Set $z=e^{i\theta }$. The integral becomesThe roots are $z=(-3\pm 2)/(2-i)$, only $z_{+}=-1/(2-i)$ lying inside. The derivative of the quadratic at $z_{+}$ is two, so the residue is $1/(2i)$. Consequently $\boxed{{\int }_0^{2\pi }d\theta /(3+2\cos \theta +\sin \theta )=\pi }$. The denominator is positive since $3>\sqrt{5}$, so no principal-value qualification is required.

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.

The identity $w-1=-2/(z+1)$ gives $\boxed{|w-1|=2/r}$ on $|z+1|=r$. Thus the image circle has center one and radius $2/r$. Points strictly inside the original circle satisfy $|w-1|>2/r$; the center itself maps to infinity. The interior therefore becomes the exterior together with infinity on the Riemann sphere, rather than the interior of the image circle. This inversion is a useful warning against assuming conformality preserves global inside/outside assignments.

For real $x\ne -1$, $w=1-2/(x+1)$ is real. Its finite image is the entire real axis except $w=1$, which is approached only as $x\to \pm \infty$. The point $x=-1$ maps to infinity. On the extended complex plane the extended real axis maps onto itself, with $z=\infty$ mapping to one.

A conformal map preserves local oriented angles and has a nonzero local scale factor. For an orientation-preserving complex map this is equivalent to holomorphy with $f^{\prime }(z)\ne 0$ at the point concerned. Infinitesimal figures are rotated and uniformly scaled there, but finite shapes need not be preserved: derivative magnitude and direction can vary across a region. Antiholomorphic maps preserve unoriented angles while reversing orientation; the convention being used must be stated.

Separate the $n=0$ term and pair $n$ with $-n$. For small $\alpha$,Compare the constant and quadratic coefficients with the supplied expansion after subtracting ${\alpha }^{-2}$. It follows thatThe paired expansion is uniformly summable on any sufficiently small closed interval of $\alpha$, so coefficient comparison is legitimate.

On the square, $|z|\ge N$. The identity $|\sin \pi (x+\alpha +iy){|}^2={\sin }^2\pi (x+\alpha )+{\sinh }^2\pi y$ bounds the sine away from zero on the vertical sides because $x=\pm N$; on horizontal sides it grows exponentially. Therefore the whole contour integral is $O(N^{-1})$ or smaller and vanishes. The residue theorem then giveswhich proves $\boxed{\sum _{n\in \mathbb{Z}}(-1{)}^n/(n-\alpha {)}^2={\pi }^2\csc \pi \alpha \cot \pi \alpha }$. Absolute convergence justifies the limiting residue sum.

The simple poles occur at $z=n-\alpha$, $n\in \mathbb{Z}$, since the derivative of the sine there is $\pi (-1{)}^n$. ThusThere is also a double pole at zero, not included in that list because $0<\alpha <1$. Its residue, needed for the next contour calculation, is the derivative of $\csc \pi (z+\alpha )$ at zero, namely $-\pi \csc (\pi \alpha )\cot (\pi \alpha )$.

For the standard meaning of a canonical transformation, the new coordinates must retain Hamiltonian form for every Hamiltonian, with the same physical function expressed in the new coordinates. Let $z=(q,p)$, $Z=F(z)=(Q,P)$, $S=DF$ and $J=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$. With $K(Z)=H(F^{-1}(Z))$, the chain rule givesAgreement with $\dot{Z}=J{\nabla }_{Z}K$ for arbitrary $K$ forces $SJ{S}^T=J$. Its matrix entries are precisely the Poisson brackets of the new coordinates, soThe qualification for every Hamiltonian matters. Preserving one Hamiltonian flow does not imply canonicality. For example, with two coordinate pairs and $H=(q_1^2+p_1^2)/2$, the invertible change $Q_1=q_1$, $P_1=p_1$, $Q_2=2q_2$, $P_2=p_2$ preserves the equations and the same physical Hamiltonian, but $\{Q_2,P_2\}=2$. Thus a definition requiring only one chosen Hamiltonian flow is too weak to prove the claimed bracket identities. The universal definition supplies the missing condition.

