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On $|z|=3$, the perturbation has modulus $|icz+8|\le 3c+8<11<27=|z^3|$. Thus Rouché's theorem gives three roots inside $|z|<3$, counted with multiplicity.

For the quadrant information, compare instead with $P_0(z)=z^3+8$, whose roots are $r_1=-2$, $r_2=1+i\sqrt{3}$, $r_3=1-i\sqrt{3}$. Their pairwise distances are $2\sqrt{3}$. On the circle $|z-r_j|=1$,The three unit discs are disjoint, and Rouché localization of polynomial roots gives one root in each, counted with multiplicity. The disc around $1+i\sqrt{3}$ lies strictly in the first quadrant in its interior; that around $1-i\sqrt{3}$ lies in the fourth quadrant; and that around $-2$ lies in $\mathrm{Re}z<-1$. Consequently there is exactly one root in each of those two right-hand quadrants, and exactly one root in the left half-plane. The same count also shows that each root is simple.

The form of Rouché's theorem used here is as follows. Suppose $f$ and $g$ are holomorphic on a neighborhood of the closure of a bounded Jordan domain, and $|g-f|<|f|$ on its positively oriented boundary. Then $f$ and $g$ have the same number of zeros inside, counted with multiplicity.

To prove this form, put $f_t=f+t(g-f)$ for $0\le t\le 1$. The strict inequality prevents $f_t$ from vanishing on the boundary. The argument principle counts its interior zeros by $(2\pi i{)}^{-1}\oint f_t^{\prime }/f_t$. This integral varies continuously with $t$: the integrand is continuous and its denominator stays uniformly away from zero on the compact boundary times $[0,1]$. Its values are integers, so the integral is constant. Its values at $t=0,1$ are the respective zero counts. This proves Rouché's theorem with the multiplicities and boundary hypotheses required below.

Use the usual convention that a ring homomorphism preserves the identity. Such a map from the quotient ring to $\mathbb{C}$ fixes $\mathbb{Q}$: it fixes integers, and preservation of inverses fixes rational numbers. It is therefore determined by the image $\zeta$ of the class of $x$. The defining relation forces ${\zeta }^3=1$. Conversely, evaluation at any such $\zeta$ annihilates $(x^3-1)$ and hence factors through the quotient. Thus there are exactly three unital homomorphisms, namelyIf the convention permits nonunital ring homomorphisms, add the zero map. These are the only extra possibilities: the image of $1$ is an idempotent in $\mathbb{C}$, hence either zero or one, and image zero forces the whole map to vanish.

In the quotient ring, the classes of $x-1$ and $x^2+x+1$ are nonzero: neither nonzero polynomial of degree less than three belongs to the ideal generated by $x^3-1$. But their product is the zero class, sinceThe quotient thus contains zero divisors, so it is not an integral domain.

The coefficient identityholds over every field. Both factors have positive degree, including in characteristics $2$ and $3$, because their leading coefficients remain one. Therefore there is no field over which this polynomial is irreducible.

Consider the ideal $I=(2,x)$ in $\mathbb{Z}[x]$. It is proper: evaluation at zero followed by reduction modulo $2$ maps the ring onto ${\mathbb{F}}_2$ and has kernel $I$. Suppose $I=(f)$. Then $f$ divides the constant polynomial $2$, so $f$ is a constant integer dividing $2$. Up to sign, the only possibilities are $1$ and $2$. The first would make the ideal the whole ring; the second cannot divide $x$ in $\mathbb{Z}[x]$. Both are impossible. This proves the integer polynomial ring is not a principal ideal domain result: $\mathbb{Z}[x]$ is not a principal ideal domain.

In the polynomial ring $\mathbb{Z}[x]$, the ideal $(x)$ is nonzero and prime: the quotient ring $\mathbb{Z}[x]/(x)$ is isomorphic to the integral domain $\mathbb{Z}$, by evaluation at zero. It is not maximal, since that quotient is not a field. Equivalently $(x)⊊(2,x)⊊\mathbb{Z}[x]$. Thus $(x)$ is a nonzero prime ideal which is not maximal.

