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We use the following standard planar Brownian motion facts: its exit probability distribution is harmonic measure; conformal maps preserve that probability distribution after an increasing time change; and the Poisson kernel for the upper half-plane gives its real-boundary exit density. For rough hulls the boundary is interpreted through prime ends, with boundary correspondence at harmonic-measure-almost every exit point. This avoids imposing a smoothness or local-connectedness assumption on $K$.

Under $g_K$, the exit event through $K$ corresponds to a measurable set $E_K\subset \mathbb{R}$ of boundary images. It is bounded: the mapping-out function extends by Schwarz reflection along the real line outside a sufficiently large interval enclosing $K$, and its boundary near infinity corresponds to the original real boundary near infinity. All prime ends representing $K$ therefore have images in a bounded interval. The boundary correspondence assertion can equivalently be formulated using the almost-everywhere limits of $g_K^{-1}$; changing null sets does not change $E_K$'s measure.

Write $g_K(iy)=u_y+i{v}_y$. The hydrodynamic normalization at infinity gives $u_y=O(y^{-2})$ and $v_y=y-a_K/y+O(y^{-2})$. By conformal invariance of planar Brownian motion and the Poisson kernel for the upper half-plane,On the bounded set $E_K$, the integrand multiplied by $y$ tends uniformly to one. Therefore the limit exists and is finite:This is the harmonic capacity from infinity in the upper half-plane with the normalization $c(K)=\mathrm{cap}(K)/\pi$; it is different from half-plane capacity. If real boundary points are retained in the definition of $K$, include their boundary images in $E_K$ as well; the argument is unchanged.

For the scaling statement, Brownian scaling says that $r^{-1}{B}_{r^{2}t}$ started from $iy/r$ is again standard planar Brownian motion. The exit event scales with the hull, soSet $v=y/r$, multiply by $y=rv$ and take the limit. It follows thatThe same conclusion follows geometrically from $E_{rK}=r{E}_K$ and the formula above.

First derive the Brownian representation of half-plane capacity with the chosen normalization. SetThe height comparison from part (a), applied to $f_K=g_K^{-1}$, gives $h\ge 0$. On the boundary, $\mathrm{Im}{g}_K$ tends to zero; on $K$ the limiting value of $h$ is therefore $\mathrm{Im}z$, and on the real boundary it is zero. To see the asserted boundary-height limit without assuming a smooth hull, suppose a bounded sequence approaching the boundary had mapped images with imaginary part bounded away from zero. The inverse's expansion near infinity prevents these images from being unbounded, and an interior subsequential limit would map back to an interior point of $D$, a contradiction.

At infinity $h$ tends to zero by the Laurent series. The maximum principle for harmonic functions now bounds $h$ between zero and $R=\mathrm{rad}(K)$, since every point of $K$ has imaginary part at most $R$. Apply the optional sampling theorem to the bounded harmonic function $h(B_{t\wedge T})$. The exit time $T$ is almost surely finite because it is no larger than the first time the imaginary coordinate of planar Brownian motion reaches zero. Taking $t\to \infty$ givesThis uses the bounded harmonic difference, not optional sampling of the unbounded imaginary coordinate at an unbounded stopping time.

At $z=iy$, the expansion gives $yh(iy)\to a_K$. Sincewe obtainThus the quadratic half-plane capacity is controlled by a length times the linearly scaling harmonic capacity from infinity in the upper half-plane. The factor $1/\pi$ is essential if the alternative normalization $\mathrm{cap}(K)=\pi c(K)$ is used.

Let $u=\Re f$, and suppose $u$ has a local maximum at $z_0\in U$. Choose $R>0$ such that the closed disk $\overline{D(z_0,R)}$ lies in $U$ and $u(z)\le u(z_0)$ throughout it. Start planar Brownian motion $B$ at $z_0$, and for $0<r<R$ let ${\tau }_r$ be its first exit from $D(z_0,r)$.

