m_1y_1^n\in S_k(d,\varepsilon , \theta )$ . Therefore $(x_1,y_1)\in D_k(\varepsilon ,\theta ,\delta )$ and so $D_k(\varepsilon ,\theta ,\delta )$ is invariant by F. Now, take $(x,y)\in U_k(\varepsilon ,\theta )$ . Then $|y|<|x^my^n|^\gamma <\varepsilon ^{\gamma /d}\le \delta _0$ and in the same way $|x|<\delta _0$ , so $(x,y)\in D_k(\varepsilon ,\theta ,\delta _0)$ . It follows from (8) and (9) that $|y_1|<|x_1^my_1^n|^\gamma $ and analogously $|x_1|<|x_1^my_1^n|^\gamma $ . Moreover, since $x^my^n\in S_k(d,\varepsilon ,\theta )$ , $\varepsilon \le \varepsilon _F(\theta )$ , $|y|<\varepsilon ^{\gamma /d}<\delta _F(\theta )$ and $|x|<\varepsilon ^{\gamma /d}<\delta _F(\theta )$ , we have by assertion 1 of Proposition 2.2 that $x_1^my_1^n\in S_k(d,\varepsilon ,\theta )$ , which proves that $U_k(\varepsilon ,\theta )$ is invariant by F. Moreover, if $(x,y)\in U_k(\varepsilon , \delta )\cup D_k(\varepsilon ,\delta ,\theta )$ we have estimates (5) directly from assertion 1 of Proposition 2.2.

Let us prove now that $F^j\to 0$ uniformly on $U_k(\varepsilon ,\theta )$ . Take $(x,y)\in U_k(\varepsilon ,\theta )$ . Since $(x_j,y_j)\in U_k(\varepsilon ,\theta )$ for all $j\ge 0$ , by (5) we have

$$ \begin{align*} |x_j^my_j^n|^d\le \frac 2{\cos\theta}\frac{|x^my^n|^d}{1+j|x^my^n|^d}\le \frac 2{\cos\theta} \frac1j \end{align*} $$

for all $j\ge 0$ , which shows that $x_j^my_j^n\to 0$ uniformly on $U_k(\varepsilon ,\theta )$ . Then, since ${|y_j|<|x_j^my_j^n|^\gamma }$ and $|x_j|<|x_j^my_j^n|^\gamma $ , we have that $x_j\to 0$ and $y_j\to 0$ uniformly on $U_k(\varepsilon ,\theta )$ .

Consider now an orbit $(x_j,y_j)$ converging to 0 and let us show that it eventually lies in $U_k=U_k(\varepsilon ,\theta )$ for some k. Let $\tilde \delta _F(\theta )$ be the constant given by Proposition 2.2. Since $(x_j,y_j)\to 0$ , there exists $j_0\ge 0$ such that $|x_j|<\tilde \delta _F(\theta )$ and $|y_j|<\tilde \delta _F(\theta )$ for all $j\ge j_0$ . By assertion 3 of Proposition 2.2, up to increasing $j_0$ , we have that $x_j^my_j^n\in S_k(d,\varepsilon ,\theta )$ for all $j\ge j_0$ and for some $k\in \{0,\ldots , d-1\}$ so, increasing $j_0$ again if necessary, we get that $(x_j,y_j)\in D_k(\varepsilon ,\theta ,\delta _0)$ for all $j\ge j_0$ . Then, by iterated applications of (8),

$$ \begin{align*} \dfrac{|y_j|}{|x_j^my_j^n|^\gamma}&\le \dfrac{|y_{j_0}|}{|x_{j_0}^my_{j_0}^n|^\gamma}\prod\limits_{l=j_0}^{j-1} (1-\eta|x_l^my_l^n|^d) \end{align*} $$

for all $j\ge j_0$ . The estimate

$$ \begin{align*}|x_j^my_j^n|^d\ge \frac 12 \frac{|x^my^n|^d}{1+j|x^my^n|^d}\end{align*} $$

from (5) implies that $\prod _{l=j_0}^{j-1}(1-\eta |x_l^my_l^n|^d)$ tends to 0 as $j\to +\infty $ and so does $|y_j|/|x_j^my_j^n|^\gamma $ . Hence $|y_j|\le |x_j^my_j^n|^\gamma $ for j big enough and, if $n\ge 1$ , analogously $|x_j|\le |x_j^my_j^n|^\gamma $ for j big enough, which proves that eventually $(x_j,y_j)\in U_k$ . Then, to prove that $D_k(\varepsilon ,\theta ,\delta )$ is contained in the basin of attraction of $U_k(\varepsilon , \theta )$ it suffices to show that the orbit of any point in $D_k(\varepsilon ,\theta ,\delta )$ converges to 0. Take $(x,y)\in D_k(\varepsilon ,\theta ,\delta )$ . Since $(x_j,y_j)\in D_k(\varepsilon ,\theta ,\delta )$ for all $j\ge 0$ , we can apply (6) and (7) for each $(x_j,y_j)$ and we have

$$ \begin{align*}|x_j|\le |x_0|\prod\limits_{l=0}^{j-1}(1-\eta|x_l^my_l^n|^d),\quad |y_j|\le |y_0|\prod\limits_{l=0}^{j-1}(1-\eta|x_l^my_l^n|^d).\end{align*} $$

