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For an oriented three-manifold M with prescribed framed boundary endpoints P, the relative Kauffman bracket skein module is the complex vector space generated by properly embedded framed tangles, modulo framed isotopy and the local bracket relations. A crossing resolves with coefficients A and $A^{-1}$, and adjoining an unknotted circle multiplies by $\delta =-A^2-A^{-2}$. We now use the full bracket convention: the empty diagram has value one and a circle has value $\delta$. For a nonempty uncolored link this full bracket is $\delta$ times the reduced bracket used earlier. Fixing this normalization is essential for the quantum dimensions. Gluing compatible boundary pieces composes skeins; disjoint union gives tensor products. With no endpoints this is the Kauffman bracket skein module. Framing is retained, so a Reidemeister-I curl is a framing change rather than an allowed isotopy.

For $A\ne 0$, the Temperley-Lieb algebra ${\mathrm{TL}}_n(A)$ is the skein algebra of a rectangle cylinder with n marked endpoints on each horizontal side, multiplication being vertical stacking. Resolve crossings to get a basis of crossing-free matchings of its $2n$ endpoints, with every closed circle removed with factor $\delta$. There are $\frac{1}{n+1}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$ such matchings. Conversely multiplication of these matchings by joining endpoints defines an algebra satisfying the skein relations, which proves their independence. Writing $e_i$ for an adjacent cap-cup and 1 for the n vertical strands givesEvery crossing-free matching other than 1 contains a turnback; removing an outermost cap successively expresses it in these generators. Thus these diagram rules give the usual generator presentation as well.

Put $[m]=(A^{2m}-A^{-2m})/(A^2-A^{-2})$, interpreted as the Laurent sum ${\sum }_{s=0}^{m-1}{A}^{2(m-1-2s)}$ if its denominator vanishes. Set ${\Delta }_j=(-1{)}^j[j+1]$. These quantum integers satisfy ${\Delta }_0=1$, ${\Delta }_1=\delta$, and ${\Delta }_j=\delta {\Delta }_{j-1}-{\Delta }_{j-2}$. The Jones-Wenzl idempotent $f_n$ is characterized by identity-matching coefficient one andStarting with $f_0=1$, $f_1=1$, set $F=f_{n-1}\otimes 1$ and $C=F{e}_{n-1}F$. Whenever the denominators are nonzero the recursion isHere is the induction proving its properties. Closing the last strand of $f_{n-1}$ gives $({\Delta }_{n-1}/{\Delta }_{n-2})f_{n-2}$. Consequently stacking two copies of C, and straightening the middle cap-cup, gives $C^2=\mu C$, where $\mu ={\Delta }_{n-1}/{\Delta }_{n-2}$. Also $FC=CF=C$. Expanding $(F-{\mu }^{-1}C{)}^2$ proves idempotence. The earlier turnbacks are killed by F; the last is killed because the same cap-cup calculation gives $C{e}_{n-1}=\mu F{e}_{n-1}$ and its reflected identity. Only F contributes to the identity matching, with coefficient one.

For completeness, the partial-closure formula used in that induction follows from the recursion itself. Closing the last strand of F gives $\delta f_{n-1}$; closing the last strand of C straightens its cap-cup to $f_{n-1}$. Hence partial closure of $f_n$ is $(\delta -{\Delta }_{n-2}/{\Delta }_{n-1})f_{n-1}=({\Delta }_n/{\Delta }_{n-1})f_{n-1}$. Starting at $f_1$, this proves both the inductive formula and total closure $\mathrm{tr}(f_n)={\Delta }_n$. Finally, any nonidentity matching is in the turnback ideal. If two normalized projectors kill that ideal, multiplying them gives each of them, proving uniqueness. Their normalization also gives absorption of smaller projectors on consecutive substrands. At exceptional A one must check the denominators; the recursion does not assert existence of every projector there.

A positive framing twist of an n-strand projected cable consists of n individual positive curls and $n(n-1)$ positive braid crossings. Each curl contributes $-A^3$. A crossing is $A1+A^{-1}{e}_i$, so it acts on $f_n$ as A because the turnback vanishes. This proves the depicted Jones-Wenzl twist eigenvalue identity, including its exponent:A negative framing twist acts by the inverse scalar.