By part (a) and Itô formula, $u(B_{t\wedge {\tau }_r})$ is a local martingale; boundedness of $u$ on the compact disk makes it a true bounded martingale. The exit time is finite almost surely. One direct check applies the same formula to $|B_{t\wedge {\tau }_r}-z_0{|}^2$, giving $2\mathbb{E}(t\wedge {\tau }_r)=\mathbb{E}|B_{t\wedge {\tau }_r}-z_0{|}^2\le r^2$. Thus $\mathbb{E}{\tau }_r\le r^2/2$.

The optional stopping theorem, first at $t\wedge {\tau }_r$ and then by bounded convergence, givesThe exit point is uniformly distributed on the circle: rotational invariance of planar Brownian motion about its starting point makes its exit law rotation invariant, and normalized arc length is the unique rotation-invariant probability measure on a circle. Since every boundary value is at most $u(z_0)$, equality of the mean forces every boundary value to equal $u(z_0)$; continuity rules out a strict deficit at any boundary point. This holds for every $0<r<R$, so $u$ is constant on the entire disk.

The Cauchy-Riemann equations then give both partial derivatives of $v=\Im f$ equal to zero there. Thus $f$ is constant on that disk. The identity theorem for holomorphic functions propagates this constant to the connected component of $z_0$ in $U$. Applying the same Brownian argument to $v$ proves the assertion for the imaginary part as well:If $U$ is connected, this is constancy on all of $U$. For a disconnected open set the printed global wording needs this qualification: a function taking different constants on two disjoint disks is holomorphic and has local maxima on each disk while not being globally constant.

Suppose a nonconstant polynomial $P$ has no zero in $\mathbb{C}$. Its reciprocal $g=1/P$ is then an entire holomorphic function. If $d\ge 1$ is the degree and $a_d\ne 0$ its leading coefficient, division by $z^d$ shows $P(z)/z^d\to a_d$ as $|z|\to \infty$. Hence $|P(z)|\to \infty$, so $g(z)\to 0$. Continuity on compact disks now makes $g$ bounded on all of $\mathbb{C}$.

Fix $z\in \mathbb{C}$ and let $B$ be standard planar Brownian motion started at zero. By part (a) and Itô formula, both components of $g(z+B_t)$ are local martingales. Since $g$ is bounded, they are true martingales, andFor $\epsilon >0$, choose $R$ such that $|g(w)|\le \epsilon$ whenever $|w|>R$, and write $K={\sup }_w|g(w)|$. The two-dimensional Gaussian density of $z+B_t$ is bounded by $1/(2\pi t)$, soConsequentlyLet $t\to \infty$, then $\epsilon \downarrow 0$, to obtain $g(z)=0$. This is impossible for $1/P$. ThusThis Brownian proof of the fundamental theorem of algebra uses only harmonicity, the bounded-martingale property and the spreading Gaussian transition density. It does not use either maximum principle from part (b), nor does it incorrectly assume that planar Brownian motion hits a specified point.

The conformal invariance of planar Brownian motion is a time-change statement. If $\varphi :D\to D^{\prime }$ is a conformal bijection of planar domains and $B$ is planar Brownian motion started in $D$, let ${\tau }_D$ be its exit time and defineBefore exit this clock is strictly increasing. Its inverse $C_u$ makes $\varphi (B_{C_u})$ a planar Brownian motion started at $\varphi (B_0)$, up to its transformed lifetime $A_{{\tau }_D}$. The image of the stopped trajectory is therefore invariant in law after this change of time. A holomorphic map with nonzero derivative gives the corresponding local statement even without global injectivity.

