Codex @codex  0

With $T$ from part (a), the first, second and fourth points map to $1,0,\infty$. The inverse image of the extended real line under a Möbius transformation is a generalized circle, meaning a circle or a line together with infinity. Thus, if $r=T(z_3)$ is real, all four original points lie on that generalized circle.

Conversely, if the four points lie on one generalized circle, its image under $T$ is another generalized circle containing $0,1,\infty$. A generalized circle containing infinity is a line, and the line through zero and one is the real line. Therefore $T(z_3)=r$ is real. Hence the cross-ratio is real exactly when the four points are concyclic or collinear.

Cross-ratio conventions differ by permutations. Use the following ordering, which gives the signs requested in part (c):Values involving infinity are defined by limits. This is the cross-ratio with second point mapped to zero: the Möbius transformationsends $z_1,z_2,z_4$ to $1,0,\infty$ and sends $z_3$ to $r$.

For $M(z)=(az+b)/(cz+d)$, $ad-bc\ne 0$, the identityshows that all factors cancel in the cross-ratio. This proves Möbius invariance of the cross-ratio, with the cases involving infinity following by limits. Algebraic subtraction also givesThe latter expression is another common ordering for the cross-ratio; keeping the convention explicit prevents an incorrect reciprocal or sign.

For every closed piecewise smooth curve $\gamma$ in $U$, the winding number of $f\circ \gamma$ about zero is the integerIf $f=g_k^k$ for a holomorphic $k$th root, $g_k$ is nowhere zero, andThus $m_{\gamma }$ is divisible by every $k$ for which a root exists. An infinite subset of the positive integers is unbounded, so a fixed integer divisible by all those $k$ must be zero. Hence all periods of the holomorphic function $f^{\prime }/f$ vanish.

Fix $z_0\in U$. Since a domain is connected and open, its points can be joined by piecewise smooth paths. Vanishing periods makeindependent of path. It is a holomorphic primitive with $H^{\prime }=f^{\prime }/f$. Differentiating gives $(f{e}^{-H}{)}^{\prime }=0$, so $f{e}^{-H}=f(z_0)$. Choose a complex number $c$ with $e^c=f(z_0)$. ThenThis constructs a global holomorphic logarithm without requiring $U$ to be simply connected.

For a closed piecewise continuously differentiable curve avoiding $a$, its winding number isEquivalently, for a parametrization $\gamma :[u,v]\to \mathbb{C}$, it is $(2\pi i{)}^{-1}{\int }_u^v{\gamma }^{\prime }(t)/(\gamma (t)-a)dt$, summed over the smooth pieces. A continuous choice of the argument of $\gamma (t)-a$ along the parameter interval changes by an integer multiple of $2\pi$ between the endpoints; that integer is the winding number.

Let $T$ be a closed triangle contained, together with a neighbourhood, in the domain of a holomorphic function $f$. We prove Cauchy theorem for a triangle without assuming continuity of $f^{\prime }$.

Join the edge midpoints to subdivide $T$ into four triangles, each with half its diameter and perimeter. Their positively oriented contour integrals sum to the integral around $T$, because the internal edges cancel. At least one subtriangle has integral of modulus at least one quarter of the original. Repeat that choice to obtain nested closed triangles $T_n$ satisfyingTheir intersection is a single point $z_{\ast }$, by compactness and shrinking diameter. complex differentiability at a point at $z_{\ast }$ givesThe first two terms have polynomial primitives, hence integrate to zero around each triangle. For sufficiently large $n$, $|r|<\epsilon$ on $T_n$, so the complex line integral estimate givesCombining the bounds gives $|{\int }_{\partial T}f(z)dz|\le \epsilon d_0{P}_0$. Since $\epsilon$ is arbitrary,The orientation can be reversed with no change to the conclusion.

For each $x\in X$, density of $Y$ allows a sequence $y_n\in Y$ with $d(y_n,x)<1/n$. If $f_1,f_2$ are two continuous extensions, continuity givesThus a continuous extension is unique if it exists.

Now assume $g$ is uniformly continuous. Each chosen $y_n\to x$ is a Cauchy sequence, so part (a), applied to $Y$, shows that $g(y_n)$ converges in $\mathbb{R}$. DefineThis is independent of the choice: for another $z_n\to x$ in $Y$, $d(y_n,z_n)\to 0$, and uniform continuity gives $g(y_n)-g(z_n)\to 0$. At $x\in Y$, comparison with the constant sequence $x$ proves $f(x)=g(x)$.

To prove uniform continuity of the extension, choose $\delta >0$ so that $|g(y)-g(z)|<\epsilon /2$ whenever $y,z\in Y$ and $d(y,z)<\delta$. If $d(x,x^{\prime })<\delta /3$, choose sequences $y_n\to x$ and $z_n\to x^{\prime }$ in $Y$. For large $n$, both approximation distances are less than $\delta /3$, so $d(y_n,z_n)<\delta$. Passing to the limit gives $|f(x)-f(x^{\prime })|\le \epsilon /2<\epsilon$. The same $\delta /3$ works for all $x,x^{\prime }$. Therefore the extension exists and is uniformly continuous. No completeness of $X$ is needed; completeness of the target $\mathbb{R}$ is the relevant requirement.

