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Let $R$ be a nonconstant rational map of the Riemann sphere, and write $R^n$ for its $n$-fold iterate, with $R^0$ the identity. The appropriate notion of convergence is uniform convergence on compact subsets in the chordal metric, which treats poles as ordinary values at infinity. A family of meromorphic maps is a normal family if every sequence in it has a subsequence converging locally uniformly in that metric; the limit is meromorphic or constant, possibly the constant infinity.

The Fatou set $F(R)$ consists of points with a neighborhood on which the iterate family is normal. The complementary Julia set is $J(R)=\hat{\mathbb{C}}\setminus F(R)$. Thus the sphere is the disjoint union of an open stability region $F(R)$ and a closed instability region $J(R)$. These definitions concern all subsequences of iterates, rather than assuming that the iterates themselves converge on every stable region.

Equicontinuity makes this distinction precise. On a compact subset $K$ of a neighborhood, the family is equicontinuous if, for every $ϵ>0$, some $\delta >0$ makes $\chi (R^n(z),R^n(w))<ϵ$ whenever $z,w\in K$ and $\chi (z,w)<\delta$, uniformly in $n$. The Arzelà-Ascoli theorem says that equicontinuity of continuous maps from a compact domain to a compact metric target gives relative compactness for uniform convergence. Apply it to a compact exhaustion and take a diagonal subsequence to obtain local normality. Locally uniform spherical limits of meromorphic maps are again meromorphic or constant, by using finite target charts and the locally uniform limit theorem for holomorphic functions. Conversely, if equicontinuity failed, there would be maps $f_j$ and points $z_j,w_j\to z$ with $\chi (f_j(z_j),f_j(w_j))$ bounded below by a positive number. A subsequence converging uniformly on a compact neighborhood contradicts this. Hence local normality and local equicontinuity are equivalent here.

Both sets are completely invariant:For the inverse direction, if the family is normal near $R(z)$, composition with $R$ makes $R^{n+1}$ normal near $z$, and adding the identity does not affect normality. For the forward direction, suppose the iterates are equicontinuous near $z$. A nonconstant holomorphic sphere map is open. Thus each sufficiently small neighborhood of $z$ maps onto a neighborhood of $R(z)$; every nearby target point $w$ has a preimage $x$ as close to $z$ as desired. Thenis uniformly small. Applying this at each point of the image neighborhood proves equicontinuity there. This argument works at rational critical points and requires no single-valued inverse branch. Surjectivity of $R$ supplies the asserted image equalities. Also $F(R^k)=F(R)$ and $J(R^k)=J(R)$ for $k\ge 1$: one direction follows by taking a subfamily, and the other by splitting $R^n$ into the finitely many residue classes $R^r\circ (R^k{)}^m$, $0\le r<k$.

For degree $d\ge 2$, the Julia set is nonempty. Otherwise the family would be equicontinuous on the whole compact sphere, giving a subsequence $R^{n_j}$ converging uniformly to a continuous sphere map $G$. Maps sufficiently close in chordal distance are homotopic, using the unique short spherical geodesic between their values. They therefore have equal topological degrees. This is impossible because $\mathrm{deg}{R}^{n_j}=d^{n_j}\to \infty$ whereas $G$ has one fixed degree.

It cannot be a nonempty finite set either. Complete invariance would make $R$ a permutation of that finite set, with just one sphere preimage for each of its points. That preimage would have local degree of a holomorphic map $d$. Every holomorphic cycle in this set would consequently be superattracting for its return map, hence lie in $F(R)$, a contradiction. In particular $J(R)$ contains at least three points.

The Montel theorem in the form used here says that a family of meromorphic maps on a domain, each omitting the same three distinct points of the sphere, is normal for spherical convergence. It also shows that $J(R)$ is perfect. If $a$ were isolated in $J(R)$, choose a disc $U$ with $U\cap J(R)=\{a\}$ and four distinct points of $J(R)$. For each $n$, complete invariance makes $R^n(U)\cap J(R)$ have at most the single point $R^n(a)$. Partition the iterates into finitely many families according to which of the four points, if any, that single value equals. Each family omits at least three fixed points, so each is normal by Montel; their finite union is normal. This would put $a$ in $F(R)$, a contradiction. Therefore the degree-at-least-two Julia set is compact, infinite and has no isolated points.