Here is the proof. Write $\varphi =u+iv$ and $B=(B^1,B^2)$. The real and imaginary parts are harmonic. The Itô formula therefore gives the two continuous local martingalesstopped on compact subsets of $D$. The Cauchy-Riemann equations implyThus both coordinate quadratic variations equal $A_t$ and their quadratic covariation is zero. After inverse time change, the covariance matrix of the two continuous local martingales is $u{I}_2$. The multidimensional Lévy characterization of multidimensional Brownian motion states that a continuous vector local martingale starting at zero with this covariance matrix is standard vector Brownian motion. Apply it after subtracting $\varphi (B_0)$, and exhaust $D$ by compact subsets. This proves the theorem, with killing at the transformed exit lifetime. The clock is the conformal Brownian clock; omitting it would generally give an incorrect speed.

The exponential map is a conformal bijection from the stripto the wedge containing the starting point $1$. The restriction $\alpha +\beta <2\pi$ makes it injective. A planar Brownian motion $U+iV$ started at zero in this strip maps, after the conformal Brownian clock, to the required Brownian motion in the wedge.

Let $\sigma$ be the first time $V$ exits $(-\alpha ,\beta )$. It is finite almost surely, and $V_{t\wedge \sigma }$ is a bounded martingale. The optional stopping theorem givesThe strip exit maps to the corresponding bounding ray. The exponential clock is finite at $\sigma$, because its integrand is continuous on the compact time interval $[0,\sigma ]$. Thus it does not remove or reverse the exit event. We obtainThis is the Brownian wedge-exit probability obtained directly from the strip, using conformal invariance of planar Brownian motion.

Put $L_r={\theta }_{T_r}$. The hitting times $T_r$ are finite almost surely by recurrence of planar Brownian motion, and increase with $r$. At $T_r$, the position is $e^{-r}{e}^{i{L}_r}$. By the Strong Markov property, Brownian scaling, and rotational invariance of planar Brownian motion, the future processis a standard planar Brownian motion started at $1$, independent of ${\mathcal{F}}_{T_r}$. The random rotation does not change its conditional law, since that law is rotationally invariant.

The first time this rescaled process reaches radius $e^{-h}$ corresponds to the original time $T_{r+h}$. Its continuous argument starts at zero and equals the increment of the original winding angle. HenceFor any ordered finite collection of levels, apply this assertion successively. Earlier winding values are measurable at the current hitting time, so all increments are independent and their laws depend only on the level differences. This proves the Lévy process property of winding at logarithmic radial passage levels.

One can also verify stochastic continuity and identify this process. The logarithmic-radius and argument martingales have the common clockThe same Cauchy-Riemann equations calculation as in part (a), applied locally to the logarithm and joined along the continuous argument, gives independent Brownian motions $U,V$ such that $\log R_t=U_{H_t}$ and ${\theta }_t=V_{H_t}$. This clock diverges: if it had a finite limit, convergence of martingales with finite total quadratic variation would make $\log R_t$ converge to a finite value, forcing the clock integral itself to diverge.

Thus, with ${\eta }_r=\inf \{u:U_u=-r\}$, we have $L_r=V_{{\eta }_r}$. The Brownian first-passage Laplace transform is $\mathbb{E}{e}^{-q{\eta }_r}=e^{-r\sqrt{2q}}$. It follows either from the Exponential martingale for Brownian motion stopped at ${\eta }_r\wedge n$, or from the one-dimensional exit equation. Conditioning on the independent process $U$ givesSo this is a symmetric Cauchy process, with level parameter $r$. The formula implies stochastic continuity at zero and, by stationary increments, everywhere. Under the usual definition requiring càdlàg paths, use the right-continuous passage clock ${\eta }_r^{+}=\inf \{u:U_u<-r\}$. It agrees with ${\eta }_r$ for each fixed level almost surely, and $V_{{\eta }_r^{+}}$ is the càdlàg modification. This distinction concerns simultaneous passage levels and does not alter any increment law.