Choose an orthonormal basis of eigenvectors of the positive-definite operator $\alpha$ using the proved spectral theorem. Its eigenvalues ${\lambda }_j$ are strictly positive, because $⟨\alpha e_j,e_j⟩={\lambda }_j>0$. Define $\beta e_j=\sqrt{{\lambda }_j}{e}_j$. It is a positive-definite Hermitian operator and ${\beta }^2=\alpha$, establishing existence of the positive square root of an operator.

For uniqueness, let $B$ be any positive-definite Hermitian square root. The identity $B^2=\alpha$ implies $B\alpha =B^3=\alpha B$, so $B$ preserves each eigenspace $E_{\lambda }$ of $\alpha$. Its restriction to that space is Hermitian and positive-definite. Diagonalize that restriction by the same spectral theorem. Every resulting eigenvalue $\mu$ satisfies $\mu >0$ and ${\mu }^2=\lambda$, hence $\mu =\sqrt{\lambda }$. Therefore $B$ acts as $\sqrt{\lambda }I$ on the entire eigenspace, even when $\lambda$ is repeated. These eigenspaces span the space, so the positive-definite Hermitian square root is unique, with eigenvalues $\sqrt{{\lambda }_j}$.

The adjoint reverses products, so for the two Hermitian operatorsThus $\alpha \beta$ is Hermitian if and only if $\alpha \beta =\beta \alpha$. No positivity hypothesis is needed for this equivalence.

For a Hermitian operator $A$, every eigenvalue is real. Indeed, if $Av=\lambda v$ with $v\ne 0$, self-adjointness gives $\lambda ⟨v,v⟩=⟨Av,v⟩=⟨v,Av⟩=\overline{\lambda }⟨v,v⟩$. To prove the finite-dimensional spectral theorem, proceed by induction on the dimension. The fundamental theorem of algebra supplies a root of the characteristic polynomial, hence a unit eigenvector $v$. Its orthogonal complement is invariant: for $w\perp v$, $⟨Aw,v⟩=⟨w,Av⟩=0$. The restriction of $A$ to that complement is again Hermitian. Induction supplies an orthonormal basis of eigenvectors there; adjoining $v$ gives an orthonormal eigenbasis of the whole space. The zero-dimensional case is immediate. Consequently a Hermitian operator is diagonal in an orthonormal basis, with real diagonal entries.

Near zero the $n=0$ summand supplies the pole and the remaining series is holomorphic, so $f(w)=w^{-2}+F(w)$ with $F$ holomorphic near zero. Choose the local logarithm with $L(1)=0$. Writing $h=z-1$ gives $L(z)=h-h^2/2+O(h^3)$ and henceTherefore $z=1$ is a double pole, with leading coefficient ${\lim }_{z\to 1}(z-1{)}^{2}g(z)=-4{\pi }^2$. Its residue, if desired, is $-4{\pi }^2$.

Fix $z_0\ne 0,1$ and take a small disk about it that avoids zero and one. There is a holomorphic logarithm $L$ on this disk. The function $L(z)/(2\pi i)$ cannot be an integer there: if it were an integer, exponentiation would give $z=1$. Thus $f(L(z)/(2\pi i))$ is holomorphic on the disk.

Where the prescribed principal logarithm and $L$ are both evaluated, they differ by $2\pi ik$ for an integer $k$. By the period-one identity, their compositions with $f$ coincide. This remains true on either side of the principal branch cut, and at a point on that cut one may instead choose the logarithm with argument between zero and $2\pi$. The local holomorphic expressions therefore agree with the stated pointwise definition and with each other. This is periodic holomorphic descent through the exponential map, provingThe principal logarithm itself is not analytic across its cut; periodicity of the outer function removes that obstruction.

Let $K\subset \mathbb{C}\setminus \mathbb{Z}$ be compact and contained in $|z|\le R$. For $|n|>2R$, $|z-n|\ge |n|/2$, hence $|(z-n{)}^{-2}|\le 4/n^2$ uniformly on $K$. The tail is therefore absolutely and uniformly convergent by the Weierstrass M-test. Its finitely many remaining summands are holomorphic on a neighborhood of $K$.

