Morera's theorem states that if $f$ is continuous on an open set $U$ and its integral around every triangle whose interior and boundary lie in $U$ is zero, then $f$ is holomorphic on $U$.

For a proof, choose a disk $B\subseteq U$ and a base point $z_0\in B$. Define $F(z)={\int }_{[z_0,z]}f(\zeta )d\zeta$ along the straight segment. Convexity keeps every triangle with vertices $z_0,z,z+h$ inside $B$. Its zero boundary integral impliesBy continuity, the last integral tends to $f(z)$ as $h\to 0$, so $F^{\prime }(z)=f(z)$. Thus $F$ is holomorphic on $B$. The allowed result that the derivative of a holomorphic function is holomorphic now gives that $f=F^{\prime }$ is holomorphic there. Such disks cover $U$, proving

Cauchy's integral theorem states that if a function is holomorphic on a simply connected open set, its integral around every closed piecewise-smooth path in that set is zero. A local form sufficient here is: if a function is holomorphic on an open set containing a triangle together with its interior, thenFor a general domain with holes, an arbitrary closed path need not have zero integral; the triangle or simple-connectedness hypothesis supplies the required interior.

For the preliminary definition, put $S_N(z)={\sum }_{k=1}^N{f}_k(z)$. The series converges uniformly to $f$ on $U$ when, for every $ϵ>0$, some $N_0$ satisfies $|S_N(z)-f(z)|<ϵ$ for all $z\in U$ and all $N\ge N_0$. Equivalently, ${\sup }_{z\in U}|S_N(z)-f(z)|\to 0$.

Let $L(\gamma )$ be the finite length of the piecewise-smooth path. Finite linearity of the contour integral and the integral bound giveThe uniform limit is continuous by the stated allowed result, so its integral is defined. Consequently

Certainly $A\subseteq \sqrt{A}$, by taking exponent one. If $x\in \sqrt{A}$, choose $n\ge 1$ with $x^n\in A$. Primality applied to $x\cdot x^{n-1}$ gives $x\in A$ or $x^{n-1}\in A$. In the latter case repeat; induction reduces to $x\in A$. Thus $\sqrt{A}\subseteq A$ as well, provingEvery prime ideal is therefore a radical ideal.

A coset is a nilpotent element of the quotient exactly when some positive power is zero:If $J=\sqrt{J}$, this forces every nilpotent coset to equal zero. Conversely, if there is no nonzero nilpotent coset, every $x\in \sqrt{J}$ has $x+J=0$, hence $x\in J$. Since $J\subseteq \sqrt{J}$ always, equality follows. ThereforeIn standard terms, an ideal is radical exactly when its quotient is a reduced ring.

The set $\sqrt{I}$ contains zero. If $x^m\in I$, then $(-x{)}^m=(-1{)}^m{x}^m\in I$, giving closure under additive inverses; also $(rx{)}^m=r^m{x}^m\in I$ for every $r\in R$.

To prove closure under addition, let $x^m,y^n\in I$. Expand $(x+y{)}^{m+n-1}$ by the binomial theorem. In every monomial $x^j{y}^{m+n-1-j}$, either $j\ge m$ or $m+n-1-j\ge n$. Thus each term contains a factor $x^m$ or $y^n$ and belongs to $I$. Hence $(x+y{)}^{m+n-1}\in I$, so $x+y\in \sqrt{I}$. The additive subgroup and absorption properties are proved, givingThis ideal is the radical of an ideal. The finite exponent for a sum uses commutativity of the ring.

An ideal $I$ of a commutative ring is an additive subgroup such that $rx\in I$ for every $r\in R$ and $x\in I$. A prime ideal is a proper ideal $A\ne R$ such that $xy\in A$ implies $x\in A$ or $y\in A$.

If $(x+A)(y+A)=0+A$ in the quotient ring, then $xy\in A$. Primality implies $x+A=0+A$ or $y+A=0+A$. Therefore there are no two nonzero cosets whose product is zero:Properness also ensures its identity coset is not zero, so this quotient is an integral domain.

For scalars $\alpha ,\beta$ and vectors $v,w,u$, use the adjoint identity and conjugate linearity in the second argument:The difference of the two second-argument vectors is orthogonal to every $u$. Choosing $u$ to be that difference shows its squared norm is zero. ThereforeThe two conjugations are why the adjoint is a linear map, despite the conjugate-linear dependence of the representing functional on its second argument.