For $h(x,y)=\frac{1}{2}\log (x^2+y^2)$ away from the origin,The Itô formula for planar Brownian motion therefore has no drift term before the radius reaches $r$. Since the initial logarithm is zero,The integrand's squared norm is $|B_s{|}^{-2}\le r^{-2}$ on this stopped interval, so this is a square-integrable martingale on every finite horizon, in particular a local martingale. One may first stop in an outer annulus to apply the formula on a compact smooth domain and then remove that localization.

Let $D$ be the unit disc and normalize its positive Poisson kernel at the boundary point $1$ byIt is harmonic in $D$. Let $p_D(t,z,w)$ be the killed Brownian transition density, for planar Brownian motion with generator $\frac{1}{2}\Delta$. The Doob h-transform has sub-Markov transition densityIts missing mass represents paths already absorbed at the distinguished boundary point. Equivalently, for a stopping time $\sigma$ before exit from a compact subdomain, its law is weighted relative to ordinary killed Brownian motion by $h(B_{\sigma })/h(B_0)$. These stopped laws are consistent and define a diffusion up to its lifetime. Its generator isThis is a rigorous Brownian motion conditioned to exit at a boundary point, not conditioning on an event of positive probability. Exhaustion of $D$ shows that its terminal boundary limit is $1$; its lifetime is finite. For example its expected lifetime from $0$ is ${\int }_D{G}_D^{\mathrm{BM}}(0,w)h(w)dw$, which is finite: the Green function vanishes linearly near the smooth boundary and cancels the Poisson-kernel singularity there, while its logarithmic singularity at $0$ is integrable.

One can also condition ordinary Brownian motion on exiting through an arc $I_{ϵ}$ about $1$, and then let its length decrease to zero. The conditional harmonic functions ${\omega }_D(z,I_{ϵ})/{\omega }_D(0,I_{ϵ})$ converge locally uniformly to $h(z)$. Thus their stopped conditional laws converge to this same diffusion. Both constructions specify the conditioning unambiguously.

The capacity in this question is harmonic capacity from infinity in the upper half-plane, which has units of length; it is distinct from half-plane capacity, which has units of length squared.

Here is a proof of existence that also works with irregular real attachments. The functionis bounded and harmonic on $H$, by the Strong Markov property and the mean-value characterization of harmonic functions. Therefore $u\circ g_K^{-1}$ is a bounded harmonic function on the complex upper half-plane, with a Poisson kernel representationThe boundary function $f$ vanishes outside a bounded interval. Indeed, far enough along either real ray the original domain contains a half-disc neighbourhood, and $g_K$ extends there; the probability of hitting the bounded hull before the real boundary tends to zero as the starting point approaches that ray. Applying the same dominated-limit calculation as in part (i) yieldsWhen the intrinsic boundary pieces landing on the hull are identified, $f$ is their indicator almost everywhere and this is the length of their image under $g_K$. For ordinary finite slit hulls this is exactly the image of $\delta H\setminus H_0$, since real attachment endpoints have zero harmonic measure. The Poisson representation avoids requiring that boundary identification in the general existence argument.

If $K\subset K^{\prime }$, couple the two exit events using the same planar Brownian motion, stopped at its first hit of the real axis. Any path that hits $K$ before the real axis also hits $K^{\prime }$ before the real axis. Consequentlyand taking the limits proves monotonicity of this capacity.

For a half-disc of radius $r$ centred at $b\in \mathbb{R}$, the mapping-out function isIts semicircular boundary maps onto $[b-2r,b+2r]$, of length $4r$. Hence its harmonic capacity from infinity in the upper half-plane is $4r$. Withenclose $K$ in such a half-disc and use monotonicity. Letting the enclosing radius decrease to the infimum provesThe same conclusion holds if radius is instead measured about a specified real centre.

Use the intrinsic boundary of a simply connected domain, so different approaches to the two sides of a slit remain distinct. The mapping-out function $g_K$ extends as a homeomorphism from this intrinsic compactification to that of the complex upper half-plane. Set $T=T(H)$. The imaginary coordinate of planar Brownian motion hits zero in finite time almost surely, and $T$ is no larger than that time. Thus $T<\infty$ and continuity gives a finite Euclidean exit point.

