= Solution
For a nonconstant <rational map>, choose local holomorphic coordinates $u$ at $p$ and $v$ at $R(p)$, both vanishing at their centres. Its local expression has the form
$$
v\circ R\circ u^{-1}(t)=a t^e+O(t^{e+1}),\qquad a\ne0.
$$
The <local degree of a holomorphic map> is $e$. A <critical point of a rational map> is one with $e\geq2$, and its <critical multiplicity> is $e-1$. Both are unchanged by coordinate changes with nonzero first <derivatives>. At finite points where $R$ is finite, multiplicity is the <derivative>'s zero order; at <poles> and infinity it must be computed in the corresponding reciprocal coordinate.
We give an algebraic proof of the total count. Local <derivatives> are holomorphic and not identically zero, so their zeros are isolated; <compactness> makes the sphere critical set finite. Choose a regular target value and make it infinity by a target <Möbius transformation>. There are then $d$ simple <poles>. Independently choose a noncritical source point not mapping to this value and make it source infinity. In these coordinates all <poles> are finite and simple, and infinity is a regular point with finite image. Write the resulting map as $P/Q$, where $P,Q$ are coprime, $\deg Q=d$, and $\deg P\leq d$. Regularity at infinity gives
$$
R(z)=a+\frac b z+O(z^{-2}),\quad b\ne0,\qquad R'(z)=-\frac b{z^2}+O(z^{-3}).
$$
Since $R'=(P'Q-PQ')/Q^2$, its numerator $W=P'Q-PQ'$ therefore has degree exactly $2d-2$. At every <pole> $p$, $W(p)=-P(p)Q'(p)\ne0$, so the roots of $W$ are precisely the finite <rational critical points>, with their <critical multiplicities>. Infinity was chosen regular. The <fundamental theorem of algebra> now proves
$$
\boxed{\sum_{p\in\widehat{\mathbb C}}(e_p(R)-1)=2d-2.}
$$
The coordinate changes preserve the <local degrees of a holomorphic map>, so this is the count for the original map. The printed critical-point count is \b[with multiplicity]: for example $z^d$ has only two distinct <rational critical points>, zero and infinity, each of multiplicity $d-1$.
<Local degrees of a holomorphic map> multiply under composition, since substituting leading powers of orders $e$ and $f$ gives order $ef$. Thus
$$
e_p(P^n)=\prod_{j=0}^{n-1}e_{P^j(p)}(P),\qquad\boxed{\operatorname{Crit}(P^n)=\bigcup_{j=0}^{n-1}P^{-j}(\operatorname{Crit}(P)).}
$$
If $p$ is critical for $P$, the first factor is at least two, so it is critical for every positive iterate. Conversely a product of positive integers can exceed one only if one factor exceeds one, so a <rational critical point> of $P^n$ reaches a <rational critical point> of $P$ in some $j<n$ steps (including $j=0$). At finite points this is also the <derivative> <chain rule> $(P^n)'(z)=\prod_{j=0}^{n-1}P'(P^j(z))$; the local-degree argument covers infinity as well. This proves both assertions about <critical points of iterates of a rational map>.
For the given <polynomial> the finite <derivative> is $2z$, so the only finite <rational critical point> is zero. Its orbit is
$$
0\longmapsto i\longmapsto-1+i\longmapsto-i\longmapsto-1+i.
$$
The last two points form an exact two-cycle, whose <periodic-orbit multiplier> is
$$
\lambda=P'(-1+i)P'(-i)=4(-1+i)(-i)=4(1+i),\qquad|\lambda|=4\sqrt2>1.
$$
Thus $a=-1+i$ is a <repelling periodic point>. To justify its Julia membership directly, suppose the iterates were normal near $a$. A <subsequence> of $(P^2)^n$, all fixing $a$, would converge spherically on a neighborhood. Its limit takes the finite value $a$ there, so in a smaller neighborhood it and the tail of that <subsequence> lie in a finite coordinate chart. The <Cauchy integral formula> then bounds the <derivatives> at $a$. But those <derivatives> equal $\lambda^n$, unbounded, a contradiction. Therefore $a\in J(P)$. Complete backward invariance of the <Julia set> and $P^2(0)=a$ give
$$
\boxed{0\in J(z^2+i).}
$$
The <critical orbit> is bounded but preperiodic to a repelling <holomorphic cycle>, illustrating that boundedness alone does not put a point in the <Fatou set>.
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