Using the conjugate transpose and the Hermitian matrix assumptions,Thus $ABA$ is a Hermitian matrix, whether or not $A$ and $B$ commute.

Take $\boxed{c=1}$. Indeed the conjugate transpose isso the anticommutator is a Hermitian matrix. The original PDF, p. 6, supplies the factor $BA$ that is corrupted in the local OCR.

The conjugate transpose reverses the order of matrix multiplication. Since $A$ and $B$ are Hermitian matrices,Consequently $\boxed{AB\text{ is Hermitian}⟺AB=BA}$: the condition is exactly that the two matrices commute.

Iterating the Gamma function recurrence gives a meromorphic continuationNear $z=-m$, the numerator is analytic and tends to $\Gamma (1)=1$. Exactly one denominator factor, $z+m$, vanishes, while all other factors tend to the nonzero product $(-1{)}^{m}m!$. ThereforeThus $\boxed{z=0,-1,-2,\dots \text{ are all simple poles}}$. More precisely, the Gamma function residue at a nonpositive integer is $\boxed{{\mathrm{Res}}_{z=-m}\Gamma (z)=(-1{)}^m/m!}$. Taking $m$ large enough to put $z+m+1$ in the original right half-plane also shows that this continuation introduces no other finite poles.

For real $z>0$, apply integration by parts to the gamma integral with $u=t^z$ and $dv=e^{-t}dt$:The boundary term vanishes: $t^z\to 0$ at zero, and exponential decay dominates the power at infinity. This proves the Gamma function recurrence

Substituting $z=1$ into the defining gamma integral givesBoth endpoints are convergent, so no continuation of the Gamma function is needed here.

For the upper Apollonius circle, let its constant ratio be $P>1$. Matching the center and radius from part (i) givesSubtracting their squares yields $\boxed{b=\sqrt{y_0^2-R^2}}$. AlsoThe relevant inverse hyperbolic cosine is positive. Now $\log |(z+ib)/(z-ib)|$ is the real part of a local holomorphic logarithm, hence a harmonic function away from $z=\pm ib$. The point $ib$ is inside the excluded disk, because $0<y_0-b<R$, while $-ib$ is below the real axis. Thus neither singularity lies in the physical domain.

On $y=0$ the numerator and denominator have equal modulus, so the logarithm is zero. On the circular boundary its value is $\log P=\mathrm{arcosh}(y_0/R)$. Therefore the harmonic potential between a grounded line and a circular boundary isThis verifies Laplace's equation and both prescribed Dirichlet boundary conditions. It also tends to zero at infinity. A uniqueness assertion on this unbounded domain would require an additional boundedness or infinity condition; the displayed function is the natural decaying solution requested.

Write $z=x+iy$ and square the modulus equation. It becomesFor $P\ne 1$, completing the square gives the Apollonius circleHenceThe printed statement needs the exception $\boxed{P=1⟹y=0}$: equal distances from $ib$ and $-ib$ give their perpendicular bisector, a straight line rather than a finite-radius circle. For $P>1$ the center lies in the upper half-plane; this is the case needed in part (ii). The original PDF confirms the numerator $z+ib$, correcting the duplicated, corrupted OCR formula.

The sum of two odd integers plus one is odd, so the operation is closed. Both associations of a triple give $a+b+c+2$, proving associativity. Its identity element is $-1$, and the inverse element of $a$ is $-a-2$, also odd, since $a+(-a-2)+1=-1$. Thus it is an abelian group.

More explicitly, shifted addition on odd integers is ordinary integer addition in different coordinates:This bijection is a group isomorphism to $(\mathbb{Z},+)$.

The substitution $F(a)=e^a$, with $F(-\infty )=0$, is a bijection onto $[0,\infty )$ and obeysThus the log-sum-exp function transfers nonnegative addition to the given set. It is closed, associative and commutative, with identity element $-\infty$. But a finite $a$ has $e^a>0$, and adding a nonnegative number cannot give zero, so no finite $a$ has an inverse element. Therefore it is a commutative monoid but not a group.

The operation is not defined for every pair in the proposed set: choosing $a=1,b=-1$ makes its denominator zero. Thus it is not even a binary operation on $\mathbb{C}\setminus \{0\}$, and is not a group. In particular, removing zero from the domain does not prevent the sum in the denominator from vanishing.

Subtraction fails associativity: $(0\ast 0)\ast 1=-1$ but $0\ast (0\ast 1)=1$. Thus it violates the group axioms and is not a group. It has a right identity $0$, but no element can be a two-sided identity element.

