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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 formThe local degree of a holomorphic map is $e$. A critical point of a rational map is one with $e\ge 2$, 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, $\mathrm{deg}Q=d$, and $\mathrm{deg}P\le d$. Regularity at infinity givesSince $R^{\prime }=(P^{\prime }Q-P{Q}^{\prime })/Q^2$, its numerator $W=P^{\prime }Q-P{Q}^{\prime }$ therefore has degree exactly $2d-2$. At every pole $p$, $W(p)=-P(p)Q^{\prime }(p)\ne 0$, 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 provesThe 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 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 $ef$. ThusIf $p$ 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 $(P^n{)}^{\prime }(z)={\prod }_{j=0}^{n-1}{P}^{\prime }(P^j(z))$; 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 isThe last two points form an exact two-cycle, whose periodic-orbit multiplier isThus $a=-1+i$ 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 $P^2(0)=a$ giveThe critical orbit is bounded but preperiodic to a repelling holomorphic cycle, illustrating that boundedness alone does not put a point in the Fatou set.