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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.

All poles lie outside some circle $|z|=R$ with $R>1$. The rational function is holomorphic in that disc, so its Taylor series at zero converges uniformly on $|z|\le 1$. Its polynomial partial sums therefore approximate uniformly on $K$. If there are no finite poles, the function is already a polynomial.

Let $\mathcal{A}$ be the uniform closure on $K$ of polynomials. It is a closed algebra: uniform convergence on a compact set is preserved under sums and products. For $\lambda \in \Lambda$, put $q(z)=(z-\lambda {)}^{-1}\in \mathcal{A}$ and $d=\mathrm{dist}(\lambda ,K)>0$. If $|\mu -\lambda |<d/2$, then $\mu \notin K$ andThe series converges uniformly, its ratio having modulus at most one half. Each partial sum is in $\mathcal{A}$, so its limit is too. Hence $\mu \in \Lambda$ in this neighbourhood. For empty $K$, approximation is vacuous and the assertion is immediate.

A cyclic binary code is a linear subspace of ${\mathbb{F}}_2^n$ invariant under cyclic shifts. Identify a word with its coefficient polynomial in ${\mathbb{F}}_2[X]/(X^n-1)$. Multiplication by $X$ is the shift, so the code is precisely an ideal of that quotient ring. Its inverse image in the polynomial ring is a principal ideal $(g)$ containing $(X^n-1)$; its unique monic generator divides $X^n-1$. Conversely each monic divisor generates such an ideal. Thus the bijection is with monic divisors; $g=1$ gives the full code and $g=X^n-1$ the zero code.

For odd $n$, some $m$ has $n\mid 2^m-1$, so the cyclic multiplicative group of ${\mathbb{F}}_{2^m}$ contains a primitive $n$th root $\alpha$. If a nonzero word with the specified roots had weight $w<\delta$, write its nonzero positions as $r_1,\dots ,r_w$. The root equations giveTheir matrix has determinant ${\prod }_j{\alpha }^{r_j}{\prod }_{j<l}({\alpha }^{r_l}-{\alpha }^{r_j})\ne 0$, by the Vandermonde formula and distinctness of the positions. Thus all coefficients would vanish, a contradiction. Hence minimum distance is at least $\delta$.

For $n=7$, choose $\alpha$ with minimal polynomial $X^3+X+1$. The roots $\alpha ,{\alpha }^2$ force also the conjugate ${\alpha }^4$, giving this cubic as generator. Its degree gives dimension four and its weight gives distance exactly three. Expressing $c(\alpha )=0$ in a basis of ${\mathbb{F}}_8$ gives a parity-check matrix with the seven distinct nonzero vectors of ${\mathbb{F}}_2^3$ as columns. This is the binary $[7,4,3]$ Hamming code. Its sixteen disjoint radius-one balls have $16(1+7)=128$ words, so one-error correction is perfect. The other primitive cubic produces an equivalent coordinate-permuted code.

Begin with two points at distance one. A ruler draws the line through two known points; a compass draws a circle with a known centre and a constructible radius. New points are intersections of such lines and circles. A real number is constructible when it occurs as a coordinate of a point obtained by finitely many such operations; a complex point is constructible when both coordinates are.

If current coordinates lie in a field $K\subseteq \mathbb{R}$, a line-line intersection solves linear equations over $K$, while line-circle or circle-circle intersections reduce to a quadratic equation over $K$. Thus each step requires at most adjoining a square root. Consequently every constructible number lies in a towerConversely field sums and differences are obtained by laying off segments, products and quotients by similar triangles, and a positive square root by the geometric-mean construction in a semicircle. Hence every real number in such a tower is constructible. This proves the quadratic-tower criterion, not merely its necessity.

In particular a constructible algebraic number has degree a power of two. Degree alone is not sufficient in general; the exact criterion is containment in a quadratic tower. Equivalently the Galois group of its normal closure is a finite two-group: a tower's normal closure remains a two-extension, and a two-group admits a chain of subgroups with successive index two, yielding the converse by Galois correspondence.