We have as before, by estimates (5), that the product above tends to zero and so do $x_j$ and $y_j$ . This concludes the proof of assertion 1.

Let us prove assertion 2. From the expression for F we have in a neighborhood of $0\in \mathbb C^2$ that $|x_1|\le |x|(1+\nu |x^my^n|^d)$ and $|y_1|\le |y|(1+\nu |x^my^n|^d)$ for some $\nu>0$ . Let $\tilde \delta =\min \{1,\tilde \mu \}$ , where $\tilde \mu $ is given by assertion 2 of Proposition 2.2 for $\mu =\delta $ . If $x^my^n\in \widetilde D_k(\varepsilon ,\theta ,\tilde \delta )$ , by Proposition 2.2 there exists $j_0\ge 0$ such that $|x_j|<\delta $ , $|y_j|<\delta $ and $ x^m_{j}y^n_{j}\in \widetilde S_k(d,\varepsilon ,\theta )$ for all $j\le j_0$ and $x^m_{j_0}y^n_{j_0}\in S_k(d,\varepsilon ,\theta )$ . In particular, $(x_{j_0},y_{j_0})\in D_k(\varepsilon ,\theta ,\delta )$ so by assertion 1 $(x_j,y_j)\in D_k(\varepsilon ,\theta , \delta )$ for all $j\ge j_0$ so it eventually lies in $U_k(\varepsilon ,\theta )$ . Moreover, since $|x_j|<\delta $ , $|y_j|<\delta $ and $ x^m_{j}y^n_{j}\in \widetilde S_k(d,\varepsilon ,\theta )$ for all $j\ge 0$ , we have that $F^j(\widetilde D_k(\varepsilon ,\theta ,\tilde \delta ))\subset \widetilde D_k(\varepsilon ,\theta ,\delta )$ for all $ j\ge 0.$ This concludes the proof of the proposition.

Proof. Take $\tilde \varepsilon _\theta \le \varepsilon _\theta $ . Note that the domains $U_k=U_k(\tilde \varepsilon _\theta ,\theta )$ for $k\in \{0,\ldots , d-1\}$ satisfy the conclusion of Proposition 3.3.

We define $\psi _k$ as

$$ \begin{align*}\psi_k(x,y)=\lim\limits_{j\to\infty}g(x_j,y_j), \ \ (x,y)\in U_k,\end{align*} $$

where $(x_j,y_j)=F^j(x,y)$ . It is clear that this function, if well defined, will be invariant by F. Let us show that it is well defined and holomorphic. Using the expression for F and equation (4), we have that

$$ \begin{align*}\dfrac{x_1^py_1^q}{x^py^q}=1+(x^my^n)^d[ap+bq+O(x,y)]\end{align*} $$

and

for all $(x,y)$ , so

(11) $$ \begin{align} \frac{g(x_1,y_1)}{g(x,y)}=1+\ell(x,y), \quad \text{with } \ell(x,y)=(x^my^n)^dO(x,y). \end{align} $$

Since $(x_j,y_j)\in U_k$ for all j, we have that $|y_j| \le |x_j^my_j^n|^\gamma $ and $|x_j|\le |x_j^my_j^n|^{\gamma }$ so

$$ \begin{align*}|\ell(x_j,y_j)|\le K|x_j^my_j^n|^{d+\gamma}\end{align*} $$

for some $K>0$ . Therefore, by estimates (5), the product $\prod _{j\ge 0}({g(x_{j+1},y_{j+1})}/g(x_j, y_j))$ converges uniformly for $(x,y)\in {U_k}$ and defines a holomorphic function $u\in \mathcal {O}({U_k})$ . Then $g(x_j,y_j)\to u(x,y)g(x,y)$ uniformly on $U_k$ , so $\psi _k$ is well defined and holomorphic in ${U_k}$ , and we have $\psi _k=ug$ . Note that the function u is arbitrarily close to $1$ if we suppose $|x^my^n|$ to be small enough: if $n\in \mathbb {N}$ is large enough the finite product $\prod _{0\le j\le n}({g(x_{j+1},y_{j+1})}/{g(x_j,y_j)})$ is arbitrarily uniformly close to u and we have from (11) that this finite product is arbitrarily uniformly close to 1 if $|x^my^n|$ is small enough. Therefore, since $|x^my^n|<\tilde \varepsilon _\theta ^{1/d}$ in $U_k$ , we get that $|u(x,y)-1|<1/2$ for all $(x,y)\in U_k$ , provided $\tilde \varepsilon _\theta $ is small enough.

