For the Frobenius solution, differentiate term by term:The unshifted terms in the equation have coefficient $[4(n+\sigma )(n+\sigma -1)-4(n+\sigma )+3]a_n$. The extra factors of $x$ give the shifted contribution $[4(n+\sigma )(n+\sigma -1)+1]a_n{x}^{n+\sigma +1}$. Equating the coefficient of $x^{k+\sigma }$ for $k\ge 1$ gives the linear recurrence relationFactoring both coefficients is especially useful:The separate lowest-power equation is $(4{\sigma }^2-8\sigma +3)a_0=0$. Do not cancel $2k+2\sigma -3$ when it vanishes: that is the possible undetermined coefficient at Frobenius resonance.

After division by the leading coefficient, write $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$, whereThe only finite singular points of a differential equation are $\boxed{x=0\text{ and }x=-1}$. At zero, $xP=-1/(1+x)$ and $x^{2}Q=(x+3)/[4(1+x)]$ are analytic. At $-1$, $(x+1)P=-1/x$ and $(x+1{)}^{2}Q=(x+1)(x+3)/(4x^2)$ are analytic. The regular singular point criterion for a second-order equation therefore makes both points regular.

If infinity is included, it is a regular singular point at infinity as well: $P(x)=O(x^{-2})$ and $Q(x)=1/(4x^2)+O(x^{-3})$. Equivalently, with $t=1/x$ the transformed coefficients are $\tilde{P}=2/t+1/(1+t)$ and $\tilde{Q}=(1+3t)/[4t^2(1+t)]$, which satisfy the same regular-singular criterion at $t=0$.

Using the exponential form of the complex cosine, $\cos z=0$ is equivalent to $e^{2iz}=-1$. Thus the zeros are exactlyAt each point ${\cos }^{\prime }(z_n)=-\sin z_n\ne 0$, so the cosine has a simple zero. Cubing it givesThe order of a zero of a holomorphic function is therefore three at every zero. There are no finite singularities because this is an entire function.

To classify infinity, examine ${\cos }^3(1/w)$ at $w=0$. The identity ${\cos }^{3}z=(3\cos z+\cos 3z)/4$ shows that its Laurent series contains the nonzero negative-power coefficientsThere are infinitely many terms in the principal part, the criterion for an essential singularity. Therefore $\boxed{\infty \text{ is an essential singularity, not a pole.}}$ This is also the general characterization of a nonpolynomial entire function at infinity.

Insert the specified real part into the polar Cauchy-Riemann equations. The first gives $v_r=n{r}^{n-1}\sin (n\theta )$, and integration with respect to $r$ givesThe second equation then requires $v_{\theta }=n{r}^n\cos (n\theta )$, so $C^{\prime }(\theta )=0$. The remaining imaginary additive constant is fixed by $f(0)=0$. ThusThe resulting polynomial is holomorphic at the origin too, consistently extending the calculation made in polar coordinates with $r>0$.

For a holomorphic function with $z=r{e}^{i\theta }$, the chain rule givesWriting $f=u+iv$ and equating real parts and imaginary parts,For $r>0$, division by $r$ gives the Cauchy-Riemann equations in polar coordinates:The restriction $r>0$ reflects the coordinate singularity at the origin; the Cartesian equations have no such restriction.

Use $U^{†}=U^{-1}$ and take the Hermitian conjugate in the reverse product order:The factorization $I-U^{-1}=-U^{-1}(I-U)$ gives $(I-U^{-1}{)}^{-1}=-(I-U{)}^{-1}U$, while $I+U^{-1}=U^{-1}(I+U)$. ConsequentlyThe last equality uses the fact that $I+U$ commutes with $I-U$ and its inverse. Thus $B$ is a Hermitian matrix. The exclusion of the eigenvalue one is essential to define the inverse. Spectrally, an eigenvalue $e^{i\vartheta }\ne 1$ is sent to $-\cot (\vartheta /2)$, a real number. This is the inverse Cayley transform of a unitary matrix without eigenvalue one.

Part (a) makes each ${\lambda }_j$ real. In the eigenvector basis,and $|{\lambda }_j+ia{|}^2={\lambda }_j^2+a^2>0$. Hence $\boxed{A+iaI\text{ is invertible}}$.

All factors that are rational functions of $A$ commute. Taking the Hermitian conjugate reverses product order and givesThereforeso $\boxed{V\text{ is unitary}}$. This is a Cayley transform of a Hermitian matrix, with scale $a$ and the convention that an infinite real eigenvalue maps to one.