Choose $A=e^{\pi i/(2r)}$, with an integer $r\ge 3$. Then $[r]=0$ but $[1],\dots ,[r-1]$ are nonzero. The projector $f_{r-1}$ exists and has trace zero. Pass to the reduced Temperley-Lieb skein theory by setting it to zero, including every diagram containing it. The surviving colors are $j=0,\dots ,r-2$, with dimensions and twistsTheir dimensions are nonzero. The recurrence decomposes $f_j\otimes 1$ into the two orthogonal channels $f_{j+1}$ and $f_{j-1}$: the first is the new projector, while its complementary cap-cup idempotent is the latter, with inclusion/projection normalized by the nonzero partial-closure scalar. At the upper endpoint the $r-1$ channel vanishes. Iterating gives a complete decomposition of every tensor power into these colors. A matching between projected colors of different sizes has a turnback on the larger side and vanishes; on the same size only the identity matching survives. Thus their endomorphism spaces are one-dimensional and all matrix blocks split. Applying the recurrence to the decompositions proves the fusion of Jones-Wenzl colors ruleOne can verify the induction by multiplying by color 1: both sides satisfy the same adjacent-color recurrence and the same zero boundary at $r-1$. Taking closures proves $d_a{d}_b={\sum }_c{N}_{ab}^c{d}_c$. Cups and caps make every color self-dual, so these fusion coefficients are symmetric in all three labels.

We next prove the Kirby color handle-slide identity, rather than assume it. In the annular skein space let ${\hat{f}}_b$ be the closure of color b and define the Kirby color $\Omega ={\sum }_b{d}_b{\hat{f}}_b$. For a fixed color a, choose inclusion/projection maps $i_c:c\to b\otimes a$, $p_c:b\otimes a\to c$ for the fusion channels, with $p_c{i}_c=1_c$ and ${\sum }_c{i}_c{p}_c=1_{b\otimes a}$. Bend the a strand with a cup and cap, obtainingClosing the resulting diagrams and straightening the bent strand gives $\mathrm{tr}(V_c^b{U}_c^b)=\mathrm{tr}(p_c{i}_c)=d_c$. Since b is simple, $V_c^b{U}_c^b=(d_c/d_b)1_b$. The symmetric fusion rule supplies all channels of $c\otimes a$, so their normalized projections are complete. ThereforeTo slide an a-colored strand over an $\Omega$-colored component, cut at the slide band and insert the resolution ${\sum }_c{i}_c{p}_c$. Carrying the band across that loop bends its two maps into U and V. In each outgoing channel c, the boxed identity replaces the sum over the old loop color b by exactly the new loop weight $d_c$; the rest of the diagrams are isotopic as framed ribbons. This proves invariance for fixed a. Summing a with its own weight proves invariance when both surgery components are colored by $\Omega$. The band framing is carried in this argument, as required for a framed handle slide.

We must also show that the stabilization normalization is nonzero. Let $H_{ij}$ be the full bracket of the zero-framed Hopf link with colors i,j. On a fusion channel k, its double braiding has eigenvalue ${\theta }_k/({\theta }_i{\theta }_j)$: the full twist of the combined ribbon equals double braiding followed by the two separate twists. Taking the trace gives the balancing identityFor $j=1$ there are only the $i-1,i+1$ channels; substitution of their dimensions and twists gives $H_{i1}/d_i=-(A^{2i+2}+A^{-2i-2})$, with a vanished endpoint channel simply omitted. A color-1 meridian therefore acts on the simple color i by that scalar. The annular closures obey ${\hat{f}}_j=X{\hat{f}}_{j-1}-{\hat{f}}_{j-2}$, where X is a color-1 core, by the two-channel fusion decomposition. Applying this recurrence to meridians gives the Hopf pairing of Jones-Wenzl colorsIndeed the recurrence is the sine identity $\sin ((j+1)x)=2\cos x\sin (jx)-\sin ((j-1)x)$ with $x=(i+1)\pi /r$, and its initial values are $H_{i0}=d_i$ and the computed $H_{i1}$. Finite sine orthogonality now givesFor example, orthogonality follows by writing products of sines as differences of cosines and summing the finite geometric series of rth roots of unity; the result is ${\sum }_{k=1}^{r-1}\sin (ik\pi /r)\sin (jk\pi /r)=\frac{r}{2}{\delta }_{ij}$. Thus no degeneracy remains in this pairing.