Proof. Consider $\theta _1\in (0,\theta _0)$ with $\cos \theta _1>2/3$ and let $\varepsilon _1=\tilde \varepsilon _{\theta _1}$ be the constant given by Proposition 4.1. From equation (4) we have that

$$ \begin{align*} \frac{1}{(x_1^my_1^n)^d}=\frac{1}{(x^my^n)^d}+1+ O(x,y). \end{align*} $$

Thus, since $|x|\le |x^my^n|^\gamma $ and $|y|\le |x^my^n|^{\gamma }$ for all $(x,y)\in U_k(\varepsilon _1,\theta _1)$ , there exists ${K>0}$ such that

(12) $$ \begin{align} \bigg|\frac{1}{(x_1^my_1^n)^d}-\frac{1}{(x^my^n)^d}-1\bigg|<K|x^my^n|^{\gamma}<K\varepsilon_1^{\gamma/d} \end{align} $$

for all $(x,y)\in U_k(\varepsilon _1,\theta _1)$ and all $k\in \{0,\ldots , d-1\}$ . Then, up to reducing $\varepsilon _1$ if necessary, we can assume that

(13) $$ \begin{align} \bigg|\frac{1}{(x_1^my_1^n)^d}-\frac{1}{(x^my^n)^d}-1\bigg|<\frac 16 \end{align} $$

for all $(x,y)\in U_k(\varepsilon _1,\theta _1)$ and all $k\in \{1,\ldots , d-1\}$ . We now fix $\theta <\theta _1$ and $\varepsilon <\varepsilon _1$ and we denote $U_k=U_k(\varepsilon ,\theta )$ .

Let us prove assertion 1; without loss of generality we assume that $k=0$ . Let g be the function defined by (10). Using the fact that $qm-pn=1$ and $adm+bdn=-1$ , a straightforward computation shows that the map $\varphi :U_0\to \mathbb C^2$ given by

$$ \begin{align*}\varphi(x,y)=\bigg(\frac{1}{(x^my^n)^d},g(x,y)\bigg)\end{align*} $$

is injective and its inverse is given by

$$ \begin{align*}\varphi^{-1}(z,w)=(z^aw^{-n},z^bw^{m}).\end{align*} $$

Consider a point $(z_0,w_0)\in V$ , and set

$$ \begin{align*}A_{z_0}=\{(x,y)\in\mathbb C^2: x^my^n=z_0^{-1/d}, \ |\iota x|<|z_0|^{-\gamma/d}, \ |y|<|z_0|^{-\gamma/d}\}.\end{align*} $$

Notice that $A_{z_0}\subset U_0$ and $\overline {A_{z_0}}$ is a Riemann surface with boundary. If we set $x=z_0^aw_0^{-n}$ and $y=z_0^bw_0^{m}$ , then $x^my^n=z_0^{-1/d}$ and $g(x,y)=w_0$ . Moreover, $|y|=|z_0^bw_0^m|<r^m|z_0^b||z_0|^{-\mathrm{Re}\ b-\gamma/d}\le r^me^{|b|\theta }|z_0|^{-\gamma /d}$ and, if $n\ge 1$ , analogously $|x|=|z_0^aw_0^{-n}|<r^ne^{|a|\theta }|z_0|^{-\gamma /d}$ , so $(x,y)\in A_{z_0}$ if r is small enough. Therefore, since $\varphi $ is injective, $g|_{A_{z_0}}$ assumes the value $w_0$ once. If we show that

(14) $$ \begin{align} |\psi_0(x,y)-g(x,y)|<|g(x,y)-w_0|\quad \textrm{whenever} \ (x,y)\in\partial A_{z_0} \end{align} $$

then, by Rouché’s theorem, the function $\psi _0|_{A_{z_0}}$ will also assume the value $w_0$ exactly once, showing that $V\subset \phi _0(U_0)$ and that $\phi _0$ is injective in $\phi _0^{-1}(V)$ . Let us prove that inequality (14) holds. If $n\ge 1$ , the boundary $\partial A_{z_0}$ of $A_{z_0}$ is composed by two connected components,

$$ \begin{align*} \partial_1 A_{z_0}&=\{(x,y)\in\mathbb C^2:x^my^n=z_0^{-1/d}, \ |x|=|z_0|^{-\gamma/d}\},\\ \partial_2 A_{z_0}&=\{(x,y)\in\mathbb C^2:x^my^n=z_0^{-1/d}, \ |y|=|z_0|^{-\gamma/d}\}, \end{align*} $$

whereas if $n=0$ we have $\partial A_{z_0}=\partial _2 A_{z_0}$ . Consider a point $(x,y)\in \partial A_{z_0}\subset U_0$ . Since $x^my^n=z_0^{-1/d}$ , it follows from the computation of $\varphi ^{-1}$ that

$$ \begin{align*}g(x,y)^n= x^{-1}z_0^a\quad\text{and}\quad g(x,y)^m=yz_0^{-b}.\end{align*} $$

