A critical point is an equilibrium: $\boxed{f(a,b)=g(a,b)=0}$. Put $X=x-a$, $Y=y-b$. The Jacobian matrix $J$ gives the first-order linearized equation $(\dot{X},\dot{Y}{)}^{\mathsf{T}}=J(X,Y{)}^{\mathsf{T}}$. The following eigenvalue tests classify that linearized system. For a continuously differentiable nonlinear system, the cases with nonzero real parts persist as the usual local node, saddle or spiral classifications. The purely imaginary case needs extra nonlinear information, as explained in (iv).

The domain has two connected components. The zeros force the function to vanish on the upper half-plane but impose nothing on the lower one. DefineThis function is locally constant and hence holomorphic on the stated open set, and is not identically zero on that whole set. Thus the answer is yes. Connectedness is exactly the missing hypothesis for a global identity conclusion.

Here the zeros accumulate only at the boundary point zero. An explicit holomorphic, nonzero example on the upper half-plane isAt $z_n=i/(n+1)$ its argument is $\pi (n+1)$, so it vanishes. It is holomorphic because the only excluded point of its formula is zero; it is not identically zero, for instance at $z=2i$. Thus the answer is yes.

The given distinct zeros converge to $i$, which lies inside the connected upper half-plane. The identity theorem for holomorphic functions forces the function to vanish identically. Thus $\boxed{\text{no}}$.

The denominator vanishes at $z_k=1/(2\pi ik)$ for every nonzero integer $k$. These points approach zero, the denominator has a simple zero at each, and $\sin z_k\ne 0$ because a nonzero purely imaginary number is not a zero of sine. Hence there are poles arbitrarily close to zero. The origin is a nonisolated singularity, so it has no residue defined by a Laurent series on a punctured disc. In particular it must not be classified as an isolated essential singularity merely because $e^{1/z}$ has one.

Use $\sin z=z+O(z^3)$ andTheir product is $z^{-2}+(3/2)z^{-1}+O(1)$. Thus the singularity is a pole of order two, with $\boxed{{\mathrm{Res}}_{0}f=3/2}$.

Both numerator and denominator have simple zeros at zero, and $1-e^{-z}=z+O(z^2)$. Their quotient tends to one and extends holomorphically with that value. Hence the singularity is removable, and $\boxed{{\mathrm{Res}}_{0}f=0}$ because its Laurent series has no negative powers.

If a holomorphic bijection existed, its inverse would be an entire function taking values in the unit disc. Liouville theorem would make that inverse constant, contradicting bijectivity. Thus no holomorphic bijection exists.

A continuous bijection does exist, even a homeomorphism:Both maps preserve the argument and change the radius by the mutually inverse strictly increasing functions $r\mapsto r/(1-r)$ and $s\mapsto s/(1+s)$. The displayed formulas also prove continuity at zero and everywhere else.

Let $h=f^{-1}$, $w_0=f(z_0)$ and $z=h(w)$. Continuity of $h$ gives $z\to z_0$ as $w\to w_0$, and injectivity gives $z\ne z_0$ when $w\ne w_0$. ThereforeThe assumed nonvanishing derivative makes this limit well defined. It exists at every $w_0\in V$, so $h$ is holomorphic, with $\boxed{h^{\prime }(w)=1/f^{\prime }(h(w))}$.

Squaring maps the first quadrant bijectively onto the upper half-plane. For $\zeta$ in that half-plane, $|\zeta -i|<|\zeta +i|$, and the Cayley transform $(\zeta -i)/(\zeta +i)$ maps it bijectively onto the unit disc. Their composition isThe inverse is the unique first-quadrant square root of $i(1+w)/(1-w)$, and is holomorphic by the first argument. Hence $H(w)=g(F^{-1}(w))$ is holomorphic throughout the disc. Its Taylor series at zero converges for $|w|<1$, locally uniformly there. Substituting $w=F(z)$ gives the stated expansion, with $c_n=H^{(n)}(0)/n!$, convergent at every point of the quadrant and locally uniformly on it.

A Euclidean domain is an integral domain equipped with a function $\delta :R\setminus \{0\}\to \mathbb{N}$ such that for every $a\in R$ and nonzero $b\in R$ there are $q,r\in R$ with $a=qb+r$ and either $r=0$ or $\delta (r)<\delta (b)$. One may additionally require $\delta (a)\le \delta (ab)$ for nonzero $a,b$; degree on a polynomial ring also satisfies this convention.

An integral domain is a commutative ring with identity $1\ne 0$ and no zero divisors: $ab=0$ implies $a=0$ or $b=0$. Equivalently nonzero elements admit cancellation in multiplication.

For equal-length crystallographic root configurations in a plane, taking $u=v$ in the closure condition gives $S_v(v)=-v$, so every vector occurs with its negative. For fixed $v\in \Phi$, a vector $u\in \Phi$ has $⟨u,v⟩\in \{-1,-1/2,0,1/2,1\}$, because both vectors are unit vectors. Thus its angle relative to $v$ belongs to the finite list $0,\pm \pi /3,\pm \pi /2,\pm 2\pi /3,\pi$.

If all vectors are collinear, the set is exactly $\{v,-v\}$, giving two opposite points of the unit circle.