The standard impossibility results follow. Doubling a unit-volume cube requires $\sqrt[3]{2}$, whose polynomial $X^3-2$ is irreducible by Eisenstein and has degree three. Trisecting an arbitrary angle would in particular trisect $60$ degrees, requiring $2\cos 20^{\circ }$, a root of $X^3-3X-1$. This cubic has no rational root and is irreducible, so this construction is impossible, although some particular angles can be trisected. Squaring the unit circle requires side length $\sqrt{\pi }$, impossible because $\pi$ is transcendental whereas constructible numbers are algebraic.

For a regular $n$-gon, constructibility is equivalent to that of $e^{2\pi i/n}$. The cyclotomic extension is Galois of degree $\phi (n)$, so it is constructible exactly when this degree is a power of two. Writing the prime factorization of $n$ shows the equivalent conditionIndeed odd prime exponents must be one, and each $p_i-1$ must be a power of two. If $2^m+1$ is prime, $m$ must itself be a power of two by factoring when $m$ has an odd divisor. Conversely this factorization makes the cyclotomic Galois group a two-group and provides a quadratic tower. This gives both positive constructions, such as the regular seventeen-gon, and obstructions, such as the regular seven-gon.

The Chebyshev alternation theorem says that a polynomial $p$ of degree at most $n$ is a best uniform approximation to a real continuous function $f$ on a compact interval if and only if there are $n+2$ ordered points at which $f-p$ attains its uniform norm with alternating signs. For nonzero error these are equal-magnitude alternating extrema; the zero-error case is understood separately. The best approximating polynomial is unique.

Take $x_j=-1+j/4$, $0\le j\le 8$. These nine points satisfy $\cos (4\pi x_j)=(-1{)}^j$. Thus the zero polynomial has nine alternating error extrema of magnitude one and is minimax among polynomials of degree at most seven by the Chebyshev alternation theorem. Hence every such polynomial has uniform error at least one.

One can also prove the bound directly. If the error were strictly below one everywhere, $g(x_j)$ would have sign $(-1{)}^j$. The intermediate value theorem would give a distinct zero in each of the eight intervening intervals, contradicting the degree bound. Since the continuous error attains its maximum on the compact interval, there is an $x$ with $|f(x)-g(x)|\ge 1$.

Orient the circle counterclockwise and define its Cauchy transformThe inside and outside limits satisfy ${\Phi }_{+}-{\Phi }_{-}=\varphi$ and ${\Phi }_{+}+{\Phi }_{-}=(\pi i{)}^{-1}\mathrm{PV}{\int }_C\varphi (\tau )/(\tau -t)d\tau$. Also ${\Phi }_{-}(z)=O(z^{-1})$ at infinity. Applying these formulas to the two polynomial coefficients and dividing by $t^2$ gives the Riemann–Hilbert conditionThe homogeneous jump is $G=t/(t^2-1)$. A canonical factorization of a rational scalar Riemann-Hilbert jump isThe inside factor is analytic and nonvanishing; the exterior factor has no zeros or poles outside the circle and grows like $z$. The jump index is $1-2=-1$, consistently with this growth.

Put $F_{+}={\Phi }_{+}$ and $F_{-}={\Phi }_{-}/X_{-}$. Their additive jump is $R(t)=((A-1)t+1)/(t^2-1)$. Since $R$ is analytic outside the circle and decays there, the functions $F_{+}$ inside and $F_{-}+R$ outside glue to an entire function. This function tends to zero at infinity, so Liouville's theorem gives $F_{+}=0$, $F_{-}=-R$. ConsequentlyThe required decay of a Cauchy transform forces $A=1$. Then ${\Phi }_{-}=-1/z$, and the jump givesFor the homogeneous problem, ${\Phi }_{+}/X_{+}$ and ${\Phi }_{-}/X_{-}$ glue to an entire function vanishing at infinity, so only the zero solution exists; this proves uniqueness. Direct verification uses the principal-value integral $-\pi i/t$: the two terms on the left combine to $(t^4+t^3-t^2-t^4+t^3+t^2)/(2t)=t^2$, exactly the right side at $A=1$.

The polynomial $X^m-1$ is separable, since its derivative $m{X}^{m-1}$ has no common root with it when the characteristic does not divide $m$. Its roots form a cyclic multiplicative group of order $m$. For completeness, any finite subgroup of a field's multiplicative group is cyclic: if its exponent is $e$, all its elements are roots of $X^e-1$, so its size is at most $e$, while the elementary decomposition of a finite abelian group gives an element of order $e$.