We suppose first that $(x,y)\in \partial _2 A_{z_0}$ . Then

$$ \begin{align*}|g(x,y)|= |y|^{{1}/{m}}|z_0^{-b/m}| = |z_0|^{-\gamma/(dm)}|z_0^{-b/m}| \ge e^{-{|b|\theta}/m}|z_0|^{-\operatorname{\mathrm{Re}} b/m-\gamma/(dm)} \end{align*} $$

so $|g(x,y)|>e^{-{|b|\theta }/m}r^{-1} |w_0|$ . Then

$$ \begin{align*}|g(x,y)-w_0|\ge |g(x,y)|-|w_0|>(1-e^{{|b|\theta}/m}r)|g(x,y)|\ge \tfrac{1}{2}|g(x,y)|\end{align*} $$

if r is small enough. This relation, together with the fact that

$$ \begin{align*}|\psi_0(x,y)-g(x,y)|<\tfrac12|g(x,y)|\end{align*} $$

for all $(x,y)\in U_0$ , as was shown in Proposition 4.1, implies (14). Analogously, if $(x,y)\in \partial _1 A_{z_0}$ (so $n\ge 1$ ) then

$$ \begin{align*}|g(x,y)|=|x|^{-{1}/{n}}|z_0^{a/n}|= |z_0|^{\gamma/(dn)}|z_0^{{a}/{n}}| \le e^{{|a|\theta}/{n}}|z_0|^{{\operatorname{\mathrm{Re}} a}/{n}+{\gamma}/{(dn)}}\end{align*} $$

so $|g(x,y)|< e^{{|a|\theta }/{n}}r|w_0|$ and hence

$$ \begin{align*}|g(x,y)-w_0|\ge|w_0|- |g(x,y)|>(e^{-{|a|\theta}/{n}}r^{-1}-1)|g(x,y)| \ge \tfrac{1}{2}|g(x,y)|\end{align*} $$

if r is sufficiently small, which again implies (14) and assertion 1 is proved.

Now take $(x,y)\in U_k$ . If we set $(z_j,w_j)=\phi _k(x_j,y_j)$ , then clearly $|z_j|>\varepsilon ^{-1}$ and $\lvert \arg z_j\rvert <\theta $ ; moreover, since $\psi _k$ is invariant by F we have that $w_j$ is constant for all j while $z_j\to +\infty $ because of (5), so for j large enough we get $|w_j|<r|z_j|^{{-\operatorname {\mathrm {Re}} b}/{m}-{\gamma }/{(dm)}}$ and if $n\ge 1$ , since $w_j\neq 0$ , $|w_j|>r^{-1}|z_j|^{{\operatorname {\mathrm {Re}} a}/{n}+\gamma /{(dn)}}$ , so $(z_j,w_j)\in V$ and assertion 2 is proved.

Let us prove assertion 3. For $r_1\in (0,1)$ , consider the set $V_1=V(\varepsilon _1,\theta _1,r_1)$ . Proceeding exactly as in the proof of assertion 1, we can choose $r_1\in (0,1)$ such that $V_1 \subset \phi _k(U_k(\varepsilon _1,\theta _1))$ and $\phi _k:\phi _k^{-1}(V_1)\to V_1$ is a biholomorphism for every k. Then $\phi _k^{-1}$ is well defined on $V_1$ and takes values in $U_k(\varepsilon _1,\theta _1)$ , so $\tilde {F}= \phi _k\circ F\circ \phi _k^{-1}$ is well defined on $V_1$ . Since $\psi _k$ is invariant by F, we can express $\tilde {F}(z,w)=(f(z,w),w)$ . Then, if we write $\phi _k^{-1}(z,w)=(x,y)\in U_k(\varepsilon _1,\theta _1)$ and $F(x,y)=(x_1,y_1)$ , we have from equation (4) that

$$ \begin{align*}f(z,w)=\frac{1}{(x_1^my_1^n)^d}=\frac{1}{(x^my^n)^d}+1+\tilde{h}(x,y),\end{align*} $$

