= Solution
Interpret $f$ as a <conformal bijection> onto $D'$. 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 $B^1+iB^2$. 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 $D$, yields
$$
du(B_t)=u_x(B_t)dB_t^1+u_y(B_t)dB_t^2,\qquad dv(B_t)=v_x(B_t)dB_t^1+v_y(B_t)dB_t^2.
$$
Thus both coordinates are <continuous local martingales>, with
$$
[u(B)]_t=[v(B)]_t=A_t:=\int_0^{t}|f'(B_s)|^2ds,\qquad [u(B),v(B)]_t=\int_0^t(u_xv_x+u_yv_y)(B_s)ds=0\quad(t<\zeta).
$$
Since $f'$ 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 \b[<conformal invariance of planar Brownian motion> takes the form]
$$
\boxed{f(B_t)=\beta_{A_t},\qquad A_t=\int_0^t|f'(B_s)|^2ds,\quad t<\zeta.}
$$
To identify the lifetime, exhaust $D$ by relatively compact open sets $D_n$ with nested closures. The images $f(D_n)$ exhaust $D'$ because $f$ is a homeomorphism onto $D'$. 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 $D'$, including a possibly infinite lifetime. This identifies $A_{\zeta-}$ with that lifetime and needs no global extension of $f$ to the boundary.
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