Choose a primitive root $\zeta$. Then $L=K(\zeta )$, and every field automorphism sends $\zeta$ to ${\zeta }^a$ for a unique unit $a$ modulo $m$. Composition multiplies the exponents, and fixing $\zeta$ fixes all of $L$. This is the required injective homomorphismFor $K={\mathbb{F}}_q$, the least extension ${\mathbb{F}}_{q^d}$ containing a primitive $m$th root is characterized by $m\mid q^d-1$, since its multiplicative group is cyclic of order $q^d-1$. Thus $[L:K]$ is the multiplicative order of $q$ modulo $m$. The successive powers of $4$ modulo $11$ are $4,5,9,3,1$, so

For real $x$ at which the denominator is nonzero, mapping to the unit circle gives $|ax+b{|}^2=|cx+d{|}^2$. There are infinitely many such real $x$, so the resulting quadratic polynomial identity has all coefficients zero:If $c=0$, the first equality gives $a=0$, contradicting the nonzero determinant $ad-bc$. Thus $a,c\ne 0$. Put $\lambda =a/c$, $k=b/a$ and $\ell =d/c$. The equalities becomeTwo complex numbers with equal real parts and equal moduli have imaginary parts equal in magnitude. Consequently $\ell =k$ or $\ell =\overline{k}$. The first option would give $ad-bc=ac(\ell -k)=0$, so it is excluded. The second givesMoreover $k$ must be nonreal, since otherwise the two options coincide and the map would be constant. This proves the Möbius transformations from the real axis to the unit circle representation directly from coefficients.

Conversely, for real $x$, the numbers $x+k$ and $x+\overline{k}$ are complex conjugates, and their moduli agree. Nonreal $k$ ensures the denominator is nonzero on the real line; multiplication by $\lambda$ preserves modulus. At the point at infinity on the Riemann sphere, the map takes the value $\lambda$. Thus the representation also respects the extended real axis.

We use these countability facts: the integers are countable; a finite Cartesian product of countable sets is countable; a countable union of countable sets is countable; and every subset of a countable set is countable. Finite products can be enumerated by listing index tuples in increasing maximum coordinate, and countable unions by diagonal enumeration of pairs of indices.

Fix $d\ge 1$. For each degree bound $D$, there are finitely many monomials $X_1^{{\alpha }_1}\cdots X_d^{{\alpha }_d}$ with ${\alpha }_1+\cdots +{\alpha }_d\le D$. An integer polynomial of total degree at most $D$ is specified by an integer coefficient for each of them, so these polynomials form a finite Cartesian product of copies of $\mathbb{Z}$ and are countable. Every polynomial has some finite degree bound. Taking the union over $D=0,1,2,\dots$ proves the countability of integer polynomial rings:It is infinite because it contains all integer constant polynomials.

For $d=1$, each nonzero polynomial has only finitely many real roots: repeated application of the factor theorem bounds their number by its degree. The real algebraic numbers are the union of these finite root sets over the countably many nonzero integer polynomials. Thus they are countable. If the real transcendental numbers were countable too, their union with the algebraic numbers would make $\mathbb{R}$ countable, contrary to the allowed uncountability assumption. Hence there are uncountably many real transcendental numbers.

To construct an algebraically independent real sequence, choose $x_1$ outside the algebraic numbers. Suppose $x_1,\dots ,x_d$ already satisfy the required nonvanishing condition. For each nonzero $P\in \mathbb{Z}[X_1,\dots ,X_d,T]$, writeAt least one coefficient polynomial $P_j$ is nonzero, and the inductive hypothesis makes its value at $(x_1,\dots ,x_d)$ nonzero. Consequently $P(x_1,\dots ,x_d,T)$ is a nonzero one-variable polynomial and has only finitely many real roots. There are countably many choices of $P$, so the union of all these forbidden roots is countable. Choose $x_{d+1}$ outside it, which is possible because $\mathbb{R}$ is uncountable. This extends the nonvanishing condition to $d+1$. Recursion therefore givesIn particular the chosen numbers are distinct, since a relation $X_i-X_j$ is forbidden. Clearing rational denominators shows that every finite initial segment consists of algebraically independent elements over $\mathbb{Q}$.