[proofplan] We apply Ito's formula to the real and imaginary parts $u$ and $v$ of $f = u + iv$ composed with the planar Brownian motion $W$. The Cauchy-Riemann equations force $\Delta u = \Delta v = 0$ (harmonicity), which kills the drift terms and makes $u(W)$ and $v(W)$ continuous local martingales. The same equations force the gradients $\nabla u$ and $\nabla v$ to be orthogonal with equal norms $|f'(W_s)|$, which gives $\langle u(W) \rangle_t = \langle v(W) \rangle_t = \sigma(t)$ and $\langle u(W), v(W) \rangle_t = 0$. After the time change $t \mapsto \sigma^{-1}(t)$, Levy's characterization theorem identifies the time-changed process as a standard two-dimensional Brownian motion. [/proofplan]

[step:Decompose $f = u + iv$ and apply Ito's formula to $u(W)$ and $v(W)$] Write $f = u + iv$ where \begin{align*} u: U \subset \mathbb{R}^2 &\to \mathbb{R}, & v: U \subset \mathbb{R}^2 &\to \mathbb{R} \end{align*} are the real and imaginary parts of $f$, both of class $C^2$ since $f$ is holomorphic (holomorphic functions are smooth). The Brownian motion $W = (W^1, W^2)$ is a continuous semimartingale with values in $U$ up to time $\tau$. Apply [Ito's Formula](/theorems/2099) to $u \in C^2(\mathbb{R}^2)$ with the two-dimensional continuous semimartingale $W = (W^1, W^2)$. Since $W^1$ and $W^2$ are independent standard Brownian motions, we have $\langle W^i, W^j \rangle_t = \delta_{ij} t$, and Ito's formula gives \begin{align*} u(W_t) - u(W_0) = \sum_{i=1}^2 \int_0^t \frac{\partial u}{\partial x_i}(W_s) \, dW^i_s + \frac{1}{2} \sum_{i=1}^2 \int_0^t \frac{\partial^2 u}{\partial x_i^2}(W_s) \, ds, \end{align*} where the cross terms $\frac{\partial^2 u}{\partial x_i \partial x_j}$ with $i \neq j$ vanish because $\langle W^i, W^j \rangle_t = 0$ for $i \neq j$. The second sum is $\frac{1}{2}\int_0^t \Delta u(W_s) \, ds$. The identical formula holds for $v$: \begin{align*} v(W_t) - v(W_0) = \sum_{i=1}^2 \int_0^t \frac{\partial v}{\partial x_i}(W_s) \, dW^i_s + \frac{1}{2} \int_0^t \Delta v(W_s) \, ds. \end{align*} [/step]

[step:Use the Cauchy-Riemann equations to show $u(W)$ and $v(W)$ are continuous local martingales]Since $f = u + iv$ is holomorphic, $u$ and $v$ satisfy the Cauchy-Riemann equations: \begin{align*} \frac{\partial u}{\partial x_1} = \frac{\partial v}{\partial x_2}, \qquad \frac{\partial u}{\partial x_2} = -\frac{\partial v}{\partial x_1}. \end{align*} Differentiating the first equation with respect to $x_1$ and the second with respect to $x_2$: \begin{align*} \frac{\partial^2 u}{\partial x_1^2} = \frac{\partial^2 v}{\partial x_1 \partial x_2}, \qquad \frac{\partial^2 u}{\partial x_2^2} = -\frac{\partial^2 v}{\partial x_1 \partial x_2}. \end{align*} Adding: $\Delta u = \frac{\partial^2 u}{\partial x_1^2} + \frac{\partial^2 u}{\partial x_2^2} = 0$. By the same argument (differentiating the Cauchy-Riemann equations in the other order), $\Delta v = 0$. Both $u$ and $v$ are harmonic. Substituting $\Delta u = 0$ and $\Delta v = 0$ into the Ito formulas from the previous step, the drift terms vanish: \begin{align*} u(W_t) - u(W_0) &= \int_0^t \frac{\partial u}{\partial x_1}(W_s) \, dW^1_s + \int_0^t \frac{\partial u}{\partial x_2}(W_s) \, dW^2_s, \\ v(W_t) - v(W_0) &= \int_0^t \frac{\partial v}{\partial x_1}(W_s) \, dW^1_s + \int_0^t \frac{\partial v}{\partial x_2}(W_s) \, dW^2_s. \end{align*} Both processes are stochastic integrals against continuous local martingales, hence they are themselves continuous local martingales.[/step]

[guided]The vanishing of the drift is the key consequence of holomorphicity. In Ito's formula applied to $u(W)$, the drift is $\frac{1}{2}\Delta u(W_s) \, ds$. For a general $C^2$ function $u$, this drift would be non-zero, and $u(W)$ would be a semimartingale but not a local martingale. The Cauchy-Riemann equations force harmonicity ($\Delta u = 0$), which eliminates the drift. This is the stochastic analogue of the classical fact that a harmonic function composed with Brownian motion is a local martingale. The connection runs deep: the generator of two-dimensional Brownian motion is $\frac{1}{2}\Delta$, and a function $u$ is in the kernel of the generator if and only if $\Delta u = 0$, which is precisely the condition for $u(W)$ to be a local martingale.[/guided]