where $\tilde {h}(x,y)=O(x,y)$ . That is,

$$ \begin{align*}f(z,w)=z+1+{h}(z,w),\end{align*} $$

where by (12) and (13)

$$ \begin{align*}|h(z,w)|<K|z|^{-\gamma/d} \quad \text{and} \quad |h(z,w)|<\tfrac 16\end{align*} $$

for all $(z,w)\in V_1$ . Thus, since $\cos \theta _1>2/3$ , we have

$$ \begin{align*} |f(z,w)|^2&=|z+1+h(z,w)|^2= |z|^2+2\operatorname{\mathrm{Re}} z +2\operatorname{\mathrm{Re}}(z\overline{h}(z,w))+|1+h(z,w)|^2 \\ &\ge |z|^2+2\cos\theta_1|z|-2|z||h(z,w)|+|1+h(z,w)|^2\\ &> |z|^2+4|z|/3-|z|/3+(5/6)^2>(|z|+1/2)^2, \end{align*} $$

so $|f(z,w)|> |z| +1/2$ for all $(z,w)\in V_1$ .

We now fix r satisfying assertion 1 and such that $r<r_1$ . Consider a point $(z,w)\in V$ . Since $(f(z,w),w)\in \phi _k(U_k)$ , it is clear that $|f(z,w)|>\varepsilon ^{-1}$ and $\lvert \arg (f(z,w))\rvert <\theta $ and, since $|w|<r|z|^{{-\operatorname {\mathrm {Re}} b}/{m}-{\gamma }/{(dm)}}$ and $|f(z,w)|>|z|$ , we also have that

$$ \begin{align*}|w|<r|f(z,w)|^{{-\operatorname{\mathrm{Re}} b}/{m}-{\gamma}/{(dm)}}.\end{align*} $$

Analogously, if $n\ge 1$ , since $|w|>r^{-1}|z|^{{\operatorname {\mathrm {Re}} a}/{n}+\gamma /{(dn)}}$ and $|f(z,w)|>|z|$ , we have $|w|>r^{-1}|f(z,w)|^{{\operatorname {\mathrm {Re}} a}/{n}+\gamma /{(dn)}}$ , so $\tilde F(z,w)\in V$ .

To prove the bound for ${\partial h}/{\partial z}$ let us first show that there exists $\rho>0$ such that if $(z_0,w_0)\in V$ and $|z-z_0|<\rho |z_0|$ then $(z,w_0)\in V_1$ . Consider $(z_0,w_0)\in V$ and assume that $|z-z_0|<\rho |z_0|$ . Then

$$ \begin{align*}|z|>(1-\rho )|z_0|>(1-\rho)\varepsilon^{-1},\end{align*} $$

so $|z|>\varepsilon _1^{-1}$ for $\rho $ sufficiently small. Since $|z/z_0-1|<\rho $ , we have $\lvert \arg (z/z_0)\rvert <\arcsin \rho $ , so

$$ \begin{align*}\lvert\arg z\rvert\le \lvert\arg z_0 \rvert+\arcsin\rho<\theta +\arcsin\rho,\end{align*} $$

hence $\lvert \arg z\rvert <\theta _1$ if $\rho $ is small enough. Since $|z_0|<(1-\rho )^{-1}|z|$ , it follows that

$$ \begin{align*}|w_0|<r|z_0|^{{-\operatorname{\mathrm{Re}} b}/{m}-{\gamma}/{(dm)}}<r[(1-\rho)^{-1}|z|]^{{-\operatorname{\mathrm{Re}} b}/{m}-{\gamma}/{(dm)}},\end{align*} $$

so $|w_0|<r_1|z|^{{-\operatorname {\mathrm {Re}} b}/{m}-{\gamma }/{(dm)}}$ if $\rho $ is small enough and, if $n\ge 1$ ,

$$ \begin{align*}|w_0|>r^{-1}|z_0|^{{\operatorname{\mathrm{Re}} a}/{n}+\gamma/{(dn)}}>r^{-1}[(1-\rho)^{-1}|z|]^{{\operatorname{\mathrm{Re}} a}/{n}+\gamma/{(dn)}},\end{align*} $$

so $|w_0|>r_1^{-1}|z|^{{\operatorname {\mathrm {Re}} a}/{n}+\gamma /{(dn)}}$ if $\rho $ is small enough. Hence, $(z,w_0)\in V_1$ if $\rho $ is small enough. Now, take a point $(z_0,w_0)\in V$ . As we have seen, if $D\subset \mathbb {C}$ is the disk of radius $\rho |z_0|$ centered at $z_0$ , then $D\times \{w_0\}$ is contained in $V_1$ , so the function

$$ \begin{align*}h_{w_0}\colon z\in D \mapsto h(z,w_0)\end{align*} $$

is well defined and

$$ \begin{align*}|h_{w_0}(z)|< K|z|^{-\gamma/d}< K(1-\rho)^{-\gamma/d}|z_0|^{-\gamma/d}.\end{align*} $$

Thus, it follows from Cauchy’s inequality that

$$ \begin{align*} \bigg|\frac{\partial h}{\partial z}(z_0,w_0)\bigg|=|(h_{w_0})'(z_0)|\le K(1-\rho)^{-\gamma/d}|z_0|^{-\gamma/d}(\rho|z_0|)^{-1}=K'|z_0|^{-1-\gamma/d}, \end{align*} $$

which finishes the proof of assertion 3.

