Away from $y=\pm 1$, divide by the coefficient of $x^{\prime \prime }$ to obtain the normal formThus $\boxed{y=1\text{ and }y=-1}$ are the only finite singular points. For a second-order linear differential equation, $y_0$ is a regular singular point if $(y-y_0)P(y)$ and $(y-y_0{)}^{2}Q(y)$ extend analytically to $y_0$. At $y_0=1$ these functions areand at $y_0=-1$ they areEach pair is analytic at its respective point. Therefore $\boxed{\text{both finite singular points are regular}}$.

Put $w=z-1$ and $a=1-i$. Partial fraction decomposition givesBoth terms admit a geometric series at $w=0$. The local Laurent series is actually a Taylor series, since the center is regular:The nearest pole is $z=i$, a distance $|1-i|=\sqrt{2}$ from the center; the other pole is at distance $4$. Thus the requested local radius of convergence is $\boxed{R=\sqrt{2}}$.

For completeness, specifying a Laurent series without a region can also mean either of the other maximal annuli. Expanding the first fraction in inverse powers and the second in positive powers givesExpanding both fractions in inverse powers instead givesThese are different Laurent series on different annuli, rather than extensions of the local Taylor series beyond its radius of convergence.

For the first function, the complex exponential has no zeros, while the denominator has three distinct simple zeros. Thus there are $\boxed{\text{simple poles at }z=0,2i,-2i;\text{no finite zeros}}$. To examine infinity, substitute $z=1/w$:The factor $e^{1/w}$ has an essential singularity at zero, and multiplication by a nonzero finite-order factor cannot remove it. Therefore $\boxed{\infty \text{ is an isolated essential singularity}}$ of the first function.

For the second function, $\sin z$ has simple zeros at $z=n\pi$. At zero, its Taylor series givesso $\boxed{z=0\text{ is removable, with extended value }1}$. Every other $z=n\pi$ is a double pole, since the numerator is nonzero there. There are no finite zeros, including after removal at zero.

At infinity the double poles $n\pi$ accumulate. Equivalently, after $z=1/w$, the function $1/[w^2{\sin }^2(1/w)]$ has double poles at $w=1/(n\pi )$ tending to zero. Hence $\boxed{\infty \text{ is a non-isolated singularity}}$. The removable/pole/essential trichotomy for isolated singularities does not apply there; calling this an isolated essential singularity would be incorrect.

The Cauchy-Riemann equations determine the harmonic conjugate. Firstso $v_y=u_x$ integrates with respect to $y$ toNowand the second Cauchy-Riemann equation, $v_x=-u_y$, forces $h^{\prime }(x)=0$. HenceThe normalization at zero gives $C=0$, so the required entire function is

For a holomorphic function, the derivative supplied by complex differentiability must have the same value along real and imaginary increments. Those two computations giveEquating real and imaginary parts proves the Cauchy-Riemann equations:Conversely, if the first partial derivatives are continuous near a point and obey these equations there, real differentiability giveswhich proves complex differentiability under this regularity assumption.

The gradients of the real and imaginary parts have dot productand $|\nabla u{|}^2=|\nabla v{|}^2=|f^{\prime }(z){|}^2$. At every point with $f^{\prime }(z)\ne 0$, the two level sets are regular curves, their gradients are nonzero normal vectors, and their tangent directions are therefore perpendicular.

The printed phrase "at every point" needs this regular-point qualification. At a critical point with $f^{\prime }(z)=0$, the level sets may cross or fail to have a unique tangent. For example, $f(z)=z^2$ gives $u=x^2-y^2$ and $v=2xy$: at the origin the zero level sets are respectively the diagonals and the coordinate axes. Thus the valid conclusion is $\boxed{\text{orthogonality at every regular intersection, where }{f}^{\prime }(z)\ne 0}$.

The anticommutator condition and invertibility give the matrix similaritiesSince matrix trace is invariant under matrix similarity,Hence $\boxed{\mathrm{Tr}A=\mathrm{Tr}B=0}$. Taking determinants in the first matrix similarity also givesSince $A$ is an invertible matrix, $\det A\ne 0$, and $\boxed{n\text{ is even}}$. In terms of eigenvalues, the same matrix similarity pairs each nonzero eigenvalue $\lambda$ with $-\lambda$, with the same algebraic multiplicity, so an odd number of them is impossible.

The cyclic property of the trace impliesTaking the matrix trace of the asserted equality therefore gives $nc=0$. For a positive integer $n$ over the complex numbers this requires $\boxed{c=0}$. It is also sufficient: choose $M=N=0$, or any two commuting matrices. Thus a finite-dimensional operator commutator cannot be a nonzero scalar multiple of the identity matrix.

For every vector $v$, the Hermitian inner product givesThe norm vanishes only on the zero vector, so $Cv=0$ for every $v$ and $\boxed{C=0}$. In contrast,is a nonzero nilpotent matrix. Multiplying a matrix by its conjugate transpose detects column norms; multiplying it by itself has no corresponding positivity property.

The matrix sending the standard basis vectors to the given columns isWith the Hermitian inner product, the squared column norms are $3/6+1/6+2/6=1$, $3/6+1/6+2/6=1$, and $4/6+2/6=1$. The off-diagonal products areThus the columns form an orthonormal basis, so $\boxed{U^{†}U=I}$ and $U$ is a unitary matrix. Here the operation maps the old basis vectors to the new vectors; the inverse coordinate change of basis uses $U^{†}$.

The Translation property of the Fourier transform givesFor the remaining integral, the same Translation property of the Fourier transform identifies $e^{20i{k}^{\prime }}\tilde{g}(k^{\prime })$ as the Fourier transform of $g(x+20)$. By the product form of the convolution theorem, with inverse normalization $1/(2\pi )$,But $g(x+20)$ is supported in $[-21,-19]$, whereas the four translated triangular functions making up $r$ are all supported in $[-8,8]$. Their product is identically zero. Therefore the requested integral is $\boxed{0}$ for every real $k$.

The convolution $(g\ast g)(x)$ is the length of the overlap of $(-1,1)$ with $(x-1,x+1)$. This overlap has length $2-|x|$ when $|x|<2$ and zero otherwise, so $u=g\ast g$. The convolution theorem now givesThe value at zero is also the area of the triangular function, providing a normalization check on its Fourier transform.

Direct evaluation of the Fourier transform gives, for $k\ne 0$,At $k=0$ the defining integral is the interval length, so $\boxed{\tilde{g}(0)=2}$. This also equals the continuous limit of the displayed formula.