= Solution
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 becomes
$$
dU_t=u_x(Z_t)dX_t+u_y(Z_t)dY_t,\qquad dV_t=-u_y(Z_t)dX_t+u_x(Z_t)dY_t.
$$
Stop 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>. Thus
$$
\boxed{[U]_t=[V]_t=\int_0^t|f'(Z_s)|^2ds,\qquad [U,V]_t=0.}
$$
These 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 $n\ge1$, 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>:
$$
W=ic\,\frac{1+e^{i\Theta}}{1-e^{i\Theta}}=-c\cot(\Theta/2).
$$
The Jacobian of this change of variable gives the centered <Cauchy distribution> of scale $c$:
$$
p_W(w)=\frac{c}{\pi(w^2+c^2)},\qquad w\in\mathbb R.
$$
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 law
$$
\boxed{\mathbb P(Z_T\in dr)=\frac{4a^2r}{\pi(r^4+4a^4)}dr,\qquad \mathbb P(Z_T\in i\,dr)=\frac{4a^2r}{\pi(r^4+4a^4)}dr\quad(r>0).}
$$
Here $dr$ 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, and
$$
\boxed{\mathbb P(|Z_T|\le r)=\frac2\pi\arctan\!\left(\frac{r^2}{2a^2}\right),\quad r\ge0.}
$$
This is <Brownian exit law from a quadrant>, obtained as a <harmonic measure> and not as an absolutely continuous planar density.
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