The cyclic matrix acts by $(v_1,v_2,v_3,v_4)\mapsto (v_2,v_3,v_4,v_1)$. For each fourth root of unity $\lambda$, the vector $v_{\lambda }=(1,\lambda ,{\lambda }^2,{\lambda }^3{)}^T$ satisfies $A{v}_{\lambda }=\lambda v_{\lambda }$, because ${\lambda }^4=1$. This supplies the four distinct eigenvalues and nonzero eigenvectors:Eigenvectors for distinct eigenvalues are linearly independent. Thus, with these columns in the indicated order,They form a basis over $\mathbb{C}$, so there are no other eigenvalues. Equivalently the characteristic polynomial is ${\lambda }^4-1$.

For a complex square matrix, define its determinant byThe cofactor is $C_{ij}=(-1{)}^{i+j}\det A^{(ij)}$, where $A^{(ij)}$ deletes row $i$ and column $j$. The adjugate is the transposed cofactor matrix, $\mathrm{adj}(A{)}_{ij}=C_{ji}$.

The $(i,j)$ entry of $\mathrm{adj}(A)A$ is ${\sum }_k{C}_{ki}{a}_{kj}$. For $j=i$, expansion along column $i$ gives $\det A$. For $j\ne i$, this same sum is the determinant of the matrix obtained by replacing column $i$ by column $j$: deleting the replaced column leaves all its cofactors unchanged. The new matrix has two equal columns, so its determinant is zero. ThusThis proof does not assume invertibility and therefore also covers singular matrices.

A particle pathline follows the time-dependent velocity field, so its coordinates satisfy $\dot{x}-x=\sin t$ and $\dot{y}=-y$. The second equation and the initial data give $y=e^{-t}$. Using the integrating factor $e^{-t}$ in the first,Therefore the complete pathline isIf $x_0+\frac{1}{2}\ne 0$, its first coordinate grows without bound in magnitude. If $x_0=-\frac{1}{2}$, the first coordinate remains bounded and periodic while $y\to 0$. The unique release coordinate giving a bounded particle path is $x_0=-1/2$.

Integrating ${\psi }_y=x+\sin t$ and imposing $-{\psi }_x=-y$ gives the streamfunctionup to an arbitrary function of time. At the specified time, the streamlines areFor $C\ne 0$, the connected branches are rectangular hyperbolas. The zero level consists of the lines $y=0$ and $x=-1$, with the stagnation point $(-1,0)$ at their intersection; their half-lines and the stationary point are the distinct streamlines.

The PDF has the doubly exponential denominator $e^{e^n}$. For its positive summands, the successive-term ratio isThus the ratio test proves absolute convergence. Concretely, the tail is eventually dominated by a geometric series of ratio $1/2$; the polynomial numerator cannot offset the doubly exponential decay.

Let $a_n$ denote the summand. Its denominator is positive for every $n\ge 1$: odd $n$ gives $n^4+n^2$, while even $n\ge 2$ gives $n^4-n^2$. Dividing numerator and denominator by their highest powers givesThe terms are positive for sufficiently large $n$, so the limit comparison test with the divergent harmonic series gives divergence to $+\infty$. More explicitly, for $n\ge 16$ the numerator is at least $n^3/2$ and the denominator is at most $2n^4$, giving $a_n\ge 1/(4n)$. Any finitely many early negative terms change partial sums by only a finite constant and cannot prevent divergence of this positive tail.

For odd $n$, the denominator is $(n+1{)}^2$; for even $n$, it is $(n-1{)}^2$. There is no zero denominator in the printed index range. For all $n\ge 2$,The comparison test for series with the convergent p-series gives absolute convergence. In fact the consecutive pair $n=2k-1,2k$ contributes $1/(2k{)}^2+1/(2k-1{)}^2$, so the sum is the same as ${\sum }_{m=1}^{\infty }{m}^{-2}$. Positivity justifies this paired reordering; an alternating-sign appearance in the denominator does not make this an alternating series.

For $n\ge 2$, $0<n^{-2}\le [n(n-1){]}^{-1}=1/(n-1)-1/n$. The upper bound has telescoping partial sums, bounded by one. The positive partial sums of the original series are consequently bounded by two and increasing, so they converge as a monotone bounded sequence. The series converges absolutely. This also supplies the $p=2$ case of a p-series without requiring its exact sum.