For $L_i={ϵ}_{ijk}{q}_j{p}_k$, the canonical Poisson brackets give $\{L_i,L_j\}={ϵ}_{ijk}{L}_k$. Consequently projections onto fixed axes $n,m$ obeyIf the two original projections are conserved and the axes are linearly independent, the preceding Jacobi identity argument makes the projection along $n\times m$ conserved as well. The vectors $n,m,n\times m$ form a basis of three-dimensional space, so every fixed-axis projection is a linear combination of these conserved quantities. Angular momentum is therefore conserved along every axis.

The Jacobi identity is $\{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\}=0$. If $f,g$ have no explicit time dependence and are conserved, $\{f,H\}=\{g,H\}=0$. Taking $h=H$ in that identity yields $\{H,\{f,g\}\}=0$, orThus the Poisson bracket of the two conserved functions is another conserved function; it need not be independent of the original ones.

For a phase-space function with no explicit time dependence, the chain rule and Hamilton's equations giveAn explicitly time-dependent function would instead satisfy $df/dt={\partial }_{t}f+\{f,H\}$.

The canonical Poisson bracket isThe coordinate derivatives are Kronecker deltas, so direct substitution gives

Let $\mathcal{E}=\sqrt{m^2{c}^4+c^2{\pi }^2}$. Hamilton's equations giveSubtracting $qd{A}_i/dt$ from the second equation gives the familiar Lorentz force, $\dot{\pi }=q(E+\dot{r}\times B)$, with $E=-\nabla \varphi -{\partial }_{t}A$ and $B=\nabla \times A$. This also checks the signs of both potential terms.

Minimal electromagnetic coupling changes the canonical momentum to $p=\gamma m\dot{r}+qA$. The mechanical momentum is $\pi =p-qA$. The Legendre transform yieldsThe cancellation of $q\dot{r}\cdot A$ in the Legendre transform does not remove magnetic effects: $A$ remains in the relation between canonical and mechanical momenta. The expression is the relativistic kinetic-plus-rest energy with an electrostatic potential-energy term, expressed in canonical variables.

The canonical momentum is $p=\gamma m\dot{r}$, where $\gamma =(1-{\dot{r}}^2/c^2{)}^{-1/2}$. Eliminating velocity givesIts Hamilton equations are $\dot{r}=c^{2}p/H$ and $\dot{p}=0$. In the limit $|p|\ll mc$, the Taylor expansion is $H=m{c}^2+p^2/(2m)+O(p^4/(m^3{c}^2))$, and $\dot{r}\simeq p/m$. The additive rest energy does not change the classical free-particle equations.

The canonical momentum is $p_i=\partial L/\partial {\dot{q}}_i$. If the velocity-momentum relation can be locally inverted, the Legendre transform defines $H(q,p,t)={\sum }_i{p}_i{\dot{q}}_i-L$, with the velocities expressed in terms of $q,p,t$. Its differential isThe Euler-Lagrange equations then yield Hamilton's equations,A singular velocity Hessian instead requires a constrained-Hamiltonian treatment.

In $D$ dimensions the antisymmetric angular momentum tensor has componentsThere are $D(D-1)/2$ independent rotation generators. Together with $D$ translation generators and one time-translation generator, the standard Noether charge enumeration is thereforeThe word independent here counts symmetry generators and tensor components; it needs qualification if it is meant as functional independence of all charges for only two particles in arbitrarily high dimension. In centre-of-mass coordinates, $L=R\wedge P+r\wedge p$. Consequently $P\wedge L\wedge L=0$, a nontrivial identity from dimension five onwards, since $(r\wedge p)\wedge (r\wedge p)=0$. Thus the displayed number is the symmetry-charge count; it is not an assertion that these functions have unrestricted values for a two-particle realization in every $D$.