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A holomorphic function on the open set $\Omega$ is complex differentiable at every point of $\Omega$: the limitexists independently of the direction of approach. Write $f(x+iy)=u(x,y)+iv(x,y)$. The Cauchy-Riemann equations areTo prove necessity, approach first with real $h$, obtaining $f^{\prime }(z)=u_x+i{v}_x$. Approach instead with $h=it$ for real $t$, obtainingEquality of real and imaginary parts yields the displayed Cauchy-Riemann equations.

A sufficient condition is that $u$ and $v$ have continuous first partial derivatives throughout $\Omega$ and satisfy the Cauchy-Riemann equations there. Indeed, real differentiability gives, for $h=a+ib$,The Cauchy-Riemann equations imply $f_y=i{f}_x$, so this is $f_{x}h+o(|h|)$. Dividing by $h$ proves complex differentiability and $f^{\prime }(z)=f_x(z)$ at every point. Merely having first partial derivatives and the equations at a point, without a suitable differentiability hypothesis, would not justify this argument.

Put $F={\mathbb{R}}^n\setminus \Omega$. This is a closed set. First assume $F$ is nonempty. The Heine-Borel theorem makes the nonempty bounded closed set $K$ a compact set. By the preceding parts, $x\mapsto d(x,F)$ is a continuous function and is strictly positive at every $x\in K$, because $K\subset \Omega$.

The extreme value theorem gives $x_{\ast }\in K$ attaining its minimum. ThereforeThe equality with the distance between two sets follows by taking the two infima successively. Crucially, compactness supplies a point attaining the minimum; positivity at each point alone would not make an arbitrary infimum positive. This is the principle that a compact subset of an open set has positive distance from its complement.

If $\Omega ={\mathbb{R}}^n$, its complement is empty. With the usual extended-real convention $\inf \varnothing =+\infty$, the stated inequality holds in that case too.

A subset $\Omega \subseteq {\mathbb{R}}^n$ is an open set if every $x\in \Omega$ admits a radius $r>0$ such that the open ballis contained in $\Omega$. A subset $F\subseteq {\mathbb{R}}^n$ is a closed set if its complement ${\mathbb{R}}^n\setminus F$ is an open set. Equivalently, $F$ contains the limit of every convergent sequence of its points.

Set $X=0$ to define a surjective ring homomorphismIts kernel of a ring homomorphism is exactly $(X)$: the constant coefficient in $X$ must vanish, so every remaining monomial is divisible by $X$. The first isomorphism theorem for rings therefore givesThe polynomial ring $\mathbb{Q}[Y]$ is an integral domain, because the product of two nonzero polynomials has nonzero leading coefficient. It is not a field: $Y$ has no polynomial multiplicative inverse, since a nonzero product $YQ(Y)$ has positive degree of a polynomial. Consequently $(X)$ is prime and is not maximal. Concretely, $(X)⊊(X,Y)⊊\mathbb{Q}[X,Y]$ is a strict chain of ideals.

If the quotient ring $R/I$ is an integral domain, then $I\ne R$, because an integral domain has $1\ne 0$. For $ab\in I$,There are no nonzero zero divisors in an integral domain, so $a+I=0+I$ or $b+I=0+I$. Thus $a\in I$ or $b\in I$, proving $I$ is a prime ideal.

If $R/I$ is a field, again $I$ is proper. Let $J$ be an ideal properly containing $I$, and choose $a\in J\setminus I$. The nonzero element $a+I$ has a multiplicative inverse $b+I$, so $ab-1\in I\subset J$. Since $ab\in J$, subtraction gives $1\in J$, hence $J=R$. Therefore $I$ is a maximal ideal. These arguments establish the relevant prime ideal quotient criterion and maximal ideal quotient criterion.

A prime ideal is a proper ideal $I\ne R$ such that $ab\in I$ implies $a\in I$ or $b\in I$. A maximal ideal is a proper ideal $I\ne R$ such that there is no ideal strictly between $I$ and $R$.

Every ideal of the integers has the form $(n)=n\mathbb{Z}$ for a unique nonnegative integer $n$. Its prime ideals and maximal ideals are respectivelyThe zero ideal is a prime ideal because the integers form an integral domain. For a prime number $p$, divisibility of $ab$ by $p$ forces divisibility of one factor, so $(p)$ is a prime ideal. Also any ideal $(d)$ containing $(p)$ has $d\mid p$, so it equals $(p)$ or $(1)$; thus $(p)$ is a maximal ideal.