The theorem that a locally uniform limit of holomorphic functions is holomorphic now proves that the sum is analytic on $\mathbb{C}\setminus \mathbb{Z}$. Absolute convergence also permits a shift of the integer index:Thus the function has period one.

In the integer basis $u,v,e_3$, direct substitution gives $w=-7u+5v+18e_3$. ThereforeThe first two basis coordinates disappear in the quotient and the third is taken modulo eighteen. Equivalently the Smith normal form has diagonal entries $1,1,18$.

The matrix with rows $u,v,e_3$ has determinantIts inverse has integer entries by the adjugate formula, so these rows form an integer basis of $L$. The unimodular basis test for an integer direct summand now givesThus $N$ is a direct summand, with the displayed complement. This proves integral splitting, not just real or rational linear independence.

A proper finite-index subgroup of a torsion-free group cannot be a direct summand when the complement is a subgroup of that same free abelian group. Indeed, if $L=M\oplus K$, every $k\in K$ has $18k\in M$ because the quotient has order eighteen. But also $18k\in K$, so $18k=0$ by the trivial intersection. The ambient group ${\mathbb{Z}}^3$ is torsion-free, forcing $k=0$ and hence $K=0$. This would give $M=L$, contrary to index eighteen. Thus $M$ is not a direct summand.

Put the generators into the rows of the integer matrixIts determinant is $5-2-21=-18$. A full-rank integer lattice generated by its rows has index equal to the absolute determinant, henceFor a direct verification, $u,v,e_3$ form a unimodular basis because their row determinant is $-1$, and $w=-7u+5v+18e_3$. In these integer coordinates $M=\mathbb{Z}u\oplus \mathbb{Z}v\oplus 18\mathbb{Z}{e}_3$, visibly leaving eighteen cosets.

An example in dimension four isBoth are invertible, so each has rank four. Both have characteristic polynomial $(x-1{)}^4$, and each has minimal polynomial $(x-1{)}^2$, since its nonzero nilpotent part squares to zero. However $\mathrm{rank}(A-I)=2$ and $\mathrm{rank}(B-I)=1$. If $B=PA{P}^{-1}$, then $B-I=P(A-I)P^{-1}$, which would preserve these ranks. The contradiction proves that the matrices are not similar, despite all three specified invariants agreeing.

Take a size-three Jordan block at one and a scalar block at three:Triangularity gives ${\chi }_A(x)=(x-1{)}^3(x-3)$. On the first block, $(A-I{)}^3$ is zero but $(A-I{)}^2$ is not; on the last component the eigenvalue is three. Thus both factors with exactly those powers are needed, and their product annihilates $A$. Its minimal polynomial is therefore also $(x-1{)}^3(x-3)$.

Let $m_{\alpha }(x)$ be the monic minimal polynomial. The Cayley-Hamilton theorem says ${\chi }_{\alpha }(\alpha )=0$. Divide ${\chi }_{\alpha }$ by $m_{\alpha }$; its remainder also annihilates $\alpha$, and a nonzero remainder of smaller degree would contradict minimality. Thus $m_{\alpha }$ divides ${\chi }_{\alpha }$.

Every characteristic root ${\lambda }_i$ is an eigenvalue, with a nonzero eigenvector $v_i$. Since $m_{\alpha }(\alpha )v_i=m_{\alpha }({\lambda }_i)v_i=0$, it must be a root of the minimal polynomial as well. This proves there are no possibilities beyondEvery such choice really occurs. For each $i$, take one Jordan block of size $r_i$ at ${\lambda }_i$, together with $n_i-r_i$ scalar blocks at that eigenvalue, and form their direct sum. The characteristic polynomial has the prescribed multiplicities. The nilpotent part of a block of size $r_i$ has its $r_i$th power zero and its $(r_i-1)$st power nonzero, so this component's minimal exponent is exactly $r_i$. A polynomial annihilates the direct sum precisely when it annihilates every component, establishing the displayed minimal polynomial without any extra restrictions.