A conformal map preserves angles between intersecting regular curves, including their orientation. If $f$ is holomorphic with $f^{\prime }(a)\ne 0$, the image tangent vectors are $f^{\prime }(a)$ times the original tangent vectors. Multiplication by this nonzero complex number gives a common positive scaling and rotation, so their oriented angle is unchanged. This proves conformality throughout the stated domain.

At a zero derivative this can fail: $f(z)=z^2$ at zero sends rays of arguments $0$ and $\pi /4$ to rays of arguments $0$ and $\pi /2$. The angle is doubled, so this holomorphic function is not conformal there.

For the disk map $C(z)=i(1+z)/(1-z)$, computeIts inverse is $z=(w-i)/(w+i)$; for $\mathrm{Im}w>0$, the inequality $|w-i|<|w+i|$ puts this inverse in the unit disk. Both rational maps are holomorphic on their respective domains. ThusFor the final region, writing $z=x+iy$ gives $|e^z|=e^x\in (0,1)$ and $\arg (e^z)=y\in (0,\pi )$. Hence its exponential image is exactly the upper half of the unit disk,The exponential is injective on this strip, since two equal exponential values differ by $2\pi i$ times an integer while the imaginary parts here differ by less than $\pi$. Its derivative never vanishes and its inverse on $V$ is the logarithm with argument in $(0,\pi )$.

Now $L(w)=(1+w)/(1-w)$ maps $V$ conformally onto the first quadrant. Indeedand the inverse $(s-1)/(s+1)$ has modulus below one and positive imaginary part for every first-quadrant $s$. Squaring maps that quadrant bijectively onto $\mathbb{H}$, with nonzero derivative. Therefore a required composition isEvery component is a conformal bijection between the stated domains, so the composition is one as well.

The residue at $a$ is the coefficient $c_{-1}$ of $(z-a{)}^{-1}$ in the Laurent series. A simple pole means $f(z)=c_{-1}/(z-a)+h(z)$ near $a$, with $c_{-1}\ne 0$ and $h$ holomorphic across $a$.

For the specified positively oriented arc, direct parametrization gives ${\int }_{{\gamma }_{ϵ}}(z-a{)}^{-1}dz=i(\beta -\alpha )$. The remaining integral is bounded by $ϵ(\beta -\alpha )\sup |h|$, which tends to zero because $h$ is bounded near $a$. ThereforeThe simple-pole hypothesis matters. For $f(z)=(z-a{)}^{-2}$, with $\alpha =0$, $\beta =\pi$, the residue is zero butFor the particular function, expanding the exponential givesThus $\boxed{\mathrm{res}(f;0)=1/2}$, and the pole is simple. On the large upper semicircle, $|e^{iz}|\le 1$, givingClose the real-axis path from $-R$ to $-ϵ$ and $ϵ$ to $R$ by this upper arc and a clockwise upper indentation at zero. The function is holomorphic in the indented region, so the sum of all four integrals is zero. The clockwise indentation tends to $-i\pi /2$. For real $x$,The real part is odd and cancels on the symmetric real pieces; the imaginary part is even. Its integral converges since $(x-\sin x)/x^3\to 1/6$ at zero and is $O(x^{-2})$ at infinity. Letting $ϵ\to 0$ and $R\to \infty$ therefore yields $2iI-i\pi /2=0$, where $I$ is the requested real integral. Hence

For an azimuth-independent product $u=R(r)\Theta (\theta )$, multiply Laplace's equation by $r^2/(R\Theta )$ where the factors are nonzero. The radial and angular expressions must equal opposite constants:The resulting differential identities extend across isolated zeros by continuity. Put $\rho =\cos \theta$. Then ${\Theta }^{\prime }=-\sin \theta P^{\prime }$ andThis gives $((1-{\rho }^2)P^{\prime }{)}^{\prime }+\lambda P=0$ on $(-1,1)$.

A nonzero angular factor must be regular at both poles: it must be bounded as $\rho \to \pm 1$ and extend smoothly as an axisymmetric function on the sphere. Equivalently one may use the regular endpoint solution at each end; the logarithmic endpoint solutions are excluded. These two endpoint conditions select $\boxed{\lambda =n(n+1),n=0,1,2,\dots }$ and the multiples of $P_n$. It is not sufficient to impose no condition at the endpoints, since local solutions exist for arbitrary $\lambda$. One way to see the quantization is the regular Legendre series about one endpoint: unless it terminates at $\lambda =n(n+1)$, continuation to the other endpoint has a logarithmic singularity.

Direct differentiation verifiesIndeed $R=r^m$ gives $m(m+1)=\lambda$, so the two powers for degree $n$ are $r^n$ and $r^{-n-1}$.