Proof. By Proposition 5.2, we have that $\phi _k:\phi _k^{-1}(V)\to V$ is a biholomorphism conjugating F with $\widetilde F=\phi _k\circ F\circ \phi _k^{-1}\colon V\to V$ . Thus, it is enough to find a biholomorphism $\Phi \colon V\to W\subset \mathbb C^2$ , with $W\subset \mathbb C\times \mathbb C^*$ if $n\ge 1$ , conjugating $\tilde {F}$ with the map $(z,w)\mapsto (z+1,w)$ . Since $\tilde {F}\colon V\to V$ is written $\tilde {F}(z,w)=(f(z,w),w)$ , each function $f_{w}\colon z\mapsto f(z,w)$ maps the domain $V_w=\{z\in \mathbb C:(z,w)\in V\}$ into itself. Thus, we start considering $w\in \mathbb C$ fixed ( $w\in \mathbb {C}^*$ if $n\ge 1$ ) and we will find, following the ideas in [Reference Milnor13, Lemma 10.10], a map $\beta _w:V_w\to \mathbb C$ conjugating $f_w$ with $z\mapsto z+1$ . We will also show, arguing as in [Reference Milnor13, Lemma 10.11], that

(16) $$ \begin{align}\bigcup_{j\in\mathbb{N}} (\beta_w(V_w)-j)=\mathbb{C}. \end{align} $$

Finally, from the maps $\beta _w$ we will construct a global map $\beta :V\to \mathbb C$ such that ${\beta \circ \tilde F(z,w)=\beta (z,w)+1}$ , so the function $\Phi :V\to \mathbb C^2$ given by $\Phi (z,w)=(\beta (z,w),w)$ is a Fatou coordinate for $\tilde F$ .

From the definition of V we have

$$ \begin{align*}V_w=\{z\in\mathbb{C}\colon |z|>R_w,\; \lvert\arg(z)\rvert<\theta \},\end{align*} $$

where

$$ \begin{align*}R_w=\max \{\varepsilon^{-1},\; (r^{-1}|w|)^{-dm/(d\operatorname{\mathrm{Re}} b+\gamma)}, \; \iota(r|w|)^{dn/(d\operatorname{\mathrm{Re}} a+\gamma)}\}\end{align*} $$

(recall that $\iota =0$ if $n=0$ and $ \iota =1$ if $n\ge 1$ ).

Take a base point $p\in V_w$ . The map $\beta _w$ conjugating $f_w$ with $z\mapsto z+1$ will be constructed as the limit of the functions

$$ \begin{align*}\beta_j(z)= f^j_w(z)-f^{j}_w(p),\quad j\in\mathbb{N}.\end{align*} $$

In order to simplify the proof of the convergence of these functions we assume p to be large enough so that for all $z\in V_w$ the Euclidean segment $[z,p]$ is contained in $V_w$ , which is possible because $\theta <\pi /2$ . Since $|f_w(p)|\ge |p|+1/2$ by Proposition 5.2, the sequence $|f_w^j(p)|$ is increasing, hence the property above also holds for $f_w^j(p)$ . In particular we have, for all $z\in V_w$ and all $j\in \mathbb {N}$ ,

$$ \begin{align*}[f_w^j(z),f_w^j(p)]\subset V_w.\end{align*} $$

Since $f_w(z)=z+1+h(z,w)$ and $|{\partial h}/{\partial z}(z,w)|<K'|z|^{-1-\gamma /d}$ by Proposition 5.2, it follows from the mean value inequality that, if $[z_1,z_2]\subset V_w$ , then

$$ \begin{align*} \bigg|\frac{f_w(z_1)-f_w(z_2)}{z_1-z_2}-1\bigg|= \bigg|\frac{h(z_1,w)-h(z_2,w)}{z_1-z_2}\bigg|\le \max\limits_{z\in[z_1,z_2]}\frac{K'}{|z|^{1+\gamma/d}}. \end{align*} $$