Applying it to each eigenvector yieldsThe eigenvectors are unchanged, and $|{\nu }_j|=1$ because numerator and denominator are complex conjugates. No ${\nu }_j$ equals one when $a>0$. The scalar map is injective on real $\lambda$, so the eigenspaces, including any degeneracies, are preserved exactly.

A complex square matrix $U$ is a unitary matrix if $\boxed{U^{†}U=U{U}^{†}=I}$. Equivalently, it preserves the complex inner product and hence the norm. If $Uv=\mu v$ with $v\ne 0$, thenDivision by the positive squared norm gives $\boxed{|\mu |=1}$. Thus every eigenvalue lies on the unit circle.

Let $A=A^{†}$ be a Hermitian matrix and $Av=\lambda v$ with $v\ne 0$. The scalar $v^{†}Av$ equals its own complex conjugate, becauseSince $v^{†}Av=\lambda v^{†}v$ and the squared norm $v^{†}v$ is positive, $\boxed{\lambda \in \mathbb{R}}$.

Now take $Au=\lambda u$ and $Av=\mu v$. The inner product satisfiesSubtracting gives $(\mu -\lambda )u^{†}v=0$. Thus distinct eigenvalues imply $\boxed{u^{†}v=0}$, proving orthogonality of the corresponding eigenvectors. Within a degenerate eigenspace, an orthonormal basis may be chosen, but arbitrary distinct vectors in that same eigenspace need not be orthogonal.

At the angles in part (f),Let the two diagonal matrices be $D=\mathrm{diag}(d_1,d_2,d_3)$ and $D^{\prime }=\mathrm{diag}(e_1,e_2,e_3)$. Multiplication givesThe vector condition is exactly $c{e}_2{d}_3=b$. For $c\ne 0$, the general form is thereforeEvery pair $p,q$ is attainable, for example with $D=I$ and $D^{\prime }=\mathrm{diag}(-p,b/c,-q)$. The transformation need not be orthogonal or invertible; the question asks only for diagonal rescalings with this one vector image. If $c=0$ and $b\ne 0$, no linear map can satisfy the condition. If $c=b=0$, the $(2,3)$ entry is arbitrary as well.

Applying the rightmost rotation matrix first givesEquality with ${\mathbf{e}}_y$ requires $\cos \theta =0$, so $\theta =\pi /2$ in the specified range. Then $\sin \varphi =1$ and $\cos \varphi =0$ giveIn the reverse order, $R_z$ first fixes ${\mathbf{e}}_z$, so $R_2{\mathbf{e}}_z=R_y(\pi /2){\mathbf{e}}_z={\mathbf{e}}_x$. Therefore $\boxed{R_2\mathbf{v}\ne {\mathbf{v}}^{\prime }}$. This explicitly demonstrates that rotations about different axes do not generally commute.

The Hermitian conjugate of a real matrix is its transpose, soTherefore $\boxed{M\text{ is anti-Hermitian for all }\theta ,\varphi .}$ It is also a real skew-symmetric matrix. It is Hermitian as well only in the exceptional case $M=0$, precisely when $R_1$ is symmetric. This includes the identity and half-turn cases from part (d), so the zero-matrix overlap should not be excluded.

With the same right-handed active rotation convention as $R_z$, the $y$ direction is fixed, ${\mathbf{e}}_z$ rotates towards ${\mathbf{e}}_x$, and ${\mathbf{e}}_x$ towards $-{\mathbf{e}}_z$. ThereforeIts columns give the images of the three Cartesian unit vectors, which also makes the signs of this rotation matrix explicit.

The $z$ direction is fixed, giving the eigenvalue one. The planar block has characteristic polynomialThus the full list of eigenvalues isThe last two are real precisely when $\sin \varphi =0$, so all three are real exactly for $\boxed{\varphi \in \pi \mathbb{Z}}$. At even multiples of $\pi$ the matrix is the identity and every vector is fixed. At odd multiples it is a half-turn: the axial eigenvalue remains $+1$, while the whole perpendicular plane has eigenvalue $-1$.

For other angles, the only real eigenvector direction is the rotation axis; the perpendicular plane is rotated rather than preserving an individual real direction. Its complex eigenvectors can be taken as $(1,-i,0{)}^T$ and $(1,i,0{)}^T$ for $e^{i\varphi }$ and $e^{-i\varphi }$, respectively. These represent the two complex conjugate planar eigenvalues.

The columns of $R_z$ are an orthonormal basis, since ${\cos }^2\varphi +{\sin }^2\varphi =1$ and the first two columns have zero dot product. Thus $R_z^T{R}_z=I$: it is an orthogonal matrix. Consequentlyand hence $\boxed{|{\mathbf{v}}^{\prime }|=|\mathbf{v}|}$. This is the preservation of the Euclidean norm by a rotation matrix.