The two bounding circles meet at zero and one, which the Möbius transformation sends to zero and infinity. They therefore become straight lines through zero. More explicitly write $w=f(z)$, so $z=w/(1+w)$. The two disk inequalities becomeSince $\zeta =1/2+i\sqrt{3}/2$, these say $\mathrm{Re}w>\sqrt{3}|\mathrm{Im}w|$. ThusThe positive real axis belongs to the image because $f(1/2)=1$, fixing which sector is intended. Cubing maps this sector bijectively onto the right half-plane: its argument interval of width $\pi /3$ becomes $(-\pi /2,\pi /2)$. The Cayley transform between the half-plane and disk $v\mapsto (v-1)/(v+1)$ then maps that half-plane bijectively onto the unit disk. Hence a holomorphic bijection isEvery denominator is nonzero on the corresponding domain, and the sector width ensures the power map is injective. This is the conformal map of a symmetric circular lens.

For $f=u+iv$, the Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$. Holomorphy implies these relations. Conversely, if the first partial derivatives exist and are continuous near a point, satisfying the equations there gives complex differentiability; satisfying them on an open set gives a holomorphic function on that set.

Let $r=\sqrt{x^2+y^2}$. On the cut plane, $r+x>0$: equality could occur only at a nonpositive real point, which is excluded. Thus $u=\sqrt{(r+x)/2}$ is strictly positive and $v=y/(2u)$ is well defined and smooth. Since $y^2=(r-x)(r+x)$,ConsequentlyOne can verify holomorphy directly:Here $r>0$ throughout the domain, so all partial derivatives are continuous. Therefore $f$ is holomorphic, specifically the principal square root on the cut plane, with derivative $f^{\prime }(z)=1/[2f(z)]$.

A cubic over a field is reducible exactly when it has a root in that field. For $p(X)=X^3+2X^2+1$ modulo five, direct evaluation givesNo value is zero, so $p$ is irreducible in ${\mathbb{Z}}_5[X]$.

The integer polynomial in the second part reduces to precisely $p$, because $-2008\equiv 2$, $2010\equiv 0$ and $-2009\equiv 1(\mathrm{mod}5)$. It is monic, hence primitive, and its degree is preserved by reduction. The irreducibility by degree-preserving reduction modulo a prime criterion states that a primitive integer polynomial with irreducible reduction of the same degree is irreducible over the rationals. This follows from Gauss's lemma: a rational factorization would give integer factors whose leading coefficients remain nonzero modulo five, contradicting the irreducible reduction. Thus

A self-adjoint operator satisfies $⟨Tx,y⟩=⟨x,Ty⟩$ for every $x,y$. The finite-dimensional finite-dimensional spectral theorem gives an orthonormal basis of eigenvectors and real eigenvalues; eigenvectors for distinct eigenvalues are orthogonal. Consequently a matrix represents a self-adjoint transformation in some, possibly nonorthonormal, basis exactly when it is diagonalizable with real eigenvalues.

For $A_0$, the eigenvalues are $2,1,0$, all distinct and real. Eigenvectors are $e_1$, $e_2$, and $(0,-1,1{)}^T$. Thus $A_0$ can represent a self-adjoint transformation after choosing a suitable basis.

For $A_1$, the eigenvalue one has algebraic multiplicity two but its eigenspace has dimension one. The lower $2\times 2$ block is a nontrivial Jordan block, so the matrix is not diagonalizable. Hence $A_1$ cannot represent a self-adjoint transformation in any basis.

For $A_2$, the eigenspace for two is spanned by $e_1$ and $(0,1,1{)}^T$, and the eigenvalue-one eigenspace is spanned by $e_2$. These three vectors form an eigenbasis. Therefore $A_2$ can represent a self-adjoint transformation. To see sufficiency on the given inner product space, first define the real diagonal transformation in an orthonormal basis and then apply the inverse of a diagonalizing basis change to obtain the requested matrix. Symmetry of the displayed matrix itself would be required only for an orthonormal representing basis.

The zero polynomial belongs to $I_v$. If $f,g\in I_v$, linearity gives $(f-g)(T)v=0$. If $a\in K[X]$ and $f\in I_v$, polynomial evaluation obeys $(af)(T)=a(T)f(T)$, so $(af)(T)v=0$. Thus $I_v$ is an ideal of the polynomial ring $K[X]$.

The $n+1$ vectors $v,Tv,\dots ,T^{n}v$ are linearly dependent in the $n$-dimensional vector space. Their dependence supplies a nonzero polynomial of degree at most $n$ in $I_v$. Hence this ideal is nonzero. A polynomial ring over a field is a Euclidean domain under degree, and every ideal in a Euclidean domain is principal. It follows that $I_v=(m_v)$ for a nonzero polynomial $m_v$, which can be scaled uniquely to be monic.

