A holomorphic function with $f^{\prime }(z)\ne 0$ at every point is locally a conformal map. Write $a=z_0$. Since $|a|<1$ and $|z|<1$, $1-\overline{a}z\ne 0$, andThus the specified Möbius transformation is conformal throughout the disk. Expanding the two squared absolute values givesso $f$ maps the open unit disk into itself. Solving $w=(a-z)/(1-\overline{a}z)$ gives $z=(a-w)/(1-\overline{a}w)=f(w)$. Equivalently $f(f(z))=z$. Hence $f$ is its own inverse, proving both injectivity and surjectivity on the disk. Nonvanishing derivative establishes the local angle-preserving property; the explicit inverse supplies the separate global bijection claim.

Apply the residue theorem to $F(z)=(z^2+4{)}^{-2}$ on the upper semicircle of radius $R>2$, traversed counterclockwise. Its sole enclosed pole is the double pole at $2i$, andOn the semicircular arc, $|z^2+4|\ge R^2-4$, so the ML inequality bounds its contribution by $\pi R/(R^2-4{)}^2\to 0$. The real integral is absolutely convergent, since its integrand is $O(|x{|}^{-4})$ at infinity. Taking $R\to \infty$ in the contour identity yields

A closed curve has $\gamma (a)=\gamma (b)$. A simple curve has no repeated points apart from the permitted equality of its two endpoints when it is closed. A smooth curve, in the regular contour-integration convention, has a continuous nonzero tangent ${\gamma }^{\prime }$; piecewise smooth contours are treated by adding the integrals over their smooth pieces. For a continuous integrand, its contour integral is defined byFor the positively oriented circle, ${\gamma }^{\prime }(t)=ir{e}^{it}$ and $(\gamma (t)-z_0{)}^n=r^n{e}^{int}$. Direct parameter integration therefore givesFor $n\ne -1$, an elementary antiderivative is $e^{i(n+1)t}/[i(n+1)]$ and its endpoint values agree because $n$ is an integer. For $n=-1$, the integrand of the parameter integral is the constant $i$. No integral theorem is needed.

Construct the vectors inductively by the Gram-Schmidt process, proving its needed properties directly. Suppose $w_1,\dots ,w_{j-1}$ form an orthonormal basis of $U_{j-1}$, and putFor every $i<j$, taking an inner product with $w_i$ gives $⟨r_j,w_i⟩=0$. Also $r_j\ne 0$, because otherwise $u_j$ would lie in the linear span of its predecessors, contrary to linear independence. DefineIt lies in $U_j$, has unit norm, and is orthogonal to every $u_i$ with $i<j$, since those vectors lie in $U_{j-1}$. Writing $u_j=r_j+{\sum }_{k<j}⟨u_j,w_k⟩w_k$ gives $⟨w_j,u_j⟩=\Vert r_j\Vert >0$. The same identity shows that $w_1,\dots ,w_j$ span $U_j$, completing the induction, including the empty sum at $j=1$.

For uniqueness, any candidate $w\in U_j$ expands in the basis $w_1,\dots ,w_{j-1},r_j$. Orthogonality to every earlier $u_i$, hence to every earlier $w_i$, makes its first $j-1$ coefficients zero. Thus $w=t{r}_j$. Unit norm forces $t=\pm 1/\Vert r_j\Vert$, and $⟨w,u_j⟩>0$ forces the positive sign. This proves uniqueness in the Gram-Schmidt process with positive normalization without assuming a projection theorem or any other major orthogonalization result.

Integrating $e^{-pt}\sin (ct)$, or taking the imaginary part of ${\int }_0^{\infty }{e}^{-(p-ic)t}dt$, givesThe Laplace transform convolution theorem states $\mathcal{L}(f\ast g)=\bar{f}\bar{g}$ whenever the transforms and convolution are defined. Applying it to the Volterra integral equation gives $\bar{x}+\bar{x}/(p^2+1)=\bar{F}$, soSince ${\mathcal{L}}^{-1}[(p^2+2{)}^{-1}]=\sin (\sqrt{2}t)/\sqrt{2}$, a second application of the convolution theorem yields the sine-kernel Volterra resolventThe transform argument assumes suitable growth for the transforms to exist. The resulting causal resolvent formula also applies to locally integrable forcing on finite intervals, as can be checked by substitution and exchanging integrals over a finite triangle.

