An essential singularity is an isolated singularity which is neither removable nor a pole; equivalently its Laurent expansion has infinitely many nonzero negative-power terms.

For the final unnumbered classification, the only possible finite singularities are the zeros of the denominator, namely $z=\pm i$ and $z=k\pi$ for integers $k$. At zero,so zero is removable, with extension value one. At each nonzero $k\pi$, $\sin z$ has a simple zero, the other denominator factor is nonzero and the numerator is nonzero. These points are poles of order two, with leading coefficient $(k\pi {)}^2/((k\pi {)}^2+1)$.

At $z=\pm i$, the polynomial denominator has a simple zero. Since $\sin (\pm i)=\pm i\sinh 1\ne 0$ and the numerator is $-1$, these are simple poles. There are no essential singularities on $\mathbb{C}$ and no other finite singularities.

A pole of order $m\ge 1$ means that $(z-a{)}^{m}f(z)$ extends holomorphically to $a$ and has a nonzero value there. Equivalently, the lowest nonzero power in the Laurent expansion is $(z-a{)}^{-m}$.

A removable singularity at $a$ means that there is a holomorphic function on the whole disc which agrees with $f$ away from $a$. Equivalently, the Laurent series has no negative powers. The value at $a$ is filled in by its finite limit.

This is the Gaussian integer ring $\mathbb{Z}[i]$. It contains zero and one and is closed under addition, additive inverses and multiplication, sinceIts operations inherit the ring axioms from $\mathbb{C}$. It is a ring but not a field: the nonzero element 2 has complex inverse $1/2$, which is not a Gaussian integer.

$\mathbb{Z}/3\mathbb{Z}$ is a field, and consequently also a ring. Its nonzero elements are $\overline{1}$ and $\overline{2}$, with inverses $\overline{1}$ and $\overline{2}$ respectively. The quotient ring axioms are inherited as in the preceding part.

$\mathbb{Z}/4\mathbb{Z}$ is a ring but not a field. Quotient operations inherit associativity and distributivity from the integers, and residues have additive inverses. However the nonzero residue $\overline{2}$ satisfies $\overline{2}\overline{2}=\overline{0}$, so it cannot be invertible. A field has no nonzero zero divisors.

A ring is a set with an abelian-group addition, associative multiplication and both distributive laws; here we use rings with multiplicative identity. A field is a nonzero commutative ring in which every nonzero element has a multiplicative inverse.

The positive integers form neither a ring nor a field. They do not contain an additive identity zero, and they also lack additive inverses.

The ideal correspondence theorem identifies the ideals of $\mathbb{Z}/4\mathbb{Z}$ with the ideals of $\mathbb{Z}$ containing $4\mathbb{Z}$. The possible positive generators divide $4$, giving exactlyThe units are $\overline{1}$ and $\overline{3}$. The generators of the three displayed ideals are, respectively,Within the last pair, multiplication by the unit $\overline{3}$ exchanges the generators. The other two sets are singletons, whose members are associate elements to themselves by multiplication by $\overline{1}$. Thus equal principal ideals always have associate generators: this is an associator ring.

However, $\overline{2}\ne \overline{0}$ and ${\overline{2}}^2=\overline{0}$, so $\overline{2}$ is a nonzero zero divisor. Therefore

Suppose $R$ is an integral domain and $(a)=(b)$. If $a=0$, then $(b)=0$, so $b=0$ and $a=1\cdot b$.

Otherwise membership in the equal principal ideals supplies $r,s\in R$ with $a=rb$ and $b=sa$. Substituting gives $a=rsa$, or $(1-rs)a=0$. There are no nonzero zero divisors in an integral domain, and $a\ne 0$, so $rs=1$. Hence $r$ is a unit with inverse $s$, and $a=rb$ makes $a$ and $b$ associate elements. We have proved

If $a=ub$ with $u$ a unit in a ring, then $a$ lies in the principal ideal generated by $b$. This gives $(a)\subseteq (b)$. Since $u$ has a multiplicative inverse, $b=u^{-1}a$, giving the reverse inclusion. ThereforeThus associate elements always generate the same principal ideal; the associator ring condition asks whether equality of these ideals also forces the generators to differ by a unit.

A cubic polynomial over a field is reducible if and only if it has a root in that field: a nontrivial factorization of degree three has a linear polynomial factor, and conversely a root supplies one by the factor theorem.

The rational root theorem, or part (b), shows that a rational root of this monic polynomial must be an integer. Such an integer must divide the constant coefficient $4$, so the only candidates are $\pm 1,\pm 2,\pm 4$. Direct evaluation givesNone vanishes. Hence there is no rational root, provingThe same no-root argument alone would not decide a quartic polynomial, which could factor as two quadratic polynomials; its sufficiency here uses degree three.

Eisenstein's criterion states that an integer-coefficient polynomial is irreducible over $\mathbb{Q}$ if some prime number divides every nonleading coefficient, does not divide the leading coefficient, and its square does not divide the constant coefficient. Equivalently apply this to the primitive part when phrased for primitive polynomials.

Take $p=2$. Every nonleading coefficient of $t^5+4t+2$ is divisible by $2$ (including the zero coefficients), its leading coefficient is $1$, and its constant coefficient $2$ is not divisible by $4$. Therefore

Use the reduction criterion: a primitive integer-coefficient polynomial whose leading coefficient remains nonzero modulo a prime number $p$ is irreducible over $\mathbb{Q}$ if its reduction is irreducible over ${\mathbb{F}}_p$. This follows from Gauss lemma for polynomials, since a nontrivial factorization over the integers would reduce to a nontrivial factorization of the same degrees. For a monic polynomial the factors may be taken monic, so neither degree can drop.