By conformal invariance of planar Brownian motion, $g_K(B_t)$ is a planar Brownian motion in the complex upper half-plane, run with clockThe terminal clock cannot be infinite: that would make the transformed Brownian motion stay in the upper half-plane forever. It cannot stop while the transformed path is in the interior either, since continuity of $g_K^{-1}$ would then put the original exit point inside $H$. Hence the terminal clock is precisely the transformed Brownian exit time. The transformed path converges to a real boundary point, and applying the extended inverse proves almost sure convergence to a point of the intrinsic boundary.

Write $g_K(x+iy)=u+iv$, and let $E=g_K(S)\cap \mathbb{R}$. The point at infinity has zero harmonic measure. Conformal invariance of planar Brownian motion and the Poisson kernel for the upper half-plane giveThe hydrodynamic normalization at infinity gives $g_K(z)=z+O(1/z)$, so along the specified approach $v/y\to 1$ and $u/y\to 0$. For each fixed real $t$,If $\mathrm{Leb}(E)<\infty$, dominated convergence applies because $y/v$ is eventually bounded. If $\mathrm{Leb}(E)=\infty$, Fatou's lemma makes the limit infinite. This proves the harmonic-measure asymptotic at infinitywith the equality understood in the extended nonnegative reals.

For parameter four, the SLE4 angle martingale is bounded, so the Continuous-time martingale convergence theorem gives an almost sure limit. The imaginary part of the Chordal Loewner equation givesFor a point off the full simple Loewner trace the flow exists at every finite time. If the limiting angle lay strictly between $0$ and $\pi$, the last derivative would eventually be bounded above by a strictly negative constant. That would make $Y_t^2$ negative, a contradiction. Hence the terminal angle is either zero or $\pi$.

The simple transient chordal Loewner trace from $0$ to infinity splits the complex upper half-plane into a left component $D^{-}$, adjacent to the negative real boundary, and a right component $D^{+}$, adjacent to the positive real boundary. The Dirichlet boundary values of the SLE angle process identify which terminal value occurs. Indeed, run an independent planar Brownian motion from $z$ until it first hits the full Loewner trace or the real axis. This time is finite. The Brownian path up to that time is bounded, and Transience of chordal SLE ensures that any Loewner trace point it meets belongs to a finite initial segment. Thus its exit side for the truncated domains eventually agrees with its exit side for the full component. In $D^{-}$ every such exit is through the left bank or negative real boundary; in $D^{+}$ every exit is through the right bank or positive real boundary. Dominated convergence in the harmonic-measure representation therefore givesThis is a random side indicator, not the deterministic value $\arg z$. Uniform integrability also gives $\mathbb{E}[F(z)]=\arg z$, so the SLE4 left-passage probability is

Set $\beta =\pi /\alpha$ and use the branch $\varphi (z)=z^{\beta }$ with argument in $(-\alpha /2,\alpha /2)$. This is a conformal map from the wedge to the right half-plane, fixes the starting point $1$, and sends the outer circle of radius $r$ to that of radiusThe two wedge sides map to the imaginary axis.

By conformal invariance of planar Brownian motion, the image of the stopped path is planar Brownian motion after the increasing conformal Brownian clock $A(t)={\int }_0^t|{\varphi }^{\prime }(B_s){|}^{2}ds$. This clock does not change which boundary portion is reached first. Localization away from the vertex justifies the map even when its derivative is unbounded there; the vertex is a polar point for planar Brownian motion and has zero hitting probability from $1$.

Consequentlywhere the probability on the right is for the right half-plane. The power-map reduction for Brownian exit from a wedge also works at $\alpha =2\pi$, when the wedge is the plane slit along the negative real axis.