Expanding the determinant along the second column givesIf $b\notin \{-3,2\}$, there are three distinct real eigenvalues. Part (b)(ii) gives a nonzero eigenvector for each, and part (b)(i) makes them independent. They form a basis, so the matrix is diagonalisable.

When $b=-3$,has rank one for every $a$: the third row is nonzero, and the second row is $a/5$ times the third. Therefore the eigenspace for $-3$ has dimension two. The eigenvalue $2$ provides at least one further independent eigenvector by part (b), so again there is an eigenbasis.

When $b=2$,Its rank is one if $a=0$ and two otherwise. Thus the eigenspace for $2$ has dimension two exactly when $a=0$. For $a\ne 0$ it has dimension one, while the eigenspace for $-3$ also has dimension one, since $X+3I$ then has rank two. There are no other eigenvalues by the characteristic polynomial. A diagonal matrix in an eigenbasis would need three independent eigenvectors, whereas these two eigenspaces supply only two; diagonalisation is impossible. Conversely, at $a=0$ their dimensions add to three and part (b) makes their sum direct. ThereforeThe argument uses only the determinant, rank-nullity, the established direct-sum result, and the definition of diagonalisation.

Choose a basis and let $A$ represent $T$. Define the characteristic polynomial byUnder a change of basis with invertible change matrix $S$, the new matrix is $S^{-1}AS$. HenceThus the polynomial is independent of the chosen coordinates. It is monic of degree $\dim V$.

Set $w=i$, which satisfies $\Re w=0$, and take real parts of the absolutely convergent integral. Since $x^i=\cos (\log x)+i\sin (\log x)$ and $\cos (i\pi /2)=\cosh (\pi /2)$,Absolute integrability justifies moving the real-part operation through the integral.

Use the branch in (i) and the positively oriented slit-annulus contour. Its upper ray runs from $r$ to $R$, its outer circle counterclockwise, its lower ray from $R$ back to $r$, and its inner circle clockwise. Initially keep their arguments at $\alpha$ and $2\pi -\alpha$ as in the figure, with $0<\alpha <\pi /2$; let $\alpha \downarrow 0$ for fixed $0<r<1<R$. Both poles $i,-i$ lie inside.

Write $\sigma =\Re w\in (-1,1)$. On the outer circle the absolute integral is $O(R^{\sigma -1})$, since $|z^w|\le e^{2\pi |\Im w|}{R}^{\sigma }$ and $|z^2+1|\ge R^2-1$. On the inner circle it is $O(r^{\sigma +1})$. Both tend to zero. The real integral converges absolutely because its endpoint behaviors are $x^{\sigma }$ and $x^{\sigma -2}$.

The two residues, with the chosen arguments, areThe residue theorem and the two ray orientations therefore giveFor $w\ne 0$ in this strip we may cancel $1-e^{i\pi w}$; its only possible zero here is $w=0$. ConsequentlyAt $w=0$ the same result follows directly from $I(0)={\int }_0^{\infty }(1+x^2{)}^{-1}dx=\pi /2$, or by dominated convergence. The denominator has no zero in the open strip.

On $D=\mathbb{C}\setminus [0,\infty )$ choose the continuous argument $0<\arg z<2\pi$, and putThis is holomorphic on the slit plane for every fixed $w$. The upper boundary value at $x>0$ is $x^w$, whereas the lower boundary value is $e^{2\pi iw}{x}^w$. This difference is exactly what the keyhole contour measures.

If $f=z^{-n}g$ with $g(0)\ne 0$, continuity gives a small disk on which $|g(0)|/2\le |g(z)|\le 3|g(0)|/2$. Multiplying by $|z{|}^{-n}$ gives the two required bounds.

Conversely, set $g(z)=z^{n}f(z)$ on the punctured disk. The upper bound makes $g$ bounded, so the allowed removable-singularity theorem gives a holomorphic extension. The lower bound gives $|g(0)|\ge C_1>0$ by continuity. Therefore $f=z^{-n}g$ has a pole of exact order $n$. The lower bound is essential: the upper bound alone only proves order at most $n$.