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If $f$ has no explicit $x$ dependence, the chain rule givesConsequentlyon an Euler-Lagrange equation solution. Thus the Beltrami identity supplies the first integralThe opposite sign is an equally valid convention. The conserved value can be zero for particular integrands or paths; absence of explicit $x$ alone does not guarantee a nonzero numerical value. The useful statement is that this expression is constant along an extremal.

For fixed endpoint values, a stationary differentiable path obeys the Euler-Lagrange equationIt follows by taking the first variation of the functional and integrating the term in $\delta y^{\prime }$ by parts; $\delta y$ vanishes at the endpoints.

At the smaller indicial root $\sigma =1/2$, the $k=1$ equation is $0=0$: it leaves $a_1$ free. For $k\ge 2$, the valid recurrence isChoosing $a_1=0$ makes every later coefficient zero. The resulting terminating Frobenius series is thereforeSubstitution gives $y^{\prime }=a_0/(2\sqrt{x})$ and $y^{\prime \prime }=-a_0/(4x^{3/2})$; the three terms in the equation sum to $a_0\sqrt{x}[-(1+x)-2+(x+3)]=0$.

Termination is a permitted choice, not a condition forced on every smaller-root solution. Retaining the free $a_1$ adds the solution in part (e), so the general real solution on $x>0$ is $\boxed{y=\sqrt{x}[C_1+C_2\log (1+x)]}$. This is exactly the freedom explained by undetermined coefficient at Frobenius resonance. On either negative interval avoiding $-1$ and $0$, analogous real formulas use $\sqrt{|x|}$ and $\log |1+x|$.

The Taylor series of the logarithm givesTherefore the closed form of the larger-root Frobenius solution isThe series derivation holds initially for $0<x<1$; the closed form extends as a real solution to all $x>0$. To check it without series, put $y=\sqrt{x}v(x)$ in the original equation. All undifferentiated $v$ terms cancel and the remaining expression is $4x^{5/2}[(1+x)v^{\prime \prime }+v^{\prime }]$. Hence $[(1+x)v^{\prime }{]}^{\prime }=0$, and $v=\log (1+x)$ indeed solves it. Its logarithm is analytic near zero, so this is not a forced $\log x$ companion term at the Frobenius point.

For the larger indicial root $\sigma =3/2$, the recurrence in part (b) becomesNo factor being cancelled is zero in this range, soThus the resulting Frobenius solution isThere is exactly one free parameter, $a_0$, setting the overall scale. The analytic series factor has radius of convergence one. For a real-valued expression take $x>0$; a complex version requires a local square-root branch.

Because $a_0\ne 0$, the lowest-power equation from part (b) requires the indicial equationThe two candidate indicial roots are thereforeTheir difference is the integer one, so the smaller-root coefficient equation at $k=1$ needs examination rather than automatic division.

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.