Write $g=u+iv$ and let its constant modulus be $C$. If $C=0$, then $g=0$. For $C>0$, differentiating $u^2+v^2=C^2$ with respect to $x$ and $y$, then applying the Cauchy-Riemann equations, givesThusThe coefficient matrix has determinant $-(u^2+v^2)=-C^2\ne 0$, forcing $u_x=v_x=0$. The Cauchy-Riemann equations then also give $u_y=v_y=0$. Therefore $u,v$ are constant on a connected domain andThis proves the constant-modulus holomorphic function result. If a domain is disconnected, the conclusion applies separately to each connected component.

The Cauchy-Riemann equations are $\boxed{u_x=v_y,u_y=-v_x}$. They give the gradient identityAt a regular intersection, where $f^{\prime }(z)\ne 0$, the two nonzero gradients are normal vectors to the two level sets, so their tangent directions are perpendicular. This is the orthogonality of regular real and imaginary level curves.

The regularity qualification is necessary: $|\nabla u|=|\nabla v|=|f^{\prime }|$, so at a critical point both normals vanish and the curves need not have unique tangents. For example $f=z^2$ gives diagonal lines for $u=0$ and coordinate axes for $v=0$, all meeting at the origin. Thus the conclusion is $\boxed{\text{right-angle intersection wherever }{f}^{\prime }(z)\ne 0}$, rather than an unrestricted claim at critical points.

Since both real matrices are antisymmetric, $A^{†}=-A$ and $B^{†}=-B$. Thus $U^{†}=-A+iB$. Expand the two unitary identities:Subtracting and adding give $\boxed{AB=BA,A^2+B^2=-I}$. Because $A$ has simple eigenvalues, each of its complex eigenspaces is one-dimensional. Commutation makes each one invariant under $B$: if $Av=av$, then $A(Bv)=a(Bv)$. Hence $Bv=bv$ for some scalar $b$. The complete eigenbasis of $A$ is therefore a common eigenbasis, which also diagonalizes $U$.

The skew-Hermitian decomposition of a unitary matrix is explicit:If $Uv=uv$, unitarity gives $U^{†}v=u^{-1}v=\overline{u}v$. ConsequentlyEquivalently, for $u=e^{i\theta }$ the paired eigenvalues are $\boxed{a=i\sin \theta ,b=-i\cos \theta }$. The hypotheses can only occur in even dimension: $U^T=-U$ and $U$ is invertible, whereas an odd-dimensional skew-symmetric matrix has zero determinant.

To avoid confusing the diagonalizing matrix with the exponential, write $A=WD{W}^{†}$ from part (b). The convergent matrix exponential series givesEach ${\lambda }_j$ is purely imaginary, so $|e^{{\lambda }_j}|=1$. The diagonal exponential is therefore a unitary matrix, andHence the matrix called $U=\exp (A)$ in this part is unitary. The series identity $(e^A{)}^{†}=e^{A^{†}}=e^{-A}$ gives the same conclusion directly and also covers repeated eigenvalues.

Let $A{v}_i={\lambda }_i{v}_i$ and $A{v}_j={\lambda }_j{v}_j$. The skew-Hermitian matrix property giveswhere the last equality uses part (a). Distinct eigenvalues therefore imply $\boxed{v_i^{†}{v}_j=0\text{ for }i\ne j}$. Normalize each eigenvector to length one. Since there are $n$ distinct eigenvalues, these vectors form an orthonormal basis.

Put them in the columns of $U$. Then $U$ is a unitary matrix and $AU=UD$, so

A skew-Hermitian matrix satisfies $A^{†}=-A$, where $†$ denotes the conjugate transpose. If $Av=\lambda v$ with $v\ne 0$, thenSince $v^{†}v>0$, this gives $\boxed{\overline{\lambda }=-\lambda ,\text{ so }\lambda \in i\mathbb{R}}$.

A unitary matrix satisfies $U^{†}U=U{U}^{†}=I$. It preserves the Hermitian inner product and norm. Thus if $Uv=\mu v$,and therefore $\boxed{|\mu |=1}$. Zero is allowed among the purely imaginary eigenvalues in the first statement, but not among the unitary eigenvalues.

Expand the quadratic form of this Gram matrix:For $\mathbf{c}\ne 0$, linear independence of the real vectors implies ${\sum }_i{c}_i{\mathbf{x}}_i\ne 0$, so its squared Euclidean norm is strictly positive. Thusproving the positive definiteness of a Gram matrix in the independent-vector case.

The diagonalization in part (b) gives $M^k=P\mathrm{diag}(a^k,b^k,c^k)P^{-1}$. HenceSufficiency follows from the scalar powers tending to zero; necessity also follows immediately from the diagonal entries of $M^k$. Under these conditions the matrix geometric series converges, and its sum isWriting $A_0=1/(1-a)$, $B_0=1/(1-b)$ and $C_0=1/(1-c)$, all entries are explicitlyFor example, $B_0-A_0=(b-a)/[(1-a)(1-b)]$ and $C_0-B_0=(c-b)/[(1-b)(1-c)]$. These expressions remain valid when parameter values coincide.

A single eigenbasis works for every choice of the parameters. The columns $v_1=(1,0,0{)}^T$, $v_2=(1,1,0{)}^T$, $v_3=(1,1,1{)}^T$ satisfyThey are linearly independent because their column matrix has determinant one. Thus the diagonalization isThis formula includes all coincidences among $a,b,c$, so no special repeated-eigenvalue cases need separate bases.