For completeness, uniqueness follows because any two generators divide each other and hence differ by a nonzero constant; the monic normalization forces that constant to be one. Since $m_v$ divides the polynomial of degree at most $n$ constructed above, its degree is at most $n$. The generator characterization givesThis is the minimal polynomial of a vector under a linear operator. If $v=0$, the ideal is the whole ring and the unique monic generator is the constant polynomial one.

Write the principal part as $e^{2y}{u}_{xx}-u_{yy}$. The characteristic curves satisfyor $dx/dy=\pm e^y$. Integration gives $x+e^y=\text{constant}$ and $x-e^y=\text{constant}$. Thus take $\xi =x+e^y$, $\eta =x-e^y$. Their Jacobian determinant is $-2e^y\ne 0$, and their image is $\xi >\eta$.

For $u(x,y)=U(\xi ,\eta )$, the chain rule givesThe first-derivative terms cancel exactly. The canonical form of a second-order partial differential equation isOn $y=0$, $F(x+1)+G(x-1)=x$. Differentiating this condition in $x$ and using $u_y=1$ givesHence $F^{\prime }=1$ and $G^{\prime }=0$. Substitution into the value condition fixes their combined constant and yieldsDirect substitution verifies the PDE and both initial data. The cancellation illustrates a hyperbolic equation reduced by an exponential characteristic coordinate.

Rewrite the linear equation as $(x^{2}y{)}^{\prime }=0$. The initial value gives the unique solutionThe Picard iteration is its integral equationThe proposed formula holds for $n=0$. If it holds at $n$, termwise integration of the finite sum and $d(\log t)/dt=1/t$ giveThis proves the expression by induction. The exponential series then gives $y_n(x)\to e^{-2\log x}=x^{-2}$. Convergence is also uniform on every compact subinterval of $(0,\infty )$, since $|\log x|$ is bounded there and the exponential majorant series converges. This is Picard iteration for a simple-pole linear differential equation.

Let $F(z)=e^{i\lambda z}/(z-i)$. For $\lambda >0$, close the interval $[-R,R]$ by the upper semicircle, counterclockwise. Its only enclosed pole is $z=i$, with residue $e^{-\lambda }$. The residue theorem givesOn $z=R{e}^{i\theta }$, $|e^{i\lambda z}|=e^{-\lambda R\sin \theta }$ and $|z-i|\ge R-1$. Hence the arc is bounded byby the supplied estimate, with parameter $\lambda R$.

For $\lambda <0$, close below, with clockwise orientation. The pole at $i$ lies outside, and the exponential decays there. Parametrizing the lower arc yields the same bound with $|\lambda |R$ in place of $\lambda R$. Its contribution again tends to zero, and no residue remains. Therefore

This is the contour-decay mechanism behind Jordan's lemma; the pole contributes only when the decaying closure encloses it.

The hyperbolic cosine is entire and $\cosh 0=1$. Continuity makes it nonzero in a neighborhood of zero, so its reciprocal is holomorphic there and has a convergent Taylor series. Since $\cosh$ and its reciprocal are even, all odd coefficients vanish.

Multiply the proposed expansion by $\cosh z$:Coefficient comparison gives $c_2+1/2=0$ and $c_4+c_2/2+1/24=0$, henceThe zeros of $\cosh z$ solve $e^{2z}=-1$, so they are $z=i(\pi /2+k\pi )$, $k\in \mathbb{Z}$. The nearest ones are $\pm i\pi /2$, and they are simple because $\sinh$ is nonzero there. The hyperbolic secant has genuine poles at these points and is holomorphic in the disk of radius $\pi /2$. The radius of convergence of its Taylor series is therefore

Reduction modulo the prime $p$ preserves integer addition, multiplication, and one:Every congruence class has an integer representative, so the ring homomorphism is onto. Its kernel is exactly $p\mathbb{Z}=(p)$. The quotient is a field because every class not divisible by $p$ has a multiplicative inverse by Bézout's identity.

For the coefficientwise extension to the polynomial ring, addition is preserved term by term. If $f={\sum }_i{a}_i{x}^i$ and $g={\sum }_j{b}_j{x}^j$, the coefficient of $x^k$ in their product is ${\sum }_{i+j=k}{a}_i{b}_j$. Reducing it modulo $p$ gives ${\sum }_{i+j=k}{\bar{a}}_i{\bar{b}}_j$, the coefficient in $\hat{\varphi }(f)\hat{\varphi }(g)$. Thus multiplication and one are preserved as well. Lift each coefficient of a polynomial over ${\mathbb{Z}}_p$ to an integer to prove surjectivity. The kernel isIndeed, its elements are precisely the polynomials whose coefficients are all divisible by $p$.

In the binomial theorem expansion of $(x-1{)}^p$, every intermediate coefficient $\left(\genfrac{}{}{0ex}{}{p}{j}\right)$ is divisible by $p$: the numerator contains the prime factor $p$, while $j!(p-j)!$ does not for $1\le j<p$. HenceFor $p=2$, the last equality uses $1=-1$ in characteristic two. This is an instance of the Frobenius endomorphism.