For the normalized equation, $x_0$ is an ordinary point when both $p$ and $q$ are analytic near $x_0$. A singular point is a regular singular point when $(x-x_0)p(x)$ and $(x-x_0{)}^{2}q(x)$ extend analytically to $x_0$; equivalently $p$ has at most a simple pole and $q$ at most a double pole there.

Thus ordinary points admit ordinary power-series solutions, while the Frobenius method applies at a regular singular point. The definitions refer to the coefficients after dividing by the leading coefficient of $y^{\prime \prime }$.

Put $\zeta =z-3$ and apply the geometric series:ThusThe radius is the distance from the expansion centre $3$ to the nearest singularity $2i$, not the distance of that pole from the origin. On the boundary circle the series terms all have magnitude $1/\sqrt{13}$, so they do not approach zero: this particular series diverges at every boundary point.

Recognize that $z^2=(x^2-y^2)+2ixy$ and $e^{-z}=e^{-x}(\cos y-i\sin y)$. Their product has exactly the specified real part. Its imaginary part isOne can also recover it directly from the Cauchy-Riemann equations. Differentiating the specified real part givesIntegration of $v_y=u_x$ gives $v=v_0+h(x)$; the other equation $v_x=-u_y$ then forces $h^{\prime }(x)=0$. Thus the most general answer on a connected domain isIndeed the difference of any two analytic functions with the same real part has zero real part; its two Cauchy-Riemann equations force its imaginary part to be constant. This proves that no further analytic freedom is missing.

The Cauchy-Riemann equations giveAt a regular point, each level curve has tangent perpendicular to its nonzero gradient. The two gradients, and hence the two tangents, are orthogonal. Therefore regular level curves meet at right angles.

The qualification is necessary: $|\nabla u|=|\nabla v|=|f^{\prime }(z)|$, so at a critical point both gradients vanish and this argument gives no tangent direction. For example, for $f(z)=z^2$, the zero level curves of the real and imaginary parts are diagonals and coordinate axes; they meet at $45^{\circ }$ at zero. This is the regular-point statement of orthogonal level curves of harmonic conjugates.

For $f=u+iv$ analytic in $z=x+iy$, the Cauchy-Riemann equations areThey follow by computing the same complex derivative along the real and imaginary coordinate directions.

Use the principal coordinatesPart (c) puts the equation in the canonical formIf $c>2$, all three coefficients are positive: the surface is an ellipsoid with semiaxes $1/\sqrt{6}$, $1/\sqrt{3}$ and $1/\sqrt{2c-4}$ along the three eigenvectors listed in part (c).

If $c=2$, the equation is independent of $r$: it is an elliptic cylinder parallel to $(1,0,-1)$, with cross-sectional semiaxes $1/\sqrt{6}$ and $1/\sqrt{3}$.

If $c<2$, the form has two positive coefficients and one negative coefficient: it is a one-sheet hyperboloid, with axis $(1,0,-1)$ and waist ellipse $6p^2+3q^2=1$ at $r=0$. Its axial scale is $1/\sqrt{4-2c}$.

These are the three cases of the signature transition from an ellipsoid to an elliptic cylinder and a one-sheet hyperboloid. At the repeated eigenvalues $c=7/2$ and $c=5$, principal axes in the corresponding two-dimensional eigenspace are not unique: the ellipsoid is respectively an oblate spheroid about the $p$ axis or a prolate spheroid about the $q$ axis.

The second coordinate direction is already an eigenvector with eigenvalue three. The first and third coordinates form a symmetric block, whose sum and difference directions giveFor example, the sum of the block's diagonal and off-diagonal entries is $(1+c)+(5-c)=6$, while their difference is $(1+c)-(5-c)=2c-4$. Equivalently the characteristic polynomial is $(\lambda -6)(\lambda -3)(\lambda -2c+4)$.

The displayed basis works for every real $c$. At $c=7/2$, the eigenvalue three has eigenspace spanned by the second and third listed vectors; at $c=5$, the eigenvalue six has eigenspace spanned by the first and third. Every nonzero vector in a repeated eigenspace is an eigenvector, and any orthonormal basis of it may replace the displayed choices.

A real symmetric matrix has a real orthonormal eigenbasis. Put those eigenvectors into an orthogonal matrix $Q$, so $Q^{T}AQ=\mathrm{diag}(a_1,\dots ,a_n)$. With the change of coordinates $x=Q{x}^{\prime }$, or $x^{\prime }=Q^{T}x$,Thus the coefficients are the eigenvalues and the new coordinate directions are the eigenvectors. This orthogonal diagonalization preserves Euclidean lengths and angles while removing cross terms from the quadratic form.