The Gauss sums of Jones-Wenzl colors are $p_{\pm }={\sum }_j{d}_j^2{\theta }_j^{\pm 1}$, the evaluations of the plus- and minus-one-framed $\Omega$-colored unknots. Balancing and dimension fusion proveIn the last step ${\sum }_i{d}_i{N}_{ij}^k=d_j{d}_k$, by symmetry of the fusion coefficients. In vector notation this is $HTd=p_{+}{T}^{-1}d$, with $T=\mathrm{diag}({\theta }_j)$. The matrix H and vector d are real, and each twist has modulus one, so conjugation gives $H{T}^{-1}d=p_{-}Td$. Apply H again and use its orthogonality to obtain $\boxed{p_{+}{p}_{-}={\mathcal{D}}^2}$. Hence neither factor vanishes.

We may use the surgery existence theorem and Kirby calculus: every closed oriented three-manifold has an integral framed-link surgery description, and two descriptions are related by handle slides and insertion/removal of disjoint plus- or minus-one-framed unknots. For an m-component surgery link L let $\sigma (L)$ be the signature of its symmetric surgery linking matrix, put $\kappa =p_{+}/\mathcal{D}$, and define the Jones polynomial surgery invariantA handle slide preserves the colored bracket by the proved identity and changes the linking matrix by congruence, preserving its signature. Adding a disjoint plus-one-framed unknot multiplies the bracket by $p_{+}$ and changes $(m,\sigma )$ to $(m+1,\sigma +1)$, so the net factor is $p_{+}/(\mathcal{D}\kappa )=1$. For a minus-one-framed unknot it is $\kappa p_{-}/\mathcal{D}=p_{+}{p}_{-}/{\mathcal{D}}^2=1$. Kirby calculus therefore proves that the displayed value is a well-defined oriented three-manifold invariant. Our normalization has ${\tau }_r(S^3)=1$; reversing orientation conjugates the value. Its construction uses precisely the bracket and colored projectors underlying the Jones polynomial.

For real projective 3-space the surgery diagram is a single unknot of framing two. Thus $m=1$, $\sigma =1$, and its colored evaluation is ${\sum }_j{d}_j^2{\theta }_j^2$. Since $\mathcal{D}\kappa =p_{+}$, the explicit answer isThe denominator is nonzero by the proved Gauss-sum identity. For instance at $r=3$ the twists are $1,-i$, the squared dimensions are both one, and the numerator vanishes, so this normalized invariant is zero for ${\mathbb{RP}}^3$ at that level.

With determinant-one normalization, the necessary and sufficient condition for sphere rotations as special-unitary Möbius transformations isEquivalently, the matrix $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ belongs to the special unitary group $SU(2)$. The two determinant-one representatives $M$ and $-M$ induce the same Möbius transformation, and both satisfy this condition.

Use north-pole stereographic projection in the direction from the plane to the sphere:Its chord-distance identity is $|\phi (z)-\phi (w)|=2\chi (z,w)$, where $\chi$ is the chordal metric. Direct subtraction, using $ad-bc=1$, givesThis is first calculated away from poles and then extended continuously to the Riemann sphere. A sphere rotation preserves chord distances, so for distinct finite nonpole pointsUsing three distinct points shows all these positive ratios equal one. Hence $H(z)=1+|z{|}^2$ everywhere. Comparing its coefficients givesThus $M$ is a unitary matrix. Comparing $M^{-1}=M^{\ast }$ with $M^{-1}=\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$ proves the boxed relations.

Conversely, these relations give $H(z)=1+|z{|}^2$, so the induced sphere map $T=\phi g{\phi }^{-1}$ preserves all Euclidean chord distances. Such a bijection of the unit sphere is the restriction of an orthogonal matrix: let $v_j=T(e_j)$; preservation of distances makes these an orthonormal basis, and $T(p)\cdot v_j=p\cdot e_j$ forces $T(p)={\sum }_j{p}_j{v}_j$. Finally the Möbius transformation is holomorphic with positive real Jacobian at every ordinary coordinate point. Conjugating by stereographic projection preserves orientation, so this orthogonal matrix has determinant $+1$. It is therefore a genuine rotation in three dimensions, proving sufficiency rather than just spherical isometry.