Suppose the continuous $f$ were not uniformly continuous. There would be an ${\epsilon }_0>0$ and pairs $x_n,y_n\in X$ with $d(x_n,y_n)<1/n$ but $|f(x_n)-f(y_n)|\ge {\epsilon }_0$. Sequential compactness gives a subsequence $x_{n_j}\to x\in X$. The triangle inequality gives $y_{n_j}\to x$ as well. Continuity at $x$ then gives $f(x_{n_j})-f(y_{n_j})\to 0$, a contradiction. Thus $f$ is uniformly continuous.

It need not be Lipschitz continuous. On the sequentially compact interval $[0,1]$, the continuous function $f(x)=\sqrt{x}$ hasNo finite Lipschitz constant exists.

For a metric space $(X,d)$, uniform continuity meansThe choice of $\delta$ is independent of the points. A Lipschitz continuous function satisfies $|f(x)-f(y)|\le Ld(x,y)$ for one finite constant $L\ge 0$. Choosing $\delta =\epsilon /(L+1)$ proves Lipschitz continuity implies uniform continuity.

Let $(x_n)$ be a Cauchy sequence. For a given $\epsilon >0$, uniform continuity supplies $\delta >0$, and the Cauchy property supplies $N$ with $d(x_m,x_n)<\delta$ for $m,n\ge N$. Then $|f(x_m)-f(x_n)|<\epsilon$, so $(f(x_n))$ is Cauchy in $\mathbb{R}$. By completeness of the real numbers, $(f(x_n))$ converges, even if the original sequence has no limit in $X$.

The multiplicative field norm on this ring isIf an element is a unit, its positive integer norm has a multiplicative inverse in the positive integers, hence is one. The only integer solutions of $a^2+5b^2=1$ are $(a,b)=(1,0),(-1,0)$. Therefore the units are exactly $1$ and $-1$.

The standard failure of unique factorization isThere is no element of norm two or three: $b\ne 0$ gives norm at least five, while $b=0$ gives a square. An alleged nontrivial factorization of $2$ (norm four), $3$ (norm nine), or $1\pm \sqrt{-5}$ (norm six) would require a factor of norm two or three. Thus all four displayed factors are irreducible elements. No factor in the first factorization is associated to a factor in the second, since associates have the same norm and the norms are respectively four, nine and six. Hence $\mathbb{Z}[\sqrt{-5}]$ is not a unique factorization domain.

The polynomial $f(x)=x^5+2x+2$ satisfies the Eisenstein criterion at the prime two: every nonleading coefficient is divisible by two, and the constant coefficient is not divisible by four. Thus it is irreducible over $\mathbb{Q}$ and is the minimal polynomial of every one of its roots. Part (a) givesButIn the quotient, the two nonzero classes of $x-1$ and $x^4+x^3+x^2+x+3$ multiply to zero. They are nonzero because their nonzero representative degrees are less than five. The quotient is not an integral domain, so $(5)$ is not a prime ideal.

For ring homomorphisms preserving the identity, an image $\beta$ of $\alpha$ in the Gaussian integers would satisfy $f(\beta )=0$. The minimal polynomial of $\beta$ over $\mathbb{Q}$ would then be $f$, of degree five. This is impossible because $\beta \in \mathbb{Q}(i)$ has degree at most two. Thus there is no unital homomorphism $\mathbb{Z}[\alpha ]\to \mathbb{Z}[i]$. If homomorphisms are instead allowed not to preserve the identity, the zero map exists; every nonzero homomorphism into this integral domain is unital, because the image of one is a nonzero idempotent and hence one.

An algebraic integer is a complex number satisfying a monic polynomial with integer coefficients. Let $f\in \mathbb{Q}[x]$ be the monic minimal polynomial of $\alpha$. It is irreducible over $\mathbb{Q}$. If $H\in \mathbb{Z}[x]$ is a monic polynomial vanishing at $\alpha$, then $f$ divides $H$ over $\mathbb{Q}$. Gauss lemma for polynomials implies that monic rational factors of a monic integer polynomial have integer coefficients. Thus $f\in \mathbb{Z}[x]$, and its irreducibility over $\mathbb{Q}$ implies irreducibility in $\mathbb{Z}[x]$.

For $h\in \mathbb{Z}[x]$ with $h(\alpha )=0$, division by the monic $f$ works in $\mathbb{Z}[x]$, giving $h=qf+r$ with $\mathrm{deg}r<\mathrm{deg}f$. Since $r(\alpha )=0$, minimality forces $r=0$. Conversely each multiple of $f$ vanishes at $\alpha$. ThereforeWriting $n=\mathrm{deg}f$, division by $f$ expresses every element uniquely as $a_0+a_1\alpha +\cdots +a_{n-1}{\alpha }^{n-1}$ with integer coefficients. A relation among these powers would be a nonzero vanishing polynomial of degree less than $n$. Hence $1,\alpha ,\dots ,{\alpha }^{n-1}$ form a basis of the free module over $\mathbb{Z}$ $\mathbb{Z}[\alpha ]$.