A square matrix is a diagonalizable matrix over the chosen field when there are an invertible matrix $P$ and a diagonal matrix $D$ such that $A=PD{P}^{-1}$. If $v_j$ is column $j$ of $P$ and $D_{jj}={\lambda }_j$, the identity $AP=PD$ gives $A{v}_j={\lambda }_j{v}_j$. Invertibility makes these $n$ columns linearly independent eigenvectors.

Conversely, put $n$ linearly independent eigenvectors in the columns of $P$ and their eigenvalues in $D$. Then $P$ is invertible and $AP=PD$, givingRepeated eigenvalues are allowed; the requirement is a complete eigenbasis, not distinctness of the eigenvalues.

The Parseval identity for this Fourier transform convention isApply it to $h=g$. Its squared norm is $2{\int }_0^1(1-x{)}^{2}dx=2/3$, soUnder $k=2x$, the last integrand times $dk$ becomes $2{\sin }^{4}xdx/x^4$. ConsequentlyThe integrand has continuous value $1$ at zero and decays integrably at infinity, so this is an ordinary convergent improper integral.

Direct evaluation gives the first Fourier transform:For the even triangular function, the odd sine part integrates to zero. Two elementary integrations by parts or one evaluation of the cosine integral giveThe values at zero are the respective areas, agreeing with the removable limits. Therefore $\tilde{g}={\tilde{f}}^2$. By the convolution theorem and uniqueness of the Fourier transform,There is also a direct interpretation: the convolution is the length of overlap of the intervals $(-1/2,1/2)$ and $(x-1/2,x+1/2)$, giving exactly that triangular profile.

For integrable functions $f,g$, Fubini's theorem justifies exchanging the two integrals in the Fourier transform of the convolution. Set $s=x-t$ to obtainThis proves the convolution theorem with the stated normalization:

Using the causal Green's function,Put $\xi =xs$. Since $\alpha >-5$ is equivalent to $r<3$, the improper integral converges and givesThe elementary identity ${\int }_0^1{s}^{p-1}(-\log s)ds=1/p^2$ follows by differentiating ${\int }_0^1{s}^{p-1}ds=1/p$ for $p>0$. Direct substitution checks $L(x^3)=(\alpha +5{)}^{2}x/4$, and the displayed solution and its first derivative vanish at zero.

To give all solutions allowed by the printed boundary conditions, retain the singular-endpoint qualification from part (b):Thus the causal integral gives a particular member of a two-parameter family in the first range and the unique solution in the second.

Choose the causal Green function convention $L_{x}G(x,\xi )=\delta (x-\xi )$ for each source point $\xi >0$, with $G(x,\xi )=0$ when $x<\xi$. This automatically gives the two zero-origin limits. The kernel is continuous at $x=\xi$, and integrating its equation across the source requires the unit derivative jump ${\partial }_{x}G({\xi }^{+},\xi )-{\partial }_{x}G({\xi }^{-},\xi )=1$.

For $x>\xi$, the homogeneous basis from part (a) gives a continuous kernel proportional to $x^r\log (x/\xi )$. Its derivative at $x=\xi$ is ${\xi }^{r-1}$, so the normalization is ${\xi }^{1-r}$. Hence the causal Green function for a repeated-root Euler operator isThe same formula follows from $[y_1(\xi )y_2(x)-y_2(\xi )y_1(x)]/W(\xi )$ on the right of the source, with $W(\xi )={\xi }^{-\alpha }$.

The singular endpoint matters for uniqueness. If $\alpha <-1$, then $r>1$, and both $x^r$ and $x^r\log x$, together with their first derivatives, vanish at zero. Adding any source-dependent linear combination of these homogeneous solutions to $G$ preserves the two stated origin conditions. Thus those conditions alone do not determine a unique Green's function in that parameter range. The causal requirement used here selects the displayed kernel. For $\alpha \ge -1$ no nonzero homogeneous solution satisfies both zero-origin limits, so this ambiguity is absent.

The homogeneous equation is a Cauchy-Euler equation. With the power-law ansatz $y=x^r$ on $x>0$, its indicial equation isThus $r=(1-\alpha )/2$ is a repeated indicial root. The two independent solutions are $x^r$ and $x^r\log x$, givingTheir Wronskian, with $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$, is $x^{2r-1}=x^{-\alpha }\ne 0$, verifying linear independence.

For separation of variables matching the boundary Fourier mode, take ${\psi }_0=X(x)e^{iky}$ with $X(0)=1$. Because ${\partial }_y^3{e}^{iky}=-i{k}^3{e}^{iky}$, the reduced ordinary differential equation isThereforeThe decay of a Fourier mode in a third-order spatial evolution equation requires a sign check: this satisfies the requested limit at positive infinity precisely when $\boxed{k>0}$. For $k=0$ it remains identically one, while for $k<0$ its amplitude grows exponentially. Hence for those nonpositive values there is no separated solution meeting both prescribed boundary conditions; the printed assumption that $k$ is merely real needs this additional restriction for decay.

The first-order condition $Q=0$, combined with $b_0=-i{\psi }_{0,y}$, givesDifferentiate this expression with respect to $y$ and insert it into $S=0$. The ${\psi }_1$ terms cancel, leavingThus the compatibility partial differential equation isIt is exactly ${\partial }_y(2{\psi }_{0,x}+i{\psi }_{0,yyy})=0$. Integrating with respect to $y$ therefore giveswhere the integration constant may be any function of $x$. Conversely, differentiating this last equation with respect to $y$ recovers the compatibility equation.

Insert the proposed leading relation into the two zeroth-order residuals. It givesand, since $b_{0,y}=-i{\psi }_{0,yy}$,Hence $\boxed{b_0=-i{\psi }_{0,y}⟹P=R=0}$. The same relation is forced by $P=0$ when $\omega \ne 0$; its $y$ derivative then automatically gives $R=0$.