For the circle assertion, a nondegenerate generalized circle has an equationFor $A\ne 0$ this is an ordinary circle, and for $A=0$ it is a line with infinity included. On the sphere write coordinates $(p_1,p_2,p_3)$, with $z=(p_1+i{p}_2)/(1-p_3)$ and $|z{|}^2=(1+p_3)/(1-p_3)$. Multiplication by $1-p_3$ gives the planeIf $n=(2\mathrm{Re}B,-2\mathrm{Im}B,A-C)$ and $h=-(A+C)$, then $|n{|}^2-h^2=4(|B{|}^2-AC)>0$. Thus the plane intersects the unit sphere in a proper circle. It contains the north pole exactly when $A=0$, accounting for the point at infinity on a line.

Conversely, a circle on the sphere is the intersection with a plane $n\cdot p=h$, $|h|<|n|$. Set $A=(n_3-h)/2$, $C=(-n_3-h)/2$ and $B=(n_1-i{n}_2)/2$. The preceding equation is recovered and has $|B{|}^2-AC=(|n{|}^2-h^2)/4>0$. Its inverse stereographic image is an ordinary circle or a line with infinity attached. This proves both directions of the circle-plane relation under stereographic projection.

Take $u(z)=\log |z|$. In polar coordinates its Laplacian is $u_{rr}+r^{-1}{u}_r=-r^{-2}+r^{-2}=0$, so it is harmonic on the annulus. If $u=\mathrm{Re}f$ for a single-valued holomorphic function, the Cauchy-Riemann equations would give $f^{\prime }=u_x-i{u}_y=1/z$. But for $1<\rho <2$,The first equality follows by evaluating the primitive $f$ around a closed curve. This contradiction verifies the absence of a global harmonic conjugate.

For the final function, the quotient of squared distances is $|z+1{|}^2/|z-1{|}^2$. On $|z|>1$, both $1+1/z$ and $1-1/z$ lie in the open right half-plane, so their principal logarithms are single-valued and holomorphic. DefineThe series converges locally uniformly outside the unit circle, andThus this is the required single-valued holomorphic function, even on the larger domain $|z|>1$. Its derivative is $2/(z+1)-2/(z-1)$, whose two periods around the holes cancel. The logarithmic cancellation on an exterior annulus explains why this example succeeds while $\log |z|$ fails; using independent multivalued logarithms of $z+1$ and $z-1$ would conceal that cancellation.

An explicit disk map of a slit half disk makes the two boundary approaches visible without assuming an extension theorem for the boundary. We construct the map as a succession of bijections and then determine its limits.

First $w=z^2$ maps the upper half-disc of radius two bijectively onto $|w|<4$ slit along $[0,4)$. Indeed, its inverse chooses the square root whose argument lies in $(0,\pi )$. The removed segment on the imaginary axis becomes $[-1,0)$, so the image of $D$ is preciselyThe Möbius transformation $v=4(w+1)/(w+16)$ maps the radius-four disc bijectively onto the unit disc, taking $-1$ to zero and $4$ to one. This follows also by writing it as $(w/4+1/4)/(1+w/16)$, an automorphism of the unit disk after the scaling. It takes the removed real interval to $[0,1)$, so the image is the unit disc with this radial slit removed. Choosing $0<\arg v<2\pi$ defines a single-valued square root $u=\sqrt{v}$ and maps that slit disc bijectively ontoAll these maps have nonzero derivatives on their respective domains, since their potentially problematic zero or branch point is on the removed slit.

Next the Joukowski map $J(u)=u+1/u$ is a conformal bijection from $U$ onto the lower half-plane. For $u\in U$,If $J(u)=J(t)$, then $(u-t)(1-1/(ut))=0$, and $|ut|<1$ rules out the second factor vanishing. To prove surjectivity, take $q$ in the lower half-plane and consider $u^2-qu+1=0$. Neither root lies on the unit circle, where $u+1/u$ is real. Their product is one, so exactly one has modulus less than one. For that root the displayed imaginary-part identity forces $\mathrm{Im}u>0$. Thus it lies in $U$. Also $J^{\prime }(u)=1-u^{-2}$ cannot vanish there. Finally $C(q)=(q+i)/(q-i)$ maps the lower half-plane bijectively onto the unit disc: the inequality $|q+i|<|q-i|$ is exactly $\mathrm{Im}q<0$, and its inverse is $q=i(\zeta +1)/(\zeta -1)$. Consequentlyis a conformal bijection of $D$ onto the disc.

