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Let $B$ be planar Brownian motion started at $z\in D$, let ${\tau }_D$ be its exit time, and let $\varphi :D\to D^{\prime }$ be conformal. DefineThe conformal invariance of planar Brownian motion states thatis Brownian motion started at $\varphi (z)$ and stopped when it exits $D^{\prime }$. Thus conformal maps preserve Brownian paths after this random time-change.

Put $F(z)=z+z^{-1}$. On the unit semicircle, $F(e^{i\theta })=2\cos \theta$, andThe conformal invariance of planar Brownian motion and the Poisson kernel for the upper half-plane therefore give the exit density with respect to $d\theta$:As $z\to \infty$ in $\mathbb{H}$,uniformly in $\theta \in [0,\pi ]$. Hence

Write $Z_0=r_0{e}^{i{\alpha }_0}$ with $0<r_0<1$ and choose its initial argument once. The exponential map in the source is a covering of the punctured disc, rather than a bijective map onto the full disc. Its pathwise inverse is the continuous liftFor $t<T_{\mathbb{D}}$ this lies in the upper half-plane, and at the disc exit it lies on the real line. Local conformal invariance of planar Brownian motion, or the quadratic variation of its analytic derivative $-i/Z$, shows that $W_t$ is planar Brownian motion run at the clockThe disc exit time is finite almost surely. The path avoids zero on the compact interval through its exit, so its minimum modulus there is positive and $A_{T_{\mathbb{D}}}<\infty$. Before this terminal clock time the lifted imaginary coordinate is positive, and at it the coordinate is zero. Hence it is exactly the upper-half-plane exit time of the time-changed Brownian motion, not an additional arbitrary stopping time.

The starting point of the lift is ${\alpha }_0+ib$ with $b=-\log r_0>0$. Its real exit coordinate has Cauchy distribution $b/[\pi ((s-{\alpha }_0{)}^2+b^2)]$. Since that coordinate is ${\alpha }_0+{\theta }_{T_{\mathbb{D}}}$, we obtain the stronger exact lawIts first absolute moment diverges:There is no contradiction with the terminal point lying on the compact unit circle: reducing its argument modulo $2\pi$ loses the number and direction of turns accumulated before exit.

The printed upper-tail identity cannot be a probability formula: it vanishes at zero and increases in $x$. Its correction isEquivalently the cumulative distribution function is $1/2+{\pi }^{-1}\arctan x$. Differentiating that cumulative function gives the densityThe normalization and sign can also be checked directly by conformal invariance of planar Brownian motion: the Cayley transform $(z-i)/(z+i)$ sends the starting point to the center of the unit disc. Its exit angle is uniform by rotational symmetry. The inverse on the unit circle gives $X=-\cot (\Theta /2)$, $\Theta$ uniform on $(0,2\pi )$, whose change-of-variable density is exactly the one above.

For a start $z_0=x_0+i{y}_0$, $y_0>0$, translation and positive scaling send it to $i$. Thus the exit coordinate has law $x_0+y_{0}C$, where $C$ is standard Cauchy. The corresponding formulas areThis is the upper-half-plane Poisson kernel. The printed density is correct; it is the preceding asserted tail from which it is meant to follow that needs correction.

For $0<\alpha <2$, write $z=r{e}^{i\theta }$ with $0<\theta <\pi$. The chosen power map sends this to $r^{\alpha }{e}^{i\alpha \theta }$, soAt $\alpha =2$ this is the plane slit along the nonnegative real axis. For every $0<\alpha \le 2$, the map is conformal on the upper half-plane.

To check finiteness of the physical exit time, start a Brownian motion $B$ at $i$ and let $T$ be its real-axis exit. The time change of a continuous process in conformal invariance of planar Brownian motion isWe know $T<\infty$ almost surely. The Cauchy exit law from a Brownian half-plane assigns probability zero to $B_T=0$. On such a path, $B_s\ne 0$ throughout the compact interval $[0,T]$, because it is in the open upper half-plane before exit and has nonzero endpoint. Its modulus therefore has a positive minimum and finite maximum. The displayed integrand is bounded on that interval, proving $A_T<\infty$. The image Brownian motion exits $D$ at this finite changed time.

The endpoint $\alpha =0$ printed in the source is degenerate: $z^0=1$, so its image is the singleton $\{1\}$, not an open domain, and the map is not conformal. If exit from that set is defined for Brownian motion started at one, its exit infimum is zero. It cannot be included in a nondegenerate conformal-domain argument.

Here is the martingale argument for maximal comparisons, including the starting-value issue. For $P=f+ig$, the Cauchy-Riemann equations and Itô's formula show thatare orthogonal continuous local martingales with the same quadratic variation. Localize before large disks and before exit. The Burkholder-Davis-Gundy inequalities then givewith a constant depending only on $p$; Fatou's lemma removes the localization. These are centered martingales. For the uncentered processes this impliessince $|f(i^{\alpha })|\le \sup |f(Z_t)|$.

