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.

The possible singular points are $i,-i,1$. Near $i$, put $h(z)=1/((z+i)(z-1{)}^2)$; thenThe numerator vanishes at $i$, so the apparent pole cancels and $i$ is removable. Its extension value is $h^{\prime }(i)=1/4+3i/8$, obtained from $h^{\prime }/h=-1/(z+i)-2/(z-1)$.

At $-i$, the first term has a simple pole with nonzero residueThe subtracted term is holomorphic there, so $-i$ is a simple pole. At $1$,so $1$ is a double pole. There are no other finite singularities.

A path-connected space $S$ has, for every pair $z_0,z_1\in S$, a continuous path $\gamma :[0,1]\to S$ with $\gamma (0)=z_0$ and $\gamma (1)=z_1$. Every path-connected space is a connected space, because a separation of the target would pull a joining path apart and contradict connectedness of an interval.

A connected space $S$ cannot be expressed as $S=A\cup B$ with $A,B$ disjoint nonempty relatively open sets. Equivalently, the only subsets simultaneously relatively open and closed are $\varnothing$ and $S$. A decomposition into such $A,B$ is a separation of the topological space.

In the subspace topology on $S$, a subset $A$ is open in $S$ precisely when $A=S\cap O$ for some open set $O\subseteq \mathbb{C}$. A subset $B$ is closed in $S$ precisely when $S\setminus B$ is open in $S$, equivalently $B=S\cap F$ for some closed set $F\subseteq \mathbb{C}$.

A compact space is one for which every open cover has a finite subcover. Here the cover may use relatively open subsets of $S$, or equivalently ambient open subsets covering it. The Heine-Borel theorem states thatClosedness and boundedness here refer to the ambient complex plane, not merely closedness in $S$ itself.

The union is bounded but is not a closed set: the points $-1+1/n$, $n\ge 2$, belong to the open disc and tend to $-1$, which lies in neither disc. Thus it is not compact by the Heine-Borel theorem.

Nevertheless it is connected, indeed path-connected. Every point of the open disc joins $0$ by a straight segment. The segment from $0$ to $1$ stays in the union, since its points before the endpoint belong to the open disc and the endpoint $1$ belongs to the closed disc. Every point of that closed disc joins $1$ inside it by convexity. Concatenating these paths joins any two points of the union. The fact that the two discs are disjoint does not imply disconnection: their contact point belongs to one of them. This illustrates connectedness at a one-sided contact point.

The disc is a closed set and is bounded in the complex plane, so it is compact by the Heine-Borel theorem. It is convex: the triangle inequality gives $|(1-t)z_0+t{z}_1|\le 2$ for all $t\in [0,1]$ when both endpoints lie in it. Straight segments therefore supply paths, so it is path-connected and hence connected.

The integers are unbounded in the complex plane, so the Heine-Borel theorem shows they are not compact. They are also not connected: $\{0\}$ and $\mathbb{Z}\setminus \{0\}$ are disjoint nonempty relatively open sets covering $\mathbb{Z}$. Each integer is isolated, so the induced subspace topology is discrete.

Let $N$ count everyone present, including the host. The grouping, seating and departure conditions give the modular congruencesThe first two imply $N=39+45k$, and oddness forces $k$ even. By the Chinese remainder theorem, the complete solution isBecause not every invitee attended, at most $y-1$ guests came, so $N\le y$. For $y=120$, the only positive possibility is 39 people. It is feasible: seven groups of five and one of four; three tables of ten and one of nine; nineteen departing pairs and the host.

For $y=150$, both 39 and 129 people are possible. The latter uses twenty-five groups of five and one of four, three tables of ten and eleven of nine, and sixty-four departing pairs. Thus the larger invitation count destroys uniqueness; it does not force a different attendance.

Take $R=\mathbb{Z}$, $I=a\mathbb{Z}$ and $J=b\mathbb{Z}$. Since $a,b$ are coprime integers, Bézout's identity gives $ra+sb=1$, so $I+J=\mathbb{Z}$. Their intersection is $ab\mathbb{Z}$: it consists of integers divisible by both $a$ and $b$, and coprimality makes this equivalent to divisibility by $ab$. Thus the Chinese remainder theorem givesas a ring isomorphism. This includes $a=1$ or $b=1$, where the corresponding factor is the zero ring.

The Chinese remainder theorem for two comaximal two-sided ideals $I,J$ of a ring states that the natural ring homomorphism inducesIndeed, the map $r\mapsto (r+I,r+J)$ has kernel of a ring homomorphism $I\cap J$. Choose $i\in I$, $j\in J$ with $i+j=1$. Given target representatives $a,b\in R$, the element $aj+bi$ is congruent to $a$ modulo $I$ and to $b$ modulo $J$, proving surjectivity. The first isomorphism theorem for rings gives the result. For a commutative ring, $I\cap J=IJ$: if $x\in I\cap J$, then $x=xi+xj\in IJ$, while the reverse inclusion follows from the ideal properties.

Use the inner product with $a=2$. The required first vector $v_1=(1/2,0{)}^T$ already has squared norm $4(1/2{)}^2=1$. Apply the Gram-Schmidt process to $v_1$ followed by $e_2=(0,1{)}^T$. Since $\beta (e_2,v_1)=-1$, its orthogonal remainder is $w=e_2+v_1=(1/2,1{)}^T$. Its squared norm is $\beta (w,w)=1-2+3=2$. Therefore an orthonormal basis containing the specified vector isThe second vector may also be replaced by its negative.

A real inner product must be symmetric. Here ${\beta }_a(e_1,e_2)=-2$ and ${\beta }_a(e_2,e_1)=-a$, forcing $a=2$. For this value,This is strictly positive for every nonzero vector. Bilinearity is already given, so symmetry and positive definiteness prove

The Gram-Schmidt process subtracts projections onto directions already constructed and then normalizes. Define recursivelywhere the empty sum for $k=1$ is zero. Assuming the preceding $e_j$ are orthonormal, taking the inner product of $w_k$ with $e_i$, $i<k$, gives zero. Moreover $w_k\ne 0$: otherwise $v_k$ would belong to the span of $v_1,\dots ,v_{k-1}$, contrary to linear independence. Normalization is therefore legitimate and produces an orthonormal basis. At each stage the first $k$ constructed vectors have the same span as the first $k$ original vectors.