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Write $f=u+iv$. The Cauchy-Riemann equations give $v_x=-u_y$, $v_y=u_x$ and $\Delta u=\Delta v=0$. Since the two Brownian coordinates are independent, $[X{]}_t=[Y{]}_t=t$ and $[X,Y{]}_t=0$. The Itô formula therefore becomesStop in balls on which the derivatives are bounded. The stopped integrals are martingales, proving that $U$ and $V$ are local martingales. Their coefficient rows have equal squared length and are orthogonal. ThusThese are quadratic covariations of an analytic Brownian image; orthogonality does not in general imply independence of the two transformed coordinates.

For the exit law, put $c=2a^2$. Squaring sends the positive quadrant to the upper half-plane, sends $a+ia$ to $ic$, and the Möbius transformation $w\mapsto (w-ic)/(w+ic)$ sends the upper half-plane to the unit disk with the starting point sent to zero. Its composition is the displayed conformal map. The exit time is finite: it is the minimum of the two almost surely finite one-dimensional hitting times of zero. There is no hit at the quadrant's corner, since a planar Brownian motion started away from zero avoids that point.

We can justify the uniform image exit law without assuming a conformal-time-change theorem. For each integer $n\ge 1$, the real and imaginary parts of $f(Z_{t\wedge T}{)}^n$ are bounded martingales obtained from harmonic functions. At the start their values are zero. The bounded convergence theorem at the finite exit time gives $\mathbb{E}f(Z_T{)}^n=0$. The boundary image has modulus one, and conjugation gives the vanishing negative Fourier moments as well. Trigonometric polynomials are dense among continuous functions on the unit circle, so these moments characterize the uniform distribution there.

Consequently the real variable $W=Z_T^2$ is the inverse image of a uniform point $e^{i\Theta }$ under that Möbius transformation:The Jacobian of this change of variable gives the centered Cauchy distribution of scale $c$:Positive $w$ corresponds to exit at $\sqrt{w}$ on the real axis; negative $w$ corresponds to exit at $i\sqrt{-w}$ on the imaginary axis. Pulling back by $w=\pm r^2$ therefore gives the complete boundary lawHere $dr$ on each axis denotes length along that axis, not two-dimensional area measure. Each density integrates to $1/2$, with no remaining atoms. Equivalently the chosen axis is a fair Bernoulli choice independent of the radius, andThis is Brownian exit law from a quadrant, obtained as a harmonic measure and not as an absolutely continuous planar density.