The source normalizes $P(0)=0$ at a boundary point, rather than the value of $g$ at the starting point. Consequently its displayed constant cannot be uniform in $\alpha$ with only $p$ dependence. To see this, take $P(z)=iz$ and $0<\alpha <\min (1/2,1/p)$. Then $f=-\mathrm{Im}z$ and $g=\mathrm{Re}z$. Inside this narrow wedge,Both $p$th moments are finite and the right one is positive; finiteness follows from the wedge radial-maximum tail of order $R^{-1/\alpha }$, justified below. The source inequality would force $C_p\ge {\cot }^p(\alpha \pi )$, unbounded as $\alpha \downarrow 0$. At $\alpha =0$, the immediate-exit interpretation gives the direct contradiction $1\le C_{p}0$ for the same polynomial.

For a fixed nondegenerate wedge, an uncentered inequality does hold with a constant allowed to depend on $\alpha$ as well as $p$. One rigorous outline for the polynomial class in this question is as follows. Separation of variables in a sector of radius $R$ gives, for exit through its circular arc, the expansionThus at the specified starting point the radial-maximum tail is comparable to $R^{-1/\alpha }$. Hitting any fixed interior subarc has that same power, with positive leading coefficient. If a nonzero polynomial has degree $d$, its real part has magnitude at least $c{R}^d$ on some fixed interior subarc for sufficiently large $R$. Therefore its maximal real-part $p$th moment is infinite whenever $dp\ge 1/\alpha$. In that case the desired extended-real inequality is automatic.

The remaining polynomials have $d<1/(\alpha p)$, so belong to a finite-dimensional vector space space. On that spaceis a genuine norm: if it is zero, Brownian motion's positive probability of reaching any interior neighborhood forces $\mathrm{Re}P=0$ throughout $D$, so analyticity makes $P$ a purely imaginary constant, and $P(0)=0$ makes it zero. The corresponding imaginary-part functional is bounded on this finite-dimensional vector space normed space; its finiteness follows from $dp<1/\alpha$ and the radial tail bound. equivalence of norms in finite dimensions givesThis establishes the fixed-domain interpretation and identifies exactly why the universal centered martingale inequality cannot silently be applied to the printed uncentered boundary normalization. It is the distinction in starting-value obstruction in wedge conjugate maximal inequalities.

Conformal invariance of planar Brownian motion means invariance after changing the clock, not preservation of the original speed. If $f:D\to D^{\prime }$ is a conformal bijection and $Z$ is planar Brownian motion stopped on exiting $D$, putThen $f(Z_t)$ has the law of planar Brownian motion started at $f(Z_0)$ and run at time $A_t$, up to the corresponding lifetime. Equivalently, composing with the inverse clock produces ordinary Brownian motion in $D^{\prime }$.

Indeed, the real and imaginary parts of a holomorphic function are harmonic, so Itô's formula gives zero drift. The Cauchy-Riemann equations give equal coordinate brackets $A_t$ and zero cross bracket. The Lévy characterization of multidimensional Brownian motion after time change identifies the image. In particular the harmonic measure of boundary arcs is carried to their conformal images.

If $f:D\to D^{\prime }$ is a conformal map, the image of planar Brownian motion in $D$ is Brownian motion in $D^{\prime }$ after the increasing clock changeMore precisely, $f(Z_{A^{-1}(u)})$ is planar Brownian motion, starting at $f(Z_0)$ and run up to the relevant exit time. The Itô formula and the Cauchy-Riemann equations give image-coordinate local martingales with common bracket $A$ and zero cross bracket; the Lévy characterization of multidimensional Brownian motion identifies the time-changed process. In particular, conformal invariance of planar Brownian motion transports exit locations and harmonic measure, while changing traversal speed.

By conformal invariance of planar Brownian motion, $w(Z_{T_D})$ has the uniform boundary law from part (a), because the image process starts at zero. A disk automorphism extends to an orientation-preserving boundary diffeomorphism. Choose a continuous angular lift $\varphi :\mathbb{R}\to \mathbb{R}$ satisfyingThe principal-angle version may have a branch jump, which is harmless for the density. The change of variables for probability densities yields the exit-angle density ${\varphi }^{\prime }(\theta )/(2\pi )$. Differentiating the boundary relation givesTaking absolute values proves ${\varphi }^{\prime }(\theta )=|w^{\prime }(e^{i\theta })|$. ThereforeThe angular lift explains rigorously the derivative appearing in the printed hint despite the principal-angle branch cut.

The map $z\mapsto z^{\pi /\alpha }$ takes a wedge of opening $\alpha$ to a half-plane and the circular boundary of radius $r$ to radius $r^{\pi /\alpha }$. Conformal invariance of planar Brownian motion preserves which boundary part is reached first, although it changes the clock. Use the branch defined by the wedge argument; the origin is a polar point for planar Brownian motion, so localization handles the vertex.