If $n>1$ is composite, write $n=ab$ with $1<a,b<n$. Then $ab\in (n)$ while neither factor belongs to $(n)$, so $(n)$ is not a prime ideal. A prime number $p\mid n$ gives $(n)⊊(p)⊊\mathbb{Z}$, so $(n)$ is not a maximal ideal either. The zero ideal is not maximal since $(0)⊊(2)⊊\mathbb{Z}$, and $(1)=\mathbb{Z}$ is not proper.

A diagonalizable matrix over the field $F$ admits an invertible matrix $P$ with entries in $F$ such thatEquivalently, the vector space $F^n$ has a basis of eigenvectors of $A$. Indeed, $AP=PD$ says that each column of $P$ is an eigenvector with the corresponding diagonal entry as its eigenvalue. Requiring the basis and eigenvalues to lie in $F$ makes the choice of field essential.

For a linear functional $\lambda \in V^{\prime }$, the definition of the dual map givesThus the kernel of a linear map is precisely the annihilator of a vector subspaceWrite $n=\dim U$, $m=\dim V$ and $r=\dim \mathrm{im}T$. The rank-nullity theorem, the dimension formula for an annihilator of a vector subspace, and $\dim V^{\prime }=m$ giveApplying the same dimension formula in $U$ givesFinally, if $\alpha =T^{\prime }\lambda$ and $u\in \ker T$, then $\alpha (u)=\lambda (Tu)=\lambda (0)=0$. Therefore $\mathrm{im}{T}^{\prime }\subseteq (\ker T{)}^0$. These are finite-dimensional vector subspaces of equal dimension, so inclusion is equality:This is the dual image and kernel annihilator identity.

The annihilator of a vector subspace consists of the linear functionals that vanish on every element of that vector subspace:For the dimension identity, choose a basis $w_1,\dots ,w_k$ of $W$ and use basis extension to obtain $w_1,\dots ,w_n$ for $V$. A linear functional is determined by its values on these $n$ basis vectors. Belonging to the annihilator of a vector subspace sets its first $k$ values to zero and leaves its last $n-k$ values arbitrary. Equivalently the last $n-k$ elements of the dual basis form a basis of $W^0$.

The dual space consists of the linear functionals on the original vector space, with addition and scalar multiplication defined pointwise:The dual map acts by function composition with $T$:For $u\in U$, its value is $(T^{\prime }\lambda )(u)=\lambda (Tu)$. The composite is a linear functional because both $T$ and $\lambda$ are linear maps. Moreover $T^{\prime }(a\lambda +b\mu )=a{T}^{\prime }\lambda +b{T}^{\prime }\mu$, so the dual map is itself a linear map. Its direction is reversed because a measurement on the output of $T$ becomes a measurement on its input.

A pole of order $m\ge 1$ at $a$ meansEquivalently the lowest nonzero power in its Laurent series is $(z-a{)}^{-m}$. In particular, $(z-a{)}^{m}f(z)$ tends to a finite nonzero limit.

A removable singularity at $a$ is an isolated singularity for which one can assign a value at $a$ to obtain a holomorphic function on the whole disc. Equivalently its Laurent series has no negative powers. A finite limit $L$ supplies the extension value $f(a)=L$.

An isolated singularity at $a$ means that the function is holomorphic on some punctured disc $0<|z-a|<r$, while it is undefined or fails to be holomorphic at $a$. The punctured neighbourhood contains no other singularities; this isolation is needed before applying the usual removable/pole/essential classification.

Away from zero, singularities occur exactly when $e^{1/z}=-1$, namelyFor $D(z)=z(e^{1/z}+1)$, differentiation at $z_k$ gives $D^{\prime }(z_k)=1/z_k\ne 0$. Thus every $z_k$ is a simple pole, with residue $z_k$.

As $|k|\to \infty$, these poles accumulate at zero. Therefore zero is a non-isolated singularity. Every punctured disc around it contains poles, so the function is not holomorphic on any punctured disc there. In particular zero cannot be classified as an isolated essential singularity; the isolated-singularity classification does not apply. The displayed poles and their accumulation point zero exhaust the finite singularities.