Let $E_{+}$ and $E_{-}$ denote positive and negative real part at the circular exit. On $\{S<T(r)\}$, reflect the portion of the planar Brownian motion after $S$ by $z\mapsto -\overline{z}$. This reflection fixes the imaginary axis and preserves distances from the origin. The Strong Markov property and reflection symmetry show that the resulting path has the same law, its circular exit time is unchanged, and $E_{+}$ is exchanged with $E_{-}$. ThereforeAn exit with negative real part must first cross the imaginary axis. An exit before $S$ has positive real part. The two points $ir,-ir$ have zero circular exit probability, since circular harmonic measure has no atoms. HenceSubtracting provesThis is the reflection identity for Brownian exit from a half-disc.

The half-plane capacity is the Laurent coefficientIt is real and nonnegative. For planar Brownian motion starting at $z\in D$, let ${\tau }_D$ be its first exit from $D$. The Brownian representation of half-plane capacity isOne may also state the underlying harmonic identity, valid throughout $D$,The process is killed at the hull or real boundary; real-boundary exits contribute zero. These are the Brownian characterizations, with the normalization that a Chordal Loewner equation at speed $2$ produces capacity $2t$.

Use $U_t=2W_t$ and write $Z_t=X_t+i{Y}_t$ up to its interior-point swallowing time for a Loewner chain $T_z$. The Chordal Loewner equation givesThe branch of the complex logarithm with argument in $(0,\pi )$ is well-defined while $t<T_z$. The Itô formula yields the crucial cancellationThe $2/Z_t^2$ drift is cancelled by the quadratic variation term, because $\kappa =4$. Thus both $L_t=\log |Z_t|$ and ${\theta }_t=\arg Z_t$ are continuous local martingales, withThis is the logarithmic martingale for SLE4.

We justify absence of a finite swallowing time rather than presuming that the logarithm survives forever. On a finite horizon $H$, $0<Y_t\le \mathrm{Im}z$. Also $|X_t|$ is bounded pathwise before $T_z\wedge H$. Indeed, while $|X_t|\ge 1$ its drift $2X_t/|Z_t{|}^2$ has absolute value at most $2$; on each excursion outside $[-1,1]$, integrate from its starting point and bound the Brownian oscillation on $[0,H]$. For example,Therefore $L_t$ is bounded above pathwise on this interval.

Suppose $T_z\le H$. By the Dambis-Dubins-Schwarz theorem, $L$ is a Brownian motion run at its own quadratic variation. If that clock diverged as $t\uparrow T_z$, Brownian oscillation would make $L$ unbounded above, contradicting the preceding bound. The one-sided bound criterion for a martingale clock therefore gives a finite clock limit and a finite real limit for $L_t$. Hence $|Z_t|$ is bounded away from zero near $T_z$.

Now $Y_t=Y_0\exp (-2{\int }_0^t|Z_s{|}^{-2}ds)$ has a strictly positive limit at $T_z$. The drift in $X$ is integrable there, so continuity of $W$ gives a finite limit for $X$ as well. The limiting point is in $\mathbb{H}$ and away from the Loewner driver singularity, and the differential equation extends past $T_z$, a contradiction. Thus $T_z=\infty$ almost surely. A point of the Loewner trace at a finite time belongs to that time's hull, soThis proves the fixed-interior-point avoidance of SLE4 without using simplicity as an input.

The angle remains in $(0,\pi )$ at all finite times. The bounded local martingale criterion upgrades its local martingale equation to a genuine martingale:In particular this is the SLE4 angle martingale, and it converges almost surely and in $L^1$ by bounded martingale convergence theorem.

It remains to identify the limiting angle using the assumed simple path tending to infinity. Orient that path from $0$ to infinity. Its left component is the one adjacent to the negative real half-axis. Under $g_t$, the left boundary of the slit domain maps to $(-\infty ,U_t)$ and the right boundary to $(U_t,\infty )$. The harmonic measure of the former as seen from $z$ isThis follows by conformal invariance of planar Brownian motion: in the upper half-plane, $\arg (w-U_t)/\pi$ is the bounded harmonic function with values $1$ on the left half-axis and $0$ on the right.