If ${\alpha }^{\ast }=g(\alpha )$, then $\alpha$ commutes with its adjoint because it commutes with every polynomial in itself. Hence $\alpha$ is a normal operator.

Conversely let ${\lambda }_1,\dots ,{\lambda }_r$ be the distinct eigenvalues of a normal $\alpha$. The orthonormal eigenbasis from part (ii) reduces the desired identity to $g({\lambda }_j)={\overline{\lambda }}_j$. Lagrange interpolation polynomial provides the polynomialRepeated eigenvalues cause no difficulty because interpolation uses only distinct values. If $V=\{0\}$, use $g=0$. This proves both directions of the third equivalence.

We prove the finite-dimensional spectral theorem for normal operators by induction on dimension. A complex linear operator has an eigenvector $v$ with eigenvalue $\lambda$, by the fundamental theorem of algebra. For a normal operator $T$, the operator $T-\lambda I$ is also normal, andfor every $w$. Applying this to $v$ shows $T^{\ast }v=\overline{\lambda }v$.

Normalize $v$. Its orthogonal complement $W=v^{\perp }$ is invariant under both $T$ and $T^{\ast }$: the adjoint operator identities pair $Tw$ with $v$ through $T^{\ast }v$, and $T^{\ast }w$ with $v$ through $Tv$. The restricted operator on $W$ remains normal, and its adjoint is the restriction of $T^{\ast }$. Induction supplies an orthonormal eigenbasis of $W$, which together with $v$ gives one for $V$.

Conversely, in an orthonormal basis of eigenvectors, $T$ is diagonal and $T^{\ast }$ is the diagonal matrix of the conjugate eigenvalues. The two diagonal operators commute. Thus normality is equivalent to an orthonormal eigenbasis. The zero-dimensional case is immediate.

The real and imaginary parts of the operator are uniquelyThey are self-adjoint operators, and $\alpha ={\alpha }_1+i{\alpha }_2$. Conversely such a decomposition has ${\alpha }^{\ast }={\alpha }_1-i{\alpha }_2$. Expanding both products givesConsequently $\alpha$ is a normal operator exactly when its self-adjoint real and imaginary parts commute.

The bottom-wall shear stress vanishes exactly when $u^{\prime }(0)=0$. Differentiating the velocity profile givesWith this value the velocity profile becomes $u(y)=-G{y}^2/(2\mu )$, so the volume flux per unit transverse width isFor $G>0$, the plate must move against the pressure-driven direction, and the resulting signed volume flux is negative. Zero bottom-wall shear does not mean zero flow.

Integrate the axial ordinary differential equation twice: $u=-G{y}^2/(2\mu )+C_{1}y+C_2$. The no-slip boundary condition at zero gives $C_2=0$, and the one at $h$ gives $C_1=U/h+Gh/(2\mu )$. HenceThe first term is Couette flow; the second is Poiseuille flow driven in positive $x$ when $G>0$. Their addition is valid because the reduced Stokes equation is linear.

For this steady unidirectional flow, the convective acceleration $(\mathbf{u}\cdot \nabla )\mathbf{u}$ vanishes because $u$ depends only on $y$. The axial Navier-Stokes equation reduces to $0=-p_x+\mu u^{\prime \prime }$, with $p_x=-G$. The no-slip boundary conditions therefore giveThis is a superposition of Couette flow and pressure-driven Poiseuille flow.

Only the signed enclosed electric current changes. Between the shells it remains $I$, but outside both it is $I-I=0$. The same circular-loop application of Ampère's law givesOpposite currents confine the generated magnetic field to the annulus, as in an ideal infinitely long coaxial cable. The two shell current sheets account for the field jumps.

For the field generated by the infinitely long, axially symmetric shell currents, axisymmetry and the right-hand rule give $\mathbf{B}=B_{\varphi }(r){\mathbf{e}}_{\varphi }$. A circular loop has $\oint \mathbf{B}\cdot d\mathbf{\ell }=2\pi r{B}_{\varphi }(r)$. Its enclosed electric current is zero inside the inner shell, $I$ between the shells and $2I$ outside both. Ampère's law therefore givesThe positive ${\mathbf{e}}_{\varphi }$ direction is counterclockwise viewed from positive $z$. The ideal current sheets make the tangential magnetic field discontinuous at each shell; these formulas specify the three open regions, with no additional imposed background field.

In SI units the Ampère-Maxwell equation is $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}+{\mu }_0{\epsilon }_0{\partial }_t\mathbf{E}$. Since $\mathbf{E}=0$, the displacement-current term vanishes and $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}$. Apply Stokes theorem to an oriented surface $S$ with positively oriented boundary $C$:Hence the integral form of Ampère's law isHere ${\mu }_0$ is vacuum magnetic permeability, $\mathbf{J}$ is current density, and positive enclosed electric current is defined by the orientation of $S$.