For a double pole at $z_0\in \{i,-i\}$, differentiate the remaining holomorphic factor:Since $z_0^2=-1$ and $1+z_0+z_0^2=z_0$, this simplifies toOn the selected branch $\sqrt{i}=e^{i\pi /4}$ and $\sqrt{-i}=e^{3i\pi /4}$, givingThe residue theorem therefore givesThe equality $C=2I$ uses the limiting contour integrals established in (c).

Use a counterclockwise keyhole contour around the positive real branch cut, with small radius $\epsilon$ and large radius $R$. On the upper bank $\arg z=0$, so the integral tends to $I$. On the lower bank $\arg z=2\pi$, $z^{1/2}=-\sqrt{x}$, giving the negative of the upper-bank integrand. That bank is traversed from $R$ to $\epsilon$, so its contribution also tends to $I$.

The outer circle contributes $O(R^{-3/2})$, since $f=O(R^{-5/2})$ and its length is $O(R)$. The small circle contributes $O({\epsilon }^{1/2})$, since $f=O({\epsilon }^{-1/2})$. Both vanish. HenceThe contour encloses both double poles $\pm i$ but excludes the branch point.

Factor the polynomial denominator as $(z-i{)}^2(z+i{)}^2$. The numerator has values $i$ and $-i$ at the two roots, so neither cancels. ThereforeThe factor $z^{-1/2}$ gives a branch point at zero, with local behaviour $f(z)\sim z^{-1/2}$. Infinity is also a branch point, since in the coordinate $w=1/z$ the leading behaviour is $f(1/w)\sim w^{5/2}$. A single-valued branch requires a branch cut joining these points. Away from that cut and the two poles, the selected branch is holomorphic. For the integral below, use $0<\arg z<2\pi$, putting the cut along the positive real axis.

A function is an analytic function at $z_0$ if it is holomorphic throughout some open neighbourhood of $z_0$, equivalently if it has a convergent power series there. In real components $g=u+iv$, continuous first partial derivatives in a neighbourhood together with the Cauchy-Riemann equationsthroughout that neighbourhood are sufficient, and for an analytic function they hold there. Satisfying these equations only at the single point is not enough to establish analyticity in a neighbourhood.

Apply the product rule to the vector field $\Phi \nabla \Psi -\Psi \nabla \Phi$:The gradient cross terms cancel. Integrating and applying the divergence theorem with outward unit normal $\mathbf{n}$ proves Green's second identity:Choose the Dirichlet Green function $G_D(\mathbf{r},{\mathbf{r}}^{\prime })$ with ${\Delta }_{\mathbf{r}}{G}_D=\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$ and $G_D=0$ on $S$. Taking $\Psi =G_D$ and using the Dirac delta distribution, the left side becomes $\Phi ({\mathbf{r}}^{\prime })$ while the second boundary term vanishes. HenceThe plus sign is tied to the convention $\Delta G_D=+\delta$, whose free-space fundamental solution is negative in three dimensions. The singularity may equivalently be handled by excising a small sphere about ${\mathbf{r}}^{\prime }$ before applying the smooth identity.

Treat each angular Fourier mode independently. Boundedness at the axis excludes the negative radial powers and the logarithm inside; decay at infinity excludes the positive powers, constant and logarithm outside. For the cosine mode writeContinuity at $R$ gives $B_n=A_n{R}^{2n}$. The jump of the normal derivative is thenThus $A_n=a_n{R}^{1-n}/(2n)$ and $B_n=a_n{R}^{n+1}/(2n)$. The sine mode has the same relation with $b_n$. Therefore the potential of a charged cylindrical shell with angular Fourier data isThere is no angular mean term: the given surface charge density has zero average. The interior constant is zero because its boundary value must match the vanishing exterior mean. A nonzero mean charge per unit length would instead produce a logarithmic electrostatic potential, incompatible with the specified decay at infinity.