Modulo $2$ the polynomial is $\overline{f}=t^4+t+1$. Its values at $0$ and $1$ are both $1$, so it has no linear polynomial factor. A reducible quartic polynomial without a linear polynomial factor must factor into two irreducible quadratic polynomials. The only monic irreducible quadratic polynomial over ${\mathbb{F}}_2$ is $q=t^2+t+1$: among monic quadratic polynomials a nonzero constant is necessary, and $t^2+1$ has a root, while $q$ has none.

But $t^2\equiv t+1(\mathrm{mod}q)$, so $t^3\equiv 1$ and $t^4\equiv t$. ConsequentlyThus $q$ does not divide $\overline{f}$, excluding the remaining possible factorization. We obtain

Work with the L1 norm on ${\mathbb{R}}^2$. The scalar Lipschitz hypothesis impliesWe may take $L\ge 0$. The compact set $S$ and continuity give a finite bound $M={\max }_S\Vert \mathbf{f}{\Vert }_1$. ChooseLet $I=[a-\eta ,a+\eta ]$. The space $C_{\eta }$ of continuous maps $I\to B_k(\mathbf{b})$ is complete in the supremum norm: it is a closed subset of the Banach space $C(I,{\mathbb{R}}^2)$. Define the Picard operator on a closed ball byFor any $\mathbf{g}\in C_{\eta }$, the integral triangle inequality givesThe integrand is continuous, so the result is continuous and remains in the prescribed closed ball. Thus $\mathcal{P}$ is a self-map of the complete space. Moreover,It is a contraction mapping. The contraction mapping theorem gives a unique fixed point $\mathbf{y}$. By the fundamental theorem of calculus, the fixed-point identity makes $\mathbf{y}$ continuously differentiable and gives ${\mathbf{y}}^{\prime }=\mathbf{f}(x,\mathbf{y})$ and $\mathbf{y}(a)=\mathbf{b}$. Its first component is twice continuously differentiable, because its derivative is the second component.

Conversely, integrating any solution that remains in the ball makes it a fixed point, so it equals $\mathbf{y}$. A solution starting at $\mathbf{b}$ cannot first reach the boundary of this ball during $I$: before such an exit the same estimate gives distance at most $\eta M<k$, contradicting distance $k$ at the first exit. This proves existence and uniqueness on $|x-a|\le \eta$, not merely formal convergence of an iteration. It is the local Picard-Lindelöf theorem construction.

Introduce the position and velocity variables $y_1=y$ and $y_2=y^{\prime }$. Then the required system of ordinary differential equations and initial condition areThus $f_1(x,\mathbf{y})=y_2$. Conversely, the first equation identifies $y_2=y_1^{\prime }$, and the second recovers the original second-order ordinary differential equation and both initial data.

The difference $w=v-u$ is a harmonic function, because the Laplacian is a linear differential operator:The boundary inequality says $w\ge 0$ on $\partial D$. The minimum principle from part (ii) givesThis is the comparison principle for harmonic functions. Applying it in both directions also shows why fixed Dirichlet boundary conditions determine at most one harmonic function.

For a harmonic function, $\Delta u=0$ and $\Delta (-u)=0$. Apply the preceding minimum argument to $u$ and to $-u$:These are the weak minimum and maximum principle for harmonic functions.

Use the classical regularity setting $u\in C^2(D)\cap C(\overline{D})$, with $D$ bounded. Continuity on the compact set $\overline{D}$ guarantees existence of a minimum. The inequality $\Delta u=f\le 0$ means that $u$ is superharmonic function.

A strict inequality $\Delta u<0$ would immediately exclude an interior minimum, because its Hessian matrix there would be positive semidefinite and have nonnegative trace. To handle equality, introduce the quadratic perturbation proof of the minimum principle:Its minimum must lie on $\partial D$. Choose $R$ so that $x^2+y^2\le R^2$ on $\overline{D}$. For any $p\in D$,Letting $\epsilon \downarrow 0$ gives $u(p)\ge {\min }_{\partial D}u$. ThereforeThis asserts that a minimum is attained on the boundary, not that it cannot also be attained inside; a constant function attains it everywhere.

Integrate the canonical form of a second-order partial differential equation first in $\eta$ and then in $\xi$. The first integration gives $U_{\xi }=\eta +g^{\prime }(\xi )$, and the second givesThusThe two arbitrary twice continuously differentiable functions describe the two characteristic families. Conversely, substituting this expression into $U_{\xi \eta }=1$ verifies every such solution. The formula applies on $x>0$ or $x<0$; independent choices can be made on the two half-planes. Smooth extension across the degenerate line $x=0$ requires extra compatibility and is not supplied by the logarithmic formula alone.

The characteristic curves satisfyIntegrating suggests the characteristic coordinatesTheir Jacobian determinant is $2/x$, so this is an invertible local change of variables whenever $x\ne 0$. Write $u(x,y)=U(\xi ,\eta )$. The chain rule givesSubstitution cancels both unmixed second derivatives and both first derivatives:The canonical form of a second-order partial differential equation is thereforeThe cancellation of the first derivatives is the key feature of these logarithmic characteristic coordinates.