For the shell assume $a>1$. The outer datum decomposes as ${\rho }^2=\frac{1}{3}{P}_0(\rho )+\frac{2}{3}{P}_2(\rho )$. Normalize each radial solution to vanish at $1$ and equal $1$ at $a$; superposition givesEvery term is harmonic in the shell and regular at the angular poles; substitution at each boundary gives precisely the two prescribed data. For uniqueness among functions harmonic in the shell and continuous on its closure, subtract any two solutions. Their difference is harmonic, zero on both boundary spheres, and by the maximum principle is both at most zero and at least zero on the bounded shell. It vanishes identically. Thus the classical Dirichlet solution is unique. The usual continuity at the boundary is the interpretation of the stated boundary conditions; allowing singular boundary behavior would remove that hypothesis.

Use a parameter $t$. The characteristic equations areFor a graph of a function $z=f(x,y)$ satisfying the equation, solve the first two equations after substituting $z=f(x,y)$. The chain rule and the partial differential equation give $df(x(t),y(t))/dt=f+bx-ay$, exactly the third equation. Uniqueness for the resulting linear ordinary differential equation system shows that its characteristic through the initial point agrees with this curve on the graph of a function. This proves invariance as long as the graph of a function remains defined along the curve. A claim about the entire curve additionally requires a surface continued along that flow; a restricted graph of a function patch alone cannot imply global continuation.

Let $\mathbf{k}=(a,b,c)$, $\mathbf{\ell }=(l,m,n)$ and $\mathbf{r}=(x,y,z)$. Then $\dot{\mathbf{r}}=\mathbf{r}-\mathbf{k}\times \mathbf{r}$. Since $|\mathbf{k}|=1$,Along the characteristic starting from $s\mathbf{\ell }$ these give $\mathbf{k}\cdot \mathbf{r}=e^{t}s(\mathbf{k}\cdot \mathbf{\ell })$ and $|\mathbf{r}|=e^t|s|$. Consequently the characteristic surface generated by the given line satisfiesTo see that the whole displayed cone is generated, not just contained in it, decompose $\mathbf{\ell }=\kappa \mathbf{k}+\mathbf{v}$ with $\mathbf{v}\perp \mathbf{k}$, $\kappa =\mathbf{k}\cdot \mathbf{\ell }$. The full solution isWhen $0<|\kappa |<1$, variation of $t$ gives all azimuths about the axis, while the freely chosen $s$ supplies every positive radius on both halves. This is precisely the two-sided circular cone, withThe edge cases matter: $\kappa =0$ gives the plane $ax+by+cz=0$ (semi-angle $\pi /2$), and $|\kappa |=1$ generates only the axis (semi-angle zero), not a two-dimensional surface. In the latter case the initial line is itself characteristic, so it does not determine a unique integral surface. The nondegenerate cone is smooth away from its vertex and can be used as a graph of a function only on patches whose tangent plane is not vertical. Thus the printed geometric conclusion has its ordinary characteristic-surface meaning in the nondegenerate case, with these domain and degeneracy qualifications; it cannot be asserted for every global smooth graph of a function without them.

If $Au+Bv=0$, evaluation at zero gives $A=0$ and differentiation at zero gives $B=0$. Thus the solutions are linearly independent. For $w=u{v}^{\prime }-u^{\prime }v$,so $\boxed{w(x)=\exp (-{\int }_0^{x}p(s)ds)>0}$.

Variation of parameters sets $y=A(x)u+B(x)v$ with $A^{\prime }u+B^{\prime }v=0$, $A^{\prime }{u}^{\prime }+B^{\prime }{v}^{\prime }=f$. Solving gives $A^{\prime }=-vf/w$ and $B^{\prime }=uf/w$. Zero integration constants enforce both initial conditions. ThereforeFor fixed $t$, this kernel is a linear combination of homogeneous solutions, so $K_{xx}+p{K}_x+qK=0$. On the diagonal $K(t,t)=0$ and $K_x(t,t)=(v^{\prime }(t)u(t)-u^{\prime }(t)v(t))/w(t)=1$. Differentiating the integral twice now verifies the forced equation as well as $y(0)=y^{\prime }(0)=0$. Standard uniqueness of the linear initial-value problem makes this the solution, for continuous $f$.

If $u,v$ have period $T>0$, their derivatives do also, hence $w$ is periodic. The kernel is periodic in its first variable. With $f=v$, split the integral at $x$ and use this periodicity to getAt $x=0$, the right side is $-{\int }_0^{T}v(t{)}^2/w(t)dt<0$, because $w>0$ and $v^{\prime }(0)=1$ ensures $v$ is not identically zero. Thus $\boxed{y(T)\ne y(0)}$, so $y$ cannot have period $T$. This direct sign argument also rules out any apparent cancellation between the two terms.