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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.

Suppose, for a contradiction, that $f$ has no zero in the open disk. The strict boundary inequality implies no boundary zero either. Thus $1/f$ is holomorphic on a neighborhood of the closed disk: compactness and continuity keep $f$ nonzero in a sufficiently small neighborhood. Apply the $k=0$ Cauchy estimate to $1/f$:This contradicts the assumed strict inequality. If $f(z_0)=0$, the conclusion is immediate without taking a reciprocal. Therefore $f$ has a zero in the open disk. This is a local form of the minimum modulus principle.

On the circle, the integrand in the Cauchy derivative formula has modulus at most $M/r^{k+1}$, and the circle has length $2\pi r$. The ML inequality therefore givesThe maximum $M$ exists because the modulus of the continuous function $f$ attains a maximum on the compact circle. These are the Cauchy estimate.

For the positively oriented circle $C=\{z:|z-z_0|=r\}$, the Cauchy derivative formula statesThe hypothesis that the closed disk lies in the open holomorphy domain ensures the formula applies on and inside this circle.