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The composition of hyperbolic reflections in two ultraparallel hyperbolic lines has infinite order. Map their common perpendicular to the imaginary axis in the Poincaré half-plane model. The two reflecting hyperbolic lines become semicircles centred at zero, with distinct radii $r_1,r_2$. Reflection in the radius-$r$ semicircle is $z\mapsto r^2/\overline{z}$, so their composition is $z\mapsto (r_2/r_1{)}^{2}z$. Every positive power remains nonidentity, because $(r_2/r_1{)}^{2k}\ne 1$ for $k\ge 1$.

An equilibrium is nonhyperbolic when its Jacobian matrix has at least one eigenvalue on the imaginary axis. Linearization alone then need not determine the local dynamics, and a center manifold often supplies the relevant reduction.

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 ${\epsilon }_n\downarrow 0$.

In the Poincaré half-plane model, $\mathbb{H}=\{x+iy:y>0\}$ has Riemannian metricIts orientation-preserving isometries are the Möbius transformations $z\mapsto (az+b)/(cz+d)$ with real coefficients and $ad-bc=1$, with a simultaneous sign change representing the same map. The full isometry group also contains their compositions with $z\mapsto -\overline{z}$, which reverse orientation. The hyperbolic lines are vertical Euclidean lines and upper semicircles centred on the real axis; each meets the boundary orthogonally.

A hyperbolic line with finite endpoints $p<q$ is mapped to the imaginary axis byIts coefficient determinant is $q-p>0$, so rescaling the coefficients makes this an orientation-preserving isometry. The endpoints map to $0$ and $\infty$, which characterize the imaginary-axis hyperbolic line. For a vertical line with endpoints $p,\infty$, use $T(z)=z-p$. If $T_1,T_2$ send the two prescribed lines to the imaginary axis, $T_2^{-1}{T}_1$ sends the first to the second. This proves transitivity on hyperbolic lines.

For two ultraparallel hyperbolic lines, choose their common perpendicular and map it to the imaginary axis. A hyperbolic line perpendicular to that axis is a semicircle centred at zero: orthogonality at $iy$ forces the circle centre to have real coordinate zero. Thus the reflecting lines become semicircles of distinct radii $r_1,r_2>0$. The hyperbolic reflection in the radius-$r$ semicircle is $S_r(z)=r^2/\overline{z}$; it fixes that semicircle pointwise and reverses orientation. Their composition isIts $k$th power is $z\mapsto {\lambda }^{k}z$, and no positive power is the identity since $\lambda >0$ and $\lambda \ne 1$. Conjugation by the chosen isometry preserves order of a group element, so the original composition has infinite order. Reversing the order of the reflections replaces $\lambda$ by ${\lambda }^{-1}$ and gives the same conclusion.

The triangle inequality bounds the distance sum below by the separation of the two foci, which is two. Equality holds exactly for points on the closed segment joining them. Hence the locus isBoth endpoints are included. It is the segment on the positive imaginary axis shown in the right-hand panel above.

Put $s=t^2-t^{-2}$. Part (b)(ii) gives $u=s+2i$, soThe parameter $s$ increases strictly from $-15/4$ to $15/4$. Therefore the imaginary part $Y=4s$ increases from $-15$ to $15$, while the real part is $X=s^2-4$. The endpoints areThe real axis crossing is at $s=0$, giving $\boxed{(-4,0)}$ at $t=1$. The two imaginary axis crossings satisfy $s^2=4$, givingIn order, the trajectory passes from its lower endpoint through $(0,-8)$, then the leftmost point $(-4,0)$, then $(0,8)$, to its upper endpoint. It is a right-opening parabola segment, traversed continuously upward as shown in the right panel.

Squaring givesThus $\mathrm{Im}u=2$ is constant, while $d(\mathrm{Re}u)/dt=2t+2t^{-3}>0$. The trajectory is the horizontal line segmentIt crosses the imaginary axis at $\boxed{(0,2)}$, when $t=1$, and never crosses the real axis. The middle panel shows the rightward direction.

For $z=e^{i\theta }$, $w=i\tan (\theta /2)$, which is purely imaginary; $z=-1$ supplies the point at infinity. Conversely, the inverse map is $z=(1+w)/(1-w)$. Requiring $|w|=1$ is equivalent to $|z-1|=|z+1|$, hence $\boxed{\mathrm{Re}z=0}$, including the appropriate infinite point in the extended-plane formulation. The unit circle and imaginary axis are exchanged, but the real axis is not.

Choose a continuous argument $\theta$ on a simply connected domain avoiding zero, and put $r=\sqrt{x^2+y^2}$. With $u=x\log r-y\theta$, direct differentiation givesHere ${\theta }_x=-y/r^2$, ${\theta }_y=x/r^2$; these identities specify the branch needed in the written arctangent. Integrating the Cauchy-Riemann equations givesThe literal principal $\arctan (y/x)$ works only separately on suitable half-planes. It cannot stand for a single continuous argument across the imaginary axis.

The delta function in the specified convolution kernel collapses the ${\mathbf{k}}^{\prime }$ integral to $-i\gamma G$. Set $\lambda =e\gamma$ and ${\omega }_0^2=k^2+m^2$. ThenThere is a change of assumptions here: the specified kernel is the transform of $V=-i\gamma$, which is not real. These later parts define a formal analytic continuation to an effective damped equation, $[{\partial }_{\tau }^2+2\lambda {\partial }_{\tau }+{\omega }_0^2]G=\delta (\tau )$; they cannot simultaneously describe the earlier real external potential. The following pole analysis uses that continued operator.

Completing the square gives $D(\omega )=-(\omega +i\lambda {)}^2+({\omega }_0^2-{\lambda }^2)$, soFor $e>0$, hence $\lambda >0$, the underdamped regime has two poles at $\pm \Omega -i\lambda$, with $\Omega =\sqrt{k^2+m^2-{\lambda }^2}>0$. Both lie below the real axis. In the overdamped regime they lie on the imaginary axis at $-i(\lambda \mp \kappa )$, where $\kappa =\sqrt{{\lambda }^2-k^2-m^2}<\lambda$ for $k^2+m^2>0$; both are again below the real axis. At equality they coalesce into a double pole at $-i\lambda$. If $m<|\lambda |$, the crossover occurs at $k_c=\sqrt{{\lambda }^2-m^2}$; increasing $k$ moves the two imaginary poles together and then apart horizontally. If $m>|\lambda |$, every $k$ is underdamped; if equal, only $k=0$ is critical.

The sign of $e$ is not specified: for $e<0$ all nonzero-frequency poles are instead in the upper half-plane, describing antidamping; for $e=0$ they lie on the real axis and a causal prescription is required. When $m=k=0$, one pole is at zero even for nonzero $\lambda$ and a real-axis prescription is again needed. The figure uses illustrative $\lambda =1$, not supplied numerical data.