The same omitted-value argument explains its global reach. If an open $U$ meets $J(R)$, then ${\bigcup }_{n\ge 0}{R}^n(U)$ can omit at most two sphere points, since three omissions would imply normality on $U$. If $J(R)$ had an interior and were not the whole sphere, these image sets would remain in $J(R)$ and their complement would contain the nonempty open set $F(R)$, impossible. Thus a proper Julia set has empty interior and equals the boundary of $F(R)$. If the Julia set is the whole sphere, the Fatou set is empty instead.

An attracting periodic point and its holomorphic cycle has a contracting return map in a small disc, so its basin is in the Fatou set. Repelling periodic points belong to the Julia set: return iterates fix the point but have derivatives equal to unbounded powers of their periodic-orbit multiplier, contradicting derivative bounds for a normal subsequence near that fixed value. A standard further global property is that repelling periodic points are dense in the Julia set. Fatou components can also support convergence towards a parabolic cycle on the boundary, or rotation behavior in a Siegel disc or a Herman ring; normality need not mean attraction. For example, $R(z)=z^d$ has $J(R)=\{|z|=1\}$, with the interior converging to zero and the exterior to infinity. On the unit circle, arbitrarily close radii have incompatible limiting behavior, preventing normality there.

Degree one has a different picture. A rational map of degree one is a Möbius transformation. Up to Möbius conjugacy, an elliptic map is $z\mapsto az$ with $|a|=1$; its iterates are rotations and are equicontinuous on the sphere, so $F$ is the whole sphere and $J$ is empty (including the identity). For $z\mapsto az$ with $|a|>1$, all compact subsets away from zero converge to infinity, while $z_n=a^{-n}\to 0$ and $R^n(z_n)=1$ prevent equicontinuity at zero. Thus $J=\{0\}$; for $|a|<1$ the reciprocal coordinate gives $J=\{\infty \}$. A nonidentity parabolic map is conjugate to $z\mapsto z+1$. Its iterates converge locally uniformly to infinity on the finite plane, but $z_n=-n\to \infty$ has $R^n(z_n)=0$, so $J=\{\infty \}$. Holomorphic conjugacy carries these sets to the corresponding fixed points of the original map. These finite Julia sets explain why the degree restriction in many global Julia-set properties is essential.

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.

If $F:X\to Y$ is a nonconstant holomorphic map between compact connected Riemann surfaces, its local degree at $p$ is the integer $m_F(p)\ge 1$ for which suitable local coordinates giveThe valency theorem states thatis independent of $y\in Y$. This common value is the degree of a holomorphic map, denoted $\mathrm{deg}F$.

Now let $f$ be a nonconstant rational function of degree $d$. If its distinct finite poles have orders $m_1,\dots ,m_r$, and its pole order at infinity is $m_{\infty }\ge 0$, thenThe derivative $f^{\prime }$ has a pole of order $m_j+1$ at each finite pole. When $m_{\infty }>0$, the expansion $f(z)\sim c{z}^{m_{\infty }}$ shows that $f^{\prime }(z)\sim c{m}_{\infty }{z}^{m_{\infty }-1}$ at infinity. Hence the degree of the derivative of a rational function isIn the first case $r\ge 0$, while in the second $1\le r\le d$; thereforeFor every $d\ge 1$, the lower bound is attained by $f(z)=z^d$, whose derivative has degree $d-1$ (with a constant assigned degree zero). For distinct $a_1,\dots ,a_d$, the functionhas $d$ simple poles, degree $d$, and a derivative with $d$ double poles, so $\mathrm{deg}{f}^{\prime }=2d$. Thus both rational bounds are sharp for every $d$.

Let next $g$ be a nonconstant elliptic function for the period lattice $\Lambda$. Its degree is the total order of its poles in a fundamental parallelogram; by the valency theorem, this is also the degree of the induced map $\mathbb{C}/\Lambda \to \hat{\mathbb{C}}$. If these poles have orders $n_1,\dots ,n_s$, then $d=\mathrm{deg}g={\sum }_j{n}_j$, and $g^{\prime }$ has poles of orders $n_j+1$. The degree of the derivative of an elliptic function is consequentlySince every nonconstant elliptic function has at least one pole and $1\le s\le d$,

Let $d\ge 3$ be odd. The Weierstrass elliptic function supplies the lower-bound exampleIt has one pole modulo $\Lambda$, of order $d$, so its derivative has one pole of order $d+1$. For the upper bound, choose distinct points $a_1,\dots ,a_d$ modulo $\Lambda$ and nonzero constants $c_j$ with ${\sum }_j{c}_j=0$. The quasi-periodicity of the Weierstrass zeta function makeselliptic. It has exactly $d$ simple poles, while $g^{\prime }$ has $d$ double poles. Therefore the two bounds are attained for every required odd degree.