The cubic terms are the expansion of $(z+\overline{z}{)}^3=(2x{)}^3$. Subtracting $\overline{z}=x-iy$ therefore givesHere the expression is a polynomial in $z$ and its complex conjugate, rather than a holomorphic polynomial in $z$ alone.

For $z\ne 0$, rationalizing the denominators gives $1/z=(x-iy)/(x^2+y^2)$ and $1/\overline{z}=(x+iy)/(x^2+y^2)$. Their imaginary parts cancel, soThis also follows because the summands are complex conjugates.

For $z\ne 0$, write $z=r{e}^{i\theta }$ with $r=\sqrt{x^2+y^2}>0$. The complex logarithm has values $\ln r+i(\theta +2\pi k)$, $k\in \mathbb{Z}$. ThereforeFor a specified single-valued branch, choose the corresponding continuous determination of the argument. In particular, the principal value uses $\mathrm{Arg}z\in (-\pi ,\pi ]$; it is not represented correctly by $\arctan (y/x)$ in every quadrant. The complex logarithm is undefined at zero.

Using $\cos z=(e^{iz}+e^{-iz})/2$ and collecting real and imaginary terms gives the complex cosine identity $\cos (x+iy)=\cos x\cosh y-i\sin x\sinh y$. ThusThe sign follows from the factors $e^{-y}$ and $e^y$ in the two complex exponentials.

The complex exponential function factorizes as $e^{x+iy}=e^x{e}^{iy}=e^x(\cos y+i\sin y)$. HenceThe real exponential multiplies both components; it does not change the argument.

Interchange the two dummy indices $j,k$. The product $a_j{a}_k$ is symmetric under this interchange, while the Levi-Civita symbol changes sign. Thus the contraction equals its negative and is the zero vector, equivalently $\mathbf{a}\times \mathbf{a}=0$.

Compare with $(\mathbf{b}\times \mathbf{a}{)}_i={ϵ}_{ijk}{b}_j{a}_k$. The scalar factors commute, so the answer isThe order of the contracted indices matters because the Levi-Civita symbol is antisymmetric.

The contracted Kronecker delta is ${\delta }_{ll}=3$. The other scalar contraction is $a_i{C}_{ij}{b}_j={\mathbf{a}}^{T}C\mathbf{b}$, while $C_{ik}{d}_i$ is the $k$th component of the matrix transpose applied to $\mathbf{d}$. Hence the resulting vector isBoth terms have the same single free index $k$, as required by the Einstein summation convention.

This is the incorrectly formed expression. In the Einstein summation convention, an index in a term is either free and occurs once, or summed and occurs twice. Here $i$ occurs three times in the same product, so it cannot consistently be classified as either. Renaming a dummy index before multiplying would give a valid different expression, such as ${\delta }_{kk}{\delta }_{ij}=3{\delta }_{ij}$, but does not make the printed expression valid.

The two free indices are the row and column indices of a matrix. The Kronecker delta has value one on its diagonal and zero off it, so the answer is the $3\times 3$ identity matrix $I$.

Suppose an analytic extension $F$ existed on the punctured plane. The identity theorem would extend $F^2=z$ from $D$ to that connected domain, so $F$ would never vanish and differentiation would give $2F^{\prime }/F=1/z$. Integrate around the positively oriented unit circle. The argument principle says $\int F^{\prime }/F$ is $2\pi i$ times the integer winding number of the closed curve $F(e^{it})$ about zero, whereasThis would make that winding number $1/2$, a contradiction. Hence no such analytic extension exists. Equivalently, continuation of a square root once around zero reverses its sign, preventing a single-valued holomorphic function on the punctured plane.

The slit plane is connected and contains $D$. If two holomorphic functions on it extend the same $f$, their difference vanishes on the nonempty open disc $D$. The identity theorem forces their difference to vanish on the entire slit plane. Thus there is at most one analytic extension of the chosen $f$. Choosing the opposite square root initially changes that chosen function; it does not produce two extensions of the same function.