At the slit point $z_0=i/2$ the continuous rational expression for $v$ tends to $4/21$. In a sufficiently small neighborhood of $z_0$, points of $D$ have nonzero real part. Since $\mathrm{Im}(z^2)=2\mathrm{Re}z\mathrm{Im}z$ and the real-coefficient Möbius map $w\mapsto v$ preserves the sign of the imaginary part, right-side approach gives $\arg v\to 0$, while left-side approach gives $\arg v\to 2\pi$. HenceThe two limits of $F$ are thereforeThey have modulus one and are distinct, since $L>0$ and the Möbius map $C$ is injective on the extended real line. The two-sided cluster set at an interior slit point is a geometric consequence of opening the slit into two boundary intervals.

For the given map $f$, the composition $B=f\circ F^{-1}$ is an automorphism of the unit disk. To see its precise boundary behavior, let $a=B^{-1}(0)$ and compose $B$ with the inverse of $T_a(\zeta )=(\zeta -a)/(1-\overline{a}\zeta )$. This gives a disc automorphism fixing zero. Applying Schwarz lemma to it and its inverse shows it is a rotation. ThusIts denominator never vanishes on the closed disc, and its inverse has the same property; it therefore extends to a continuous bijection of the unit circle. Put $\alpha =B({\gamma }_{+})$ and $\beta =B({\gamma }_{-})$. These are distinct points of that circle, independent of the approaching sequence.

For any sequence approaching $z_0$, every infinite right-side subsequence has image tending to $\alpha$, and every infinite left-side subsequence has image tending to $\beta$. At least one side occurs infinitely often. Conversely, any convergent image subsequence has a further subsequence confined to one side, so its limit must be the corresponding one of these two points. The cluster set is therefore exactly $\{\alpha \}$, $\{\beta \}$ or $\{\alpha ,\beta \}$, according to which sides occur infinitely often. Both sides really occur in the domain, for example along $i/2\pm {\epsilon }_n$ with positive ${\epsilon }_n\downarrow 0$.

We first derive the exterior area theorem for univalent functions directly from Green's theorem, then obtain the coefficient bound needed for the Koebe quarter theorem.

Suppose $G$ is univalent on $|\zeta |>1$ and has a Laurent seriesFor $R>1$, the image of $R{e}^{it}$, $0\le t\le 2\pi$, is a regular simple closed curve. By the Jordan curve theorem the curve has one bounded and one unbounded complementary component. The image of $|\zeta |>R$ lies in the unbounded component, since it is connected, avoids the curve by injectivity and contains points near infinity. It fills that component: any finite boundary limit of the image has preimages bounded by $G(\zeta )/\zeta \to 1$, and a convergent preimage subsequence must land on $|\zeta |=R$ or else give an interior image point. Thus no boundary of the image lies inside that complementary component. The curve is positively oriented about its bounded interior, since the exterior is on the right of $R{e}^{it}$ and a holomorphic map with nonzero derivative preserves orientation. Green's formula gives its enclosed area asFor completeness, the positive Fourier mode is $R{e}^{it}$ and the negative modes are $b_n{R}^{-n}{e}^{-int}$. On multiplying by the conjugate series and integrating, all unequal-frequency products integrate to zero by Fourier orthogonality. The equal-frequency products give $i{R}^2$ and $-in|b_n{|}^2{R}^{-2n}$, respectively; the constant term gives zero. The Laurent series and its differentiated series converge uniformly on the circle, since $R>1$, so these operations are justified. Nonnegativity of $A_R$ yields $\sum n|b_n{|}^2{R}^{-2n}\le R^2$. Letting $R\downarrow 1$, or first doing so for every finite partial sum, givesNo regularity of the map on the unit circle has been assumed.