Since the angle between $z_1\mathbb {R}^+$ and $z_2\mathbb {R}^+$ is bounded by $2\theta <\pi $ , there is a constant $\tau>0$ depending only on $\theta $ such that $\min \{|z|:z\in [z_1,z_2]\}\ge \tau \min \{|z_1|,|z_2|\}$ , so

$$ \begin{align*}\bigg|\frac{f_w(z_1)-f_w(z_2)}{z_1-z_2}-1\bigg|\le \frac{K'}{(\tau\min\{|z_1|,|z_2|\} )^{1+\gamma/d}}.\end{align*} $$

In particular, setting $z_1=f_w^j(z)$ and $z_2=f_w^j(p)$ , we obtain

$$ \begin{align*}\bigg|\frac{\beta_{j+1}(z)}{\beta_j(z)}-1\bigg|=\bigg|\frac{f_w^{j+1}(z)-f_w^{j+1}(p)}{f_w^{j}(z)-f_w^{j}(p)}-1\bigg| \le \frac{K'}{(\tau\min\{|f_w^{j}(z)|,|f_w^{j}(p)|\})^{1+\gamma/d}}\end{align*} $$

for all $z\in V_w$ and $j\in \mathbb {N}$ . Therefore, since $|f_w^{j}(z)|\ge j/2$ and $|f_w^{j}(p)|\ge j/2$ , we obtain

$$ \begin{align*}\bigg|\frac{\beta_{j+1}(z)}{\beta_j(z)}-1\bigg|\le K'(2\tau^{-1})^{1+\gamma/d}j^{-1-\gamma/d}\end{align*} $$

for all $z\in V_w$ and $j\in \mathbb {N}$ . This shows that the product $\prod ({\beta _{j+1}(z)}/{\beta _j(z)})$ is uniformly convergent in $V_w$ and therefore $\beta _j$ converges uniformly to a function $\beta _w\in \mathcal {O}(V_w)$ . Let us show that $\beta _w$ is one-to-one and conjugates $f_w$ with the map $z\mapsto z+1$ . Since $f_w(z)=z+1+h(z,w)$ and $|h(z,w)|<K|z|^{-\gamma /d}$ by Proposition 5.2, we have

$$ \begin{align*}| f_w^{j+1}(p)-f_w^j(p) -1|= | h(f_w^j(p),w)|\le \frac{K}{|f_w^j(p)|^{\gamma/d}}\to 0 \quad \textrm{as } j\to\infty.\end{align*} $$

Thus, since $\beta _j(f_w(z))=\beta _{j+1}(z)+f_w^{j+1}(p)-f_w^j(p)$ , we obtain, taking $j\to \infty $ ,

$$ \begin{align*}\beta_w(f_w(z))=\beta_w(z)+1,\quad z\in V_w.\end{align*} $$

Finally, since $\beta _j$ is injective for all j and $\beta _w$ is not constant, we conclude that $\beta _w$ is injective.

Now, as in [Reference Milnor13, Lemma 10.11], we prove that $\beta _w$ satisfies (16). We show first that $\lim \nolimits _{z\to \infty }({\beta _w(z)}/{z})=1$ . Since $\beta _j$ tends uniformly to $\beta _w$ , for some $l\in \mathbb {N}$ we have that $|\beta _w-\beta _l|$ is bounded, whence

$$ \begin{align*}|\beta_w-f_w^l|\le |\beta_w-\beta_l|+|\beta_l-f_w^l|\end{align*} $$

is bounded. Then, since $f_w(z)=z+1+h(z,w)$ and $|h(z,w)|<K|z|^{-\gamma /d}$ ,

$$ \begin{align*}\lim\limits_{z\to \infty}\frac{\beta_w(z)}{z}=\lim\limits_{z\to \infty}\frac{f_w^l(z)}{z} =1.\end{align*} $$

Consider $\zeta \in \mathbb {C}$ . In order to prove (16) we will show that for $j\in \mathbb {N}$ large enough the point $\zeta _j=\zeta +j$ belongs to $\beta _w(V_w)$ . Since $V_w$ is essentially a sector of opening $2\theta $ , if we take a positive number $\rho <\sin \theta $ , it is not difficult to see that, for j large enough, the closed disk $D_j$ of radius $r_j = \rho | \zeta _j|$ centered at $\zeta _j$ is contained in $V_w$ . By Rouché’s theorem, if

$$ \begin{align*}|\beta_w(z)-z|<r_j\end{align*} $$

for all $z\in \partial D_j$ , then $\zeta _j\in \beta _w(D_j)\subset \beta _w(V_w)$ . Since $\beta _w(z)/z\to 1$ when $z\to \infty $ , for z large enough we have $|\beta _w(z)-z|<\tau |z|$ , where $\tau>0$ is taken such that ${\tau }(1+\rho )<\rho $ . Then, if $z\in \partial D_j$ and j is large enough,

$$ \begin{align*}|\beta_w(z)-z|<\tau|z|\le\tau(|\zeta_j|+r_j)=\tau(1+\rho)|\zeta_j|<\rho|\zeta_j|=r_j\end{align*} $$

and (16) follows.