By the spectral theorem, a Hermitian matrix has real eigenvalues and an orthonormal eigenbasis. Choose column vectors $u_j$ with $H{u}_j={\lambda }_j{u}_j$ and $u_i^{†}{u}_j={\delta }_{ij}$, and form $U=(u_1\cdots u_n)$. Then $U^{†}U=I$, so $U$ is unitary, and $HU=U\Lambda$ givesThe entries of $\Lambda$ are the real eigenvalues, repeated according to their multiplicities. Within a degenerate eigenspace, any orthonormal basis may be used. Equivalently $H=U\Lambda U^{†}$.

For the positive integer $k$ in the nilpotence condition, diagonalize $B=P\Lambda P^{-1}$. ThenEvery diagonal entry must vanish, so each ${\lambda }_j=0$. Consequently $\Lambda =0$ and $\boxed{B=0}$. This proves that a diagonalizable nilpotent matrix is zero.

Without diagonalizability a counterexample isIts only eigenvalue is zero but its kernel has dimension one, so it cannot have a basis of eigenvectors. It is the simplest nontrivial nilpotent Jordan block.

The entries of $M^{T}M$ are real, and the transpose rule gives $(M^{T}M{)}^T=M^{T}M$. Thus it is a real symmetric matrix, hence also a Hermitian matrix.

For any complex vector $v$,In particular, if $M^{T}Mv=\lambda v$ with $v\ne 0$,Thus $M^{T}M$ is positive semidefinite. Its zero eigenspace is precisely the kernel of $M$, since $\Vert Mv{\Vert }^2=0$ is equivalent to $Mv=0$. This is the Gram-matrix structure behind the nonnegative eigenvalues.

An invertible matrix is a square matrix for which there is a matrix $A^{-1}$ with $A{A}^{-1}=A^{-1}A=I$. Write its assumed diagonalization as $A=P\Lambda P^{-1}$, where $P$ is invertible and $\Lambda =\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$. ThenIf the determinant is nonzero, every ${\lambda }_j$ is nonzero, so ${\Lambda }^{-1}=\mathrm{diag}({\lambda }_j^{-1})$ exists and $A^{-1}=P{\Lambda }^{-1}{P}^{-1}$. If the determinant is zero, at least one eigenvalue is zero; its nonzero eigenvector is mapped to zero, which an invertible map cannot do. HenceThe equivalence holds even without diagonalizability, but the eigenbasis proves the stated case directly.

The adjoint operator $A^{†}$, when it exists, is defined byfor all vectors in the relevant domains. This convention takes the inner product to be conjugate-linear in its first argument. Nondegeneracy makes the adjoint unique.

For the standard complex inner product $⟨x,y⟩={\overline{x}}^{T}y$, the equality becomes ${\overline{x}}^T{\overline{A}}^{T}y={\overline{x}}^T{A}^{†}y$ for every $x,y$, soThus the matrix adjoint is the Hermitian conjugate, not merely the transpose unless the entries are real. In a basis with positive Hermitian Gram matrix $G$, the same definition instead gives $A_G^{†}=G^{-1}{\overline{A}}^{T}G$, making the dependence on the chosen inner product explicit.

Let $r(z)$ be the indicator of $[-1,1]$. The finite integral is exactly the convolution $p=q\ast r$. Its additional transform factor isApplying the convolution theorem and part (d),Equivalently this is $2(\sin k/k)[\sin (k/2)/(k/2){]}^2$. The value at zero also follows from $\int p=(\int q)(\int r)=1\times 2$.

Evenness means the negative half of the function is determined as well: this is the triangular function $q(x)=1-|x|$ on $|x|\le 1$, zero outside. Its transform is real and even:Integration by parts, or integrating the two terms separately, givesThe apparent singularity at zero is removable, and its limiting value is the area under $q$. This is the Fourier transform of a triangular function.

Differentiate the defining Fourier integral $n$ times with respect to $k$:For a nonnegative integer $n$, this givesIntegrability of $x^{n}f$ together with suitable lower moments justifies differentiation under the integral. The same multiplication-derivative identity holds more generally for tempered distributions.

Assume initially that $f,g$ are integrable, so Fubini's theorem justifies interchanging the integrals. Transform the convolution and put $u=x-z$:This is the convolution theorem. No factor $2\pi$ appears in this direction because the forward transform in the question is unnormalized.

For the stated Fourier transform convention, Fourier inversion isFor example, the pointwise formula holds at continuity points under the usual integrability hypotheses; the same normalization applies to the $L^2$ or distributional inverse.