Now let $f(z)=z+a_2{z}^2+\cdots$ be a normalized univalent function. Its only zero is at zero, so $f(z)/z$, extended there by the value one, is holomorphic and nowhere zero. On the simply connected disc it has a holomorphic square root with value one at zero. Its odd square-root transform isThis is an odd function which is holomorphic and normalized. It is also univalent: $h(z)=h(w)$ gives $f(z^2)=f(w^2)$ and hence $z=\pm w$. If $z=-w$, oddness gives $h(z)=-h(z)$, so $h(z)=0$, forcing $z=w=0$. Thus $G(\zeta )=1/h(1/\zeta )$ is univalent on the exterior disc, with expansionApplying the area bound to its first negative coefficient proves the second coefficient bound for normalized univalent functions, $|a_2|\le 2$.

Let $w_0$ be any point omitted by $f$. It is nonzero. Postcomposition with a Möbius map gives another normalized univalent functionwhose denominator does not vanish. Applying the second coefficient bound both to $f$ and to $f_{w_0}$ givesEvery omitted point therefore has modulus at least $1/4$, provingThis establishes the Koebe quarter theorem from the allowed area formula.

Sharpness is witnessed by the Koebe function $k(z)=z/(1-z{)}^2$. It has the required value and derivative at zero. The equality $k(z)=k(w)$ simplifies to $(z-w)(1-zw)=0$, and $|zw|<1$ makes it injective. Moreover,Thus $-1/4$ is omitted, so no larger universal radius is possible. In fact $4k(z)+1=((1+z)/(1-z){)}^2$ shows its image is the plane slit along $(-\infty ,-1/4]$.

Without univalence take a real $\lambda >4$ andIt is entire, with $f_{\lambda }(0)=0$ and $f_{\lambda }^{\prime }(0)=1$, but it omits $-1/\lambda$, a point of modulus less than $1/4$, since the exponential never vanishes. It is genuinely not univalent on the disc: the distinct points $\pm \pi i/\lambda$ lie there and both give exponential value $-1$. Even its derivative never vanishes, illustrating why local conformality alone does not give the quarter-disc inclusion.

A countable set is a set that is finite or can be put in bijection with the natural numbers; equivalently, it admits an injective function into the natural numbers. To enumerate the rational numbers, list pairs $(p,q)$ with $p\in \mathbb{Z}$, $q\ge 1$ in increasing order of $|p|+q$, and then list $p/q$, discarding repetitions. Each level is finite and every rational eventually appears. An enumeration $q_1,q_2,\dots$ of the rationals then enumerates $\mathbb{Q}\times \mathbb{Q}$ by listing $(q_i,q_j)$ in diagonals of constant $i+j$. ThusTo prove that the real numbers are uncountable, suppose $(0,1)$ had an enumeration $x_1,x_2,\dots$. Write each $x_n$ in its decimal expansion that does not end in an infinite string of nines. Define a new decimal $y=0.d_1{d}_2\cdots$ by taking $d_n=2$ if the $n$th digit of $x_n$ is $1$, and $d_n=1$ otherwise. Then $y\in (0,1)$, has no ambiguous terminating/nines representation, and differs from $x_n$ at digit $n$. The Cantor diagonal argument contradicts the supposed enumeration. Consequently the real numbers are uncountable.

For the discs, use the enumeration of ${\mathbb{Q}}^2$ just constructed. Every positive-radius disc contains a rational-coordinate point: its interior contains a small open rectangle, and each coordinate interval contains a rational number. For each disc $D$, let $j(D)$ be the least index of an enumerated rational-coordinate point in its interior. This minimum exists and requires no arbitrary simultaneous choice. Disjoint discs cannot receive the same point, so $D\mapsto j(D)$ is an injective function into the natural numbers. This proves the countability of pairwise disjoint open disks and also the same conclusion for closed discs, by using their disjoint interiors.

For circles, take all concentric circles centred at the origin with radii $r\in (1,2)$. Two different radii give disjoint circles, and every radius is positive. The radius map is a bijection from the uncountable interval $(1,2)$ to this family. An uncountable disjoint family of nontrivial circles is therefore possible: the rational-point argument for discs relies on their nonempty interiors, which circumferences do not have.