Contract the isotropic expression from (i). Since ${\delta }_{ii}=3$ and ${\delta }_{ij}{\delta }_{ij}=3$,The value in (ii) gives $A=4\pi /105$, so the isotropic fourth moment of a ball isFor example ${\int }_B{x}_1^{4}dV=4\pi /35$ and ${\int }_B{x}_1^2{x}_2^{2}dV=4\pi /105$, illustrating the factor of three between these two fourth moments.

The contraction in the PDF is $I_{iijj}$. By Einstein notation, its integrand is $({\sum }_i{x}_i^2)({\sum }_j{x}_j^2)=r^4$. In spherical coordinates,

The integration domain is invariant under every rotation matrix, and its volume element is unchanged. The substitution $\mathbf{y}=R\mathbf{x}$ therefore provesThus $I$ is an isotropic tensor. It is also completely symmetric in its four indices, because the coordinate factors commute. In the general isotropic form, exchanging indices $j,k$ equates $\alpha$ and $\beta$, and exchanging $j,l$ equates $\alpha$ and $\gamma$. Therefore

Use three components of the proposed isotropic tensor form. Since $T_{1122}=0$, one gets $\alpha =0$. Since $T_{1212}={ϵ}_{123}^2=1$, one gets $\beta =1$. Finally $T_{1221}={ϵ}_{123}{ϵ}_{213}=-1$ gives $\gamma =-1$. Hence contraction of two Levi-Civita symbols gives

Under an orthogonal matrix $R$, the Levi-Civita symbol satisfiesUse this twice, and use $R_{mc}{R}_{mf}={\delta }_{cf}$ for the contracted index. It follows thatTherefore $\boxed{T_{ijkl}^{\prime }=R_{ia}{R}_{jb}{R}_{kc}{R}_{ld}{T}_{abcd}=T_{ijkl}}$: it is a fourth-rank isotropic tensor. Although a single alternating symbol is a pseudotensor under reflections, the two determinant factors multiply to one, so this product is an ordinary tensor even for improper orthogonal transformations.

Reflecting repeatedly in the two parallel planes creates an infinite method of images. For ${\mathbf{r}}^{\prime }=(x^{\prime },y^{\prime },z^{\prime })$, introduceThe Dirichlet Green function between parallel planes isEach bracket is $O(n^{-2})$, so the paired series converges absolutely away from the sources. At $x=-1$, reflection pairs the positive image indexed by $n$ with the negative image indexed by $-n$. At $x=1$ it pairs that positive image with the negative image indexed by $1-n$. The distances agree in each pairing, making both boundary values zero. This can be checked with symmetric finite partial sums first; the unmatched end terms tend to zero, so no subtraction of separately divergent series is needed. Only ${\mathbf{r}}_0^{+}={\mathbf{r}}^{\prime }$ lies within the slab, so ${\Delta }_{\mathbf{r}}{G}_{\mathrm{slab}}=\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$. This is the usual Green function decaying at large transverse separation; such a condition selects the physical solution in the unbounded region.

Let $\overline{{\mathbf{r}}^{\prime }}=(x^{\prime },y^{\prime },-z^{\prime })$ be the reflection of the source in the equatorial plane. For a source in the upper half-ball, the Dirichlet Green function inside a hemisphere isBoth terms vanish on the spherical boundary. On the flat boundary $z=0$, the two source distances are equal, and the distances to their spherical image points are also equal, so the difference vanishes there too. The reflected physical source is in the lower half-ball and both inverted image points are outside the sphere. Therefore the only source singularity in the desired region is the original one, with coefficient $+1$ in $\Delta G$. This combines reflection with the spherical method of images.

Write $s=|{\mathbf{r}}^{\prime }|$ and ${\mathbf{r}}^{\prime \prime }=a^2{\mathbf{r}}^{\prime }/s^2$. If $|\mathbf{r}|=a$, expansion of squared Euclidean distances givesConsequently $|\mathbf{r}-{\mathbf{r}}^{\prime \prime }|=(a/s)|\mathbf{r}-{\mathbf{r}}^{\prime }|$, and the two terms in $H$ cancel. This method of images gives the Dirichlet Green function inside a sphere:The image point is outside the sphere, so only the physical source contributes a Dirac delta distribution inside. As $s\to 0$, the image position tends to infinity and its coefficient also diverges, but their ratio has the finite limitThusThe apparent singularity of the image construction at a central source is removable; the physical source singularity at $\mathbf{r}=0$ remains.