[step:Compute the quadratic variations and covariation using the Cauchy-Riemann equations]The quadratic variation of $u(W)$ is computed using the Ito isometry for stochastic integrals against independent Brownian motions: \begin{align*} \langle u(W) \rangle_t &= \int_0^t \left(\frac{\partial u}{\partial x_1}(W_s)\right)^2 ds + \int_0^t \left(\frac{\partial u}{\partial x_2}(W_s)\right)^2 ds = \int_0^t |\nabla u(W_s)|^2 \, ds. \end{align*} Similarly, \begin{align*} \langle v(W) \rangle_t = \int_0^t |\nabla v(W_s)|^2 \, ds. \end{align*} By the Cauchy-Riemann equations, $\frac{\partial u}{\partial x_1} = \frac{\partial v}{\partial x_2}$ and $\frac{\partial u}{\partial x_2} = -\frac{\partial v}{\partial x_1}$, so \begin{align*} |\nabla u|^2 = \left(\frac{\partial u}{\partial x_1}\right)^2 + \left(\frac{\partial u}{\partial x_2}\right)^2 = \left(\frac{\partial v}{\partial x_2}\right)^2 + \left(\frac{\partial v}{\partial x_1}\right)^2 = |\nabla v|^2. \end{align*} Moreover, the complex derivative of $f$ is $f'(z) = \frac{\partial u}{\partial x_1} + i\frac{\partial v}{\partial x_1} = \frac{\partial u}{\partial x_1} - i\frac{\partial u}{\partial x_2}$, so \begin{align*} |f'(z)|^2 = \left(\frac{\partial u}{\partial x_1}\right)^2 + \left(\frac{\partial u}{\partial x_2}\right)^2 = |\nabla u|^2 = |\nabla v|^2. \end{align*} Therefore \begin{align*} \langle u(W) \rangle_t = \langle v(W) \rangle_t = \int_0^t |f'(W_s)|^2 \, ds = \sigma(t). \end{align*} For the covariation, compute \begin{align*} \langle u(W), v(W) \rangle_t &= \int_0^t \left(\frac{\partial u}{\partial x_1}(W_s) \frac{\partial v}{\partial x_1}(W_s) + \frac{\partial u}{\partial x_2}(W_s) \frac{\partial v}{\partial x_2}(W_s)\right) ds \\ &= \int_0^t \nabla u(W_s) \cdot \nabla v(W_s) \, ds. \end{align*} Substituting the Cauchy-Riemann equations: \begin{align*} \nabla u \cdot \nabla v &= \frac{\partial u}{\partial x_1} \frac{\partial v}{\partial x_1} + \frac{\partial u}{\partial x_2} \frac{\partial v}{\partial x_2} = \frac{\partial v}{\partial x_2}\left(-\frac{\partial u}{\partial x_2}\right) + \frac{\partial u}{\partial x_2} \frac{\partial v}{\partial x_2} \\ &= -\frac{\partial u}{\partial x_2} \frac{\partial v}{\partial x_2} + \frac{\partial u}{\partial x_2} \frac{\partial v}{\partial x_2} = 0. \end{align*} Here we used $\frac{\partial v}{\partial x_1} = -\frac{\partial u}{\partial x_2}$ from the Cauchy-Riemann equations. Therefore \begin{align*} \langle u(W), v(W) \rangle_t = 0. \end{align*}[/step]

[guided]The Cauchy-Riemann equations play a dual role. First, they force harmonicity ($\Delta u = \Delta v = 0$), which eliminated the drift and made $u(W)$ and $v(W)$ local martingales. Second, they force the gradients $\nabla u$ and $\nabla v$ to be orthogonal and of equal length, which gives the covariation structure needed for Levy's characterization. To see why orthogonality holds geometrically: the Cauchy-Riemann equations state that $\nabla v$ is obtained from $\nabla u$ by a $90^\circ$ rotation (specifically, $\nabla v = (-\frac{\partial u}{\partial x_2}, \frac{\partial u}{\partial x_1})$ while $\nabla u = (\frac{\partial u}{\partial x_1}, \frac{\partial u}{\partial x_2})$). Two vectors related by a $90^\circ$ rotation are orthogonal and have the same length. This rotation is the infinitesimal manifestation of the conformal property of holomorphic maps. The identity $|\nabla u|^2 = |f'|^2$ connects the Riemannian geometry ($|\nabla u|$ measures the rate of change of $u$) with the complex-analytic geometry ($|f'|$ measures the local stretching of $f$).[/guided]

[step:Apply Levy's characterization to the time-changed process]Define the time change \begin{align*} \sigma: [0, \tau] &\to [0, \infty) \\ t &\mapsto \int_0^t |f'(W_s)|^2 \, ds. \end{align*} Since $f'$ is continuous and not identically zero, $\sigma$ is a continuous strictly increasing function (at least on intervals where $f'(W_s) \neq 0$), so its inverse $\sigma^{-1}$ is well-defined. Set $\tilde{U}_t = u(W_{\sigma^{-1}(t)})$ and $\tilde{V}_t = v(W_{\sigma^{-1}(t)})$. By the Dambis-Dubins-Schwarz theorem applied to the continuous local martingale $u(W)$ with quadratic variation $\langle u(W) \rangle_t = \sigma(t)$, the time-changed process $\tilde{U}_t = u(W_{\sigma^{-1}(t)})$ is a continuous local martingale with \begin{align*} \langle \tilde{U} \rangle_t = t. \end{align*} Similarly, $\langle \tilde{V} \rangle_t