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.

Take $\lambda >0$ and let ${\sigma }_{\lambda }$ be the first time $X$ hits $\lambda$. Reflect the horizontal coordinate after ${\sigma }_{\lambda }$, leaving the vertical coordinate unchanged. The Strong Markov property and symmetry preserve the law of planar Brownian motion and leave its real-axis exit time unchanged. On paths with ${\sigma }_{\lambda }<\tau$, the reflection swaps the two terminal events $X_{\tau }<\lambda$ and $X_{\tau }>\lambda$. There is no terminal atom at $\lambda$. Hence the Brownian reflection principle givesHorizontal symmetry gives the same probability for $-\inf X>\lambda$. A union bound therefore yieldsAlso $X^{\ast }\ge |X_{\tau }|$, whose Cauchy distribution is $2{\pi }^{-1}{\int }_{\lambda }^{\infty }(1+x^2{)}^{-1}dx$. Thus the maximum tail is bounded above and below by positive constants times ${\lambda }^{-1}$ for large $\lambda$. The tail integral formula for moments for positive moments givesThe large-$\lambda$ integral is finite exactly when $0<p<1$; it diverges for $p\ge 1$. Near zero its upper bound by ${\int }_0^{1}p{\lambda }^{p-1}d\lambda$ is finite.

The vertical coordinate remains nonnegative up to exit and starts at one. The strip calculation, now with upper boundary $\lambda \ge 1$, givesThus $Y^{\ast }$ has density ${\lambda }^{-2}$ on $[1,\infty )$, and in factFor completeness, if the source's $p<1$ includes negative powers, $X^{\ast }$ also has all negative moments. Let ${\sigma }_{\eta }$ be the first horizontal exit from $(-\eta ,\eta )$. ThenThe first bound follows from the Gaussian reflection bound for the vertical hit of zero; the second follows by splitting time into blocks of length ${\eta }^2$, each having a fixed positive chance of exiting the horizontal interval. This small-value estimate integrates against every negative power. At $p=0$ both moments equal one.