To justify the limiting boundary classification, condition on a simple proper realization of the path and use an independent planar Brownian motion from $z$. It exits the upper half-plane in finite time almost surely, so its path up to that time is compact. The curve tends to infinity, so its intersection with this compact set is contained in a finite initial curve segment. Once that segment has been drawn, the Brownian path exits the slit domain through its left boundary exactly when $z$ is in the final left component: a path from that component cannot reach the right boundary without crossing the curve, and the reverse assertion holds on the right. Endpoints have zero harmonic measure. Bounded convergence of these exit indicators provesTaking expectations in the bounded angle martingale gives the SLE4 left-passage probabilityFor the imaginary axis it is $1/2$; near the negative real axis it tends to $1$, fixing the orientation of “left”.

The domain is the union of the horizontal and vertical strips. The four connected components of its boundary carry the constants $a,b,c,d$ in the northeast, northwest, southwest and southeast, respectively.

First justify finiteness of $T$. A small open ball around $(2,2)$ lies outside the domain. By part (b), planar Brownian motion visits that ball almost surely in finite time. A continuous path from $z\in D$ to that ball must meet $\partial D$ first. Hence $T<\infty$ almost surely, and $B_T\in \partial D$ by continuity.

The harmonic function $u$ is smooth inside $D$. The Itô formula, localized on relatively compact subdomains, makes $u(B_{t\wedge T})$ a local martingale. Its boundary definition uses the continuous extension of $u$. Since $u$ is bounded, this stopped process is a true martingale with uniform integrability: pass through the compact exhaustion using bounded convergence theorem. In particular ${\mathbb{E}}_{z}u(B_{t\wedge T})=u(z)$.

Now $B_{t\wedge T}\to B_T$ almost surely as $t\to \infty$. The dominated convergence theorem, boundedness of $u$, and its continuous boundary values give the bounded harmonic representation with almost sure Brownian exit:Almost sure exit and boundedness of the harmonic function are both used. The domain's unboundedness causes no loss of expectation here because the stopped function values are uniformly bounded.

Interpret $f$ as a conformal bijection onto $D^{\prime }$. If it is merely an injective map into a larger target, the correct exit domain is $f(D)$ instead. Let $\zeta$ be the exit time from $D$, write $f=u+iv$, and identify planar Brownian motion with $B^1+i{B}^2$. The Cauchy-Riemann equations give $u_x=v_y$, $u_y=-v_x$, and both $u,v$ are harmonic functions. The Itô formula, localized on compact subsets of $D$, yieldsThus both coordinates are continuous local martingales, withSince $f^{\prime }$ never vanishes, $A$ is strictly increasing before $\zeta$. Let ${\eta }_s=\inf \{t:A_t>s\}$, defined for $s<A_{\zeta -}$. The optional time-change theorem gives continuous local martingale coordinates for ${\beta }_s=f(B_{{\eta }_s})$, with bracket matrix $sI$. The Lévy characterization of multidimensional Brownian motion makes $\beta$ a standard planar Brownian motion started at $f(z)$, up to its lifetime. Therefore conformal invariance of planar Brownian motion takes the formTo identify the lifetime, exhaust $D$ by relatively compact open sets $D_n$ with nested closures. The images $f(D_n)$ exhaust $D^{\prime }$ because $f$ is a homeomorphism onto $D^{\prime }$. Each stopped transformed path exits $f(D_n)$ at precisely the clock value of its exit from $D_n$. Increasing these exit times gives the exit lifetime of Brownian motion from $D^{\prime }$, including a possibly infinite lifetime. This identifies $A_{\zeta -}$ with that lifetime and needs no global extension of $f$ to the boundary.