A single-valued potential must be $2\pi$-periodic in $\theta$. For a separated solution $\Phi =F(r)H(\theta )$, the Laplace equation in polar coordinates becomesThe periodic angular eigenvalue problem has $\mu =n^2$, $n\ge 0$: its nonconstant eigenfunctions are $\cos n\theta ,\sin n\theta$. Negative $\mu$ gives exponentials that cannot be periodic, and $\mu =0$ gives only a constant angular factor. For $n\ge 1$, the radial Cauchy-Euler equation $r^2{F}^{\prime \prime }+r{F}^{\prime }-n^{2}F=0$ gives $F=A{r}^n+B{r}^{-n}$. At $n=0$ the two radial solutions are $1$ and $\log r$. Superposition therefore givesThe expansion is interpreted on an annulus where the harmonic function is defined; particular boundary conditions determine its constants and convergence region. Single-valuedness is essential: allowing a cut in the angular domain would also permit nonperiodic angular solutions.

For any cyclic subgroup $H=⟨s⟩$ of order $n\in \{1,2,4\}$, defineBoth preserve products because scalar phases and planar rotations add their angles. Each sends the generator to an element of exact order $n$, so both have trivial kernel and are faithful representations.

For $⟨U⟩\cong C_4$, take $s=U$: the generator images are $i$ and $R(\pi /2)$. For each order-two subgroup take its nonidentity element $s$; the images are $-1$ and $-I_2$. For the trivial subgroup the unique images are $1$ and $I_2$, which are faithful on a one-element group.

The one-dimensional representation of $C_4$ here must use complex scalars. There is no faithful one-dimensional real representation of $C_4$, since a nonzero real scalar of finite order is $1$ or $-1$, and neither has order four. The two-dimensional representations above are real.

The complete subgroup list isThus there are ten subgroups: one trivial, five of order two, three of order four, and the whole group. The five order-two subgroups are mutually isomorphic. The two noncyclic order-four subgroups are mutually isomorphic Klein four-groups; neither is isomorphic to the cyclic group of order four.

For completeness, Lagrange's theorem permits only orders $1,2,4,8$. An order-four subgroup containing $U$ or $U^3$ must be $⟨U⟩$. Otherwise it can contain only $I,U^2$ and elements $U^{j}V$. The product of two distinct such elements is $U^{j-k}$. To avoid generating $U$ or $U^3$, their exponents must differ by two, giving exactly the two listed Klein four-groups. The order-two subgroups come from the five involutions $U^2$ and $U^{j}V$. This proves completeness.

The character of a representation is the trace of its representing matrix. For the rotation powers,Each $U^{j}V$ has a planar reflection block of trace zero and third diagonal entry $-1$, giving character $-1$. In the chosen element order,Equivalently its values on the conjugacy classes $\{I\}$, $\{U^2\}$, $\{U,U^3\}$, $\{V,U^{2}V\}$, $\{UV,U^{3}V\}$ are $3,-1,1,-1,-1$.

The relations in (a) reduce every word in the generators to $U^a{V}^b$, with $0\le a<4$ and $b\in \{0,1\}$. These eight matrices are distinct: the $b=0$ elements have third diagonal entry $1$ and the $b=1$ elements have entry $-1$, while the four planar blocks are distinct in each family. Their multiplication iswith exponents reduced modulo four and two. This proves closure. The identity element is $U^0{V}^0$, the inverse of $U^a$ is $U^{-a}$, and each $U^{a}V$ is its own inverse. Associativity is inherited from matrix multiplication, proving the group axioms. This is the eight-element dihedral group $D_8$.

Set $g_j=U^j$ and $g_{j+4}=U^{j}V$ for $j=0,1,2,3$, consistently with the two labels specified in the question. The row-times-column multiplication table isThe rotation powers commute with one another, and $U^a$ commutes with $U^{c}V$ exactly when $a$ is even. Two elements $U^{a}V,U^{c}V$ commute exactly when $a-c$ is even. These statements also check the symmetric portions of the table.

The upper-left block of $U$ is the planar rotation matrix $R(\pi /2)$, and $U$ fixes the third coordinate. ThereforeThe diagonal entries of $V$ square to one, so $V^2=I$. Direct conjugation gives $VUV=U^{-1}$, hence $V{U}^{p}V=U^{-p}$ for every integer $p$. ConsequentlyWhen $p\mathrm{mod}4=0$, the last exponent is four, which is equivalent to zero because $U^4=I$.