If an orthogonal vector $w$ occurs, closure supplies $\pm v,\pm w$. A unit vector making an angle of $\pi /3$ or $2\pi /3$ with $v$ would have inner product $\pm \sqrt{3}/2$ with $w$, contradicting integrality of twice that inner product. Thus these four points are the whole set: the vertices of a square.

In the remaining case a noncollinear pair makes an angle of $\pi /3$ or $2\pi /3$. Changing a sign, take their angles as $0,\pi /3$. Reflecting the second in the line perpendicular to the first supplies the angle $2\pi /3$, and negation supplies the other three hexagon vertices. No perpendicular vector can occur, by the preceding inner-product obstruction. The initial finite list now leaves only these six points. Conversely the opposite pair, square and regular hexagon each satisfy both reflection closure and the integrality condition. Hence

The functional $\ell (u)=⟨u,v⟩$ is linear and nonzero because $\ell (v)=⟨v,v⟩>0$. It therefore has rank one, and rank-nullity theorem givesNo Gram–Schmidt argument is needed. Every vector decomposes uniquely as $u=w+av$, where $w\perp v$ and $a=⟨u,v⟩/⟨v,v⟩$. Substitution gives $S_v(w+av)=w-av$. Thus the eigenspace for $-1$ is $\mathbb{R}v$, and that for $+1$ is $v^{\perp }$. The latter is an actual eigenvalue when $n\ge 2$; when $n=1$ its eigenspace is zero and only $-1$ occurs. In the plane the map is reflection in the line perpendicular to $v$.

Write $N=T-I$. Cayley-Hamilton theorem gives $N^3=0$. Its minimal polynomial is therefore $X^k$ for exactly one $k\in \{1,2,3\}$, so the possibilities for that of $T$ areLet $d=\dim \ker N$, which is the dimension of the only eigenspace. If $d=3$, then $N=0$ and $T=I$.

If $d=2$, rank-nullity theorem gives $\dim \mathrm{im}N=1$. The restriction of the nilpotent map $N$ to this invariant line is zero, so $N^2=0$. Choose $v_3$ with $v_2=N{v}_3\ne 0$, and choose $v_1\in \ker N$ independent of $v_2$. Then $v_3\notin \ker N$, so these three vectors form a basis; its matrix for $T$ is the printed $J_1(1)\oplus J_2(1)$.

If $d=1$, then $\mathrm{im}N$ has dimension two, so $N^2$ cannot vanish: otherwise $\mathrm{im}N\subseteq \ker N$. Choose $v_3$ with $N^2{v}_3\ne 0$ and set $v_2=N{v}_3$, $v_1=N^2{v}_3$. Applying $N^2$ and then $N$ to a putative linear relation proves these three vectors independent. The matrix is the printed $J_3(1)$.

These cases exhaust the possible eigenspace dimensions. They are pairwise dissimilar because similarity preserves the eigenspace dimension, equivalently the three different minimal polynomials. Hence every matrix is similar to exactly one of the three displayed forms.

Proceed by induction on $n=\dim V$. The zero-dimensional case is immediate. Over $\mathbb{C}$, the fundamental theorem of algebra supplies a root $\lambda$ of the characteristic polynomial. Since $\det (T-\lambda I)=0$, there is a nonzero vector $v_1$ with $T{v}_1=\lambda v_1$. The line $W=\mathbb{C}{v}_1$ is invariant, so $T$ induces a linear map $\overline{T}$ on $V/W$. By induction, choose an upper triangular basis for that quotient and lift its vectors to $v_2,\dots ,v_n$. Then $T{v}_j\in \mathrm{span}(v_1,\dots ,v_j)$ for every $j$. Thus $v_1,\dots ,v_n$ is the required basis and $\boxed{T\text{ is triangularizable}}$.

The same conclusion holds over $F$ when the characteristic polynomial splits over $F$. Indeed an eigenvalue then lies in $F$, and the block matrix arising from the invariant line gives ${\chi }_T(X)=(X-\lambda ){\chi }_{\overline{T}}(X)$. The quotient characteristic polynomial therefore also splits, so the induction works over $F$ without requiring algebraic closure.

Suppose a nonconstant periodic orbit exists for the continuously differentiable planar vector field. Uniqueness implies that its image over a minimal period is a simple closed curve, enclosing a region $\Omega$ of positive area. Along the orbit, the differential form $fdy-gdx$ vanishes because $dx=fdt$ and $dy=gdt$. Green's theorem, with positive orientation, therefore givesa contradiction. This is the Bendixson-Dulac criterion in the whole plane; the fact that the enclosed region also lies in the domain is essential.

For the second-order equation set $y=\dot{x}$. The equivalent system is $\dot{x}=y$, $\dot{y}=-(1+x^2)y-\cos x$. Its divergence is $\boxed{-(1+x^2)<0}$ throughout the plane. Hence there are no nonconstant periodic solutions.

The nonconstant qualification is necessary: the equilibrium solutions $x(t)\equiv \pi /2+k\pi$, $k\in \mathbb{Z}$, do exist and are periodic in the literal sense of satisfying every positive period. The source's usual dynamical interpretation of “periodic solution” here is a nonconstant periodic orbit.