The function $\beta _w$ that we have constructed depends on the choice of the base point $p\in V_w$ , but its derivative does not, as we can check from the definition of $\beta _j$ :

$$ \begin{align*}(\beta_w)'(z)=\lim\limits_{j\to\infty}(f_w^j)'(z).\end{align*} $$

It is easy to see that the same choice of p also works for any $w'$ in a neighborhood of w and the function $\beta _{w'}$ will depend holomorphically on $w'$ . That is, $\beta _{w'}(z)$ is a holomorphic function of $(z,w')$ . Thus, we can find an open covering $\mathbb {C}= \bigcup _{i\in I} W_i$ if $n=0$ or ${\mathbb {C}^*= \bigcup _{i\in I} W_i}$ if $n\ge 1$ and, for each $i\in I$ , a holomorphic function

$$ \begin{align*}\beta_i(z,w),\quad \text{for } z\in V_w,\; w\in W_i\end{align*} $$

such that, for each $w\in W_i $ , the map $z\in V_w\mapsto \beta _i(z,w)\in \mathbb {C}$ is univalent and satisfies $\beta _i(f(z,w),w)=\beta _i(z,w)+1.$ Moreover, from the observation above, the partial derivative ${\partial \beta _i}/{\partial z}$ does not depend on $i\in I$ , that is,

$$ \begin{align*}\frac{\partial \beta_i}{\partial z}(z,w)=\frac{\partial \beta_j}{\partial z}(z,w)\quad \text{for } z\in V_w,\; w\in W_i\cap W_j,\;i,j\in I.\end{align*} $$

Therefore, if $W_i\cap W_j\neq \emptyset $ , there is a function $g_{ij}\in \mathcal {O}(W_i\cap W_j)$ such that

(17) $$ \begin{align} \beta_j(z,w)-\beta_i(z,w)= g_{ij}(w) \quad \text{for } z\in V_w,\; w\in W_i\cap W_j, \end{align} $$

hence $g_{ij}+g_{jk}+g_{ki}=0$ on $W_i\cap W_j\cap W_k$ , for $i,j,k\in I$ . Then, since the first Cousin problem can be solved in $\mathbb C$ and $\mathbb {C}^*$ , there exist functions $g_i\in \mathcal {O}(W_i)$ , $i\in I$ , such that $g_{ij}=g_i-g_j$ on $W_i\cap W_j$ and it follows from (17) that

$$ \begin{align*}\beta_j(z,w)+g_j(w)=\beta_i(z,w) +g_{i}(w) \quad\text{for } z\in V_w,\; w\in W_i\cap W_j.\end{align*} $$

Therefore we can define a global function $\beta \in \mathcal {O}(V)$ by

$$ \begin{align*}\beta(z,w)=\beta_i(z,w)+g_i(w)\quad\text{for } z\in V_w,\;w\in W_i,\end{align*} $$

and we can see that for each $w\in \mathbb {C}^*$ , the map

$$ \begin{align*}z\in V_w\mapsto \beta(z,w)\in\mathbb{C}\end{align*} $$

is univalent and $\beta (f(z,w),w)=\beta (z,w)+1$ for every $(z,w)\in V$ . Now it is easy to check that the holomorphic function

$$ \begin{align*}\Phi(z,w)= (\beta(z,w),w), \quad (z,w)\in V,\end{align*} $$

is univalent and satisfies $\Phi \circ \tilde {F}(z,w)=\Phi (z,w)+(1,0)$ for every $(z,w)\in V$ . To show that $W=\Phi (V)$ satisfies (15), consider a point $(z_0,w_0)\in \mathbb C^2$ , with $(z_0,w_0)\in \mathbb {C}\times \mathbb {C}^*$ if $n\ge 1$ . If $w_0\in W_i$ , we have

$$ \begin{align*}\beta(z,w_0)=\beta_i(z,w_0)+g_i(w_0)=\beta_{w_0}(z)+g_i(w_0)\end{align*} $$

for all $z\in V_{w_0}.$ By (16) there exist $z\in V_{w_0}$ and $j\in \mathbb {N}$ such that $[z_0-g_i(w_0)]+j=\beta _{w_0}(z)$ , thus

$$ \begin{align*}\Phi(z,w_0)=(\beta(z,w_0),w_0)=(z_0+j,w_0)\end{align*} $$

and therefore $(z_0,w_0)\in W-(j,0)$ .