Put $\kappa =qB/m$, retaining its sign. The Lorentz force equations areThus $z=0$ remains invariant when $z(0)=\dot{z}(0)=0$. Also $d|\dot{\mathbf{r}}{|}^2/dt=2\dot{\mathbf{r}}\cdot \ddot{\mathbf{r}}=0$, since a vector is perpendicular to its cross product with the magnetic field. Choose the initial planar velocity along the positive $x$ axis and the initial location at $(0,0)$. Integration givesThis is a circle centred at $(0,-u_0/\kappa )$ with radius $u_0/|\kappa |$. The cyclotron frequency is $|qB|/m$. Equivalently, choosing this centre as the polar origin givesThe rotation is clockwise viewed from positive $z$ when $qB>0$, and reverses when $qB<0$. In the PDF hint, $\dot{r}=\ddot{r}=0$ are radial conditions with this centred origin; they do not require zero particle velocity. If $qB=0$, the limiting motion is a straight line instead.

With linear drag, put $a=\mu /m>0$. The governing equations for damped cyclotron motion becomeTaking the scalar product with the velocity gives $\frac{1}{2}d|\dot{\mathbf{r}}{|}^2/dt=-a|\dot{\mathbf{r}}{|}^2$, henceFor the planar path, let $w=\dot{x}+i\dot{y}$ and $\zeta =x+iy$. Then $\dot{w}=-(a+i\kappa )w$, so for arbitrary initial planar dataLet ${\zeta }_{\infty }={\zeta }_0+w_0/(a+i\kappa )$. The position relative to this limiting point isIts radius shrinks exponentially while its angle rotates uniformly: the path is a logarithmic spiral into ${\zeta }_{\infty }$, not a succession of circles about the original vacuum guiding centre. The weak-drag assumption is $\mu \ll |qB|$, so many revolutions occur before appreciable contraction.

Use the unit sphere $X^2+Y^2+Z^2=1$, its north pole $(0,0,1)$, and the equatorial complex plane $Z=0$. The line from the north pole to a sphere point meets that plane atThe north pole maps to $\infty$. The inverse stereographic projection isThe antipodal point has coordinate $w=-(X+iY)/(1+Z)$. Since $X^2+Y^2=(1-Z)(1+Z)$, this is $-1/\bar{z}$ when both sides are finite and nonzero. Zero and infinity correspond to the two poles and are exchanged by the same extended convention. The inverse formula also proves the converse, giving the antipodal stereographic coordinate relation

The requested rotation result is that sphere rotations correspond exactly to Möbius transformationsThe matrices $\left(\begin{array}{cc}a& b\\ -\bar{b}& \bar{a}\end{array}\right)$ form $SU(2)$, and opposite matrices induce the same transformation. Thus the rotation group $SO(3)$ is identified with $SU(2)/\{\pm I\}$. This is the sphere rotations as special-unitary Möbius transformations result, which need not be proved here.

For a planar circle $|z-c|=r$, $c=a+ib$ and $r>0$, substitute $|z{|}^2=(1+Z)/(1-Z)$ and $\Re (\bar{c}z)=(aX+bY)/(1-Z)$. Multiplication by $1-Z$ gives the circle-plane relation under stereographic projection:The lift lies in this plane, and conversely its intersection with the sphere projects onto the circle. It is a nondegenerate circle on the sphere. The north pole is not on the displayed plane, since substitution there gives two rather than zero.

A great circle is the intersection with a plane through the sphere's centre. Here this holds exactly whenThe unit circle lifts to the equator. If a different circle lifts to a great circle, its plane and the equatorial plane are distinct planes through the origin, so their intersection cuts the sphere at two antipodal equatorial points. Their planar coordinates are $z$ and $-z$ on $|z|=1$.

Conversely, if the planar circle meets the unit circle in $z$ and $-z$, their lifts are antipodal equatorial points. The plane of the lifted circle contains their midpoint, the origin, so that circle is great. Equivalently, adding $|z-c{|}^2=r^2$ and $|-z-c{|}^2=r^2$ gives $r^2=1+|c{|}^2$. The exclusion of the unit circle ensures the two intersection points are a genuine pair rather than coincident circles. This proves the antipodal equatorial intersection criterion for a great circle.