The characteristic equations for a first-order quasilinear equation are $\dot{x}=x$, $\dot{y}=y$, $\dot{u}=2u$, with characteristic parameter $s$. Through the prescribed point they giveFor a nonzero projected initial point, the characteristic projections are rays from the origin, not trajectories crossing through the origin to the opposite ray. If $x_0=y_0=0$, a classical solution graph requires $u_0=0$, because the original equation reduces there to $0=2u$.

Parameterize the initial line by $(x,y,u)=(a,1-a,a^3)$. Along its characteristic, $x+y=e^s>0$, $a=x/(x+y)$ and $u=a^3{e}^{2s}$. ThereforeDifferentiation verifies the equation, and substitution of $x+y=1$ verifies the initial data. Characteristic reachability for radial transport shows why the domain is this positive half-plane: each point there lies on precisely one characteristic from the data line. Rays in $x+y<0$ never meet that line, so their values are not determined by these data. Allowing a negative scaling factor would incorrectly join two different characteristic rays through a degenerate origin.

Since the exponential factor is positive, equilibria must have $y=0$ and $x^4=1$. Thus $\boxed{(x,y)=(-1,0),(1,0)}$. At either point the Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ -4x^3& 0\end{array}\right)$. At $(-1,0)$ its eigenvalues are $\pm 2$, so this is a saddle point. At $(1,0)$ they are $\pm 2i$: the linearization has a centre equilibrium.

The requested first-order approximation follows by putting $x=1+u$, $y=v$: $\dot{u}=v+O(\Vert (u,v){\Vert }^2)$ and $\dot{v}=-4u+O(\Vert (u,v){\Vert }^2)$. The linearized system conserves $4u^2+v^2$, giving the approximate ellipsesHowever, the printed nonlinear system has a weak repelling focus at $(1,0)$, rather than a centre. A centre in the linearization need not be a nonlinear centre. Here that distinction can be proved explicitly, without relying on the numerical drift in the figure.

Put $X=2(x-1)$, $Y=y$, $\tau =2t$ and $r^2=X^2+Y^2$. The weak focus of an exponential polynomial planar system has the local positive functionExpanding the vector field through cubic order and differentiating givesThe cubic correction cancels the cubic terms in the derivative of $H$, and the quartic correction leaves the positive radial term $3r^4/16$. Since $H=r^2/2+O(r^3)$ is positive and the angular velocity stays near $-1$, small nonzero trajectories spiral outward. Equivalently the reversed system has a strict local Lyapunov function and an asymptotically stable equilibrium. In particular, a periodic orbit sufficiently near the equilibrium is impossible, because $H$ would have to increase strictly and return to its starting value. This proves the failure of the printed centre assertion while retaining its correct linearized ellipse calculation.

A pole of order $m$ at $a$ means that $f(z)=h(z)/(z-a{)}^m$ near $a$, where $h$ is holomorphic and $h(a)\ne 0$. Equivalently its Laurent series has a finite principal part with leading nonzero term $a_{-m}(z-a{)}^{-m}$. The residue $c$ is the coefficient $a_{-1}$. For a simple pole, $m=1$, this coefficient is $h(a)$, soFor $\cot z=\cos z/\sin z$, the zeros of the denominator are exactly $z=n\pi$, $n\in \mathbb{Z}$: this follows from $e^{2iz}=1$. Each is simple, because ${\sin }^{\prime }(n\pi )=\cos (n\pi )=(-1{)}^n\ne 0$. ThereforeThere are no other finite singularities. If infinity is included, it is an accumulation point of these poles, so it is not an isolated singularity to which the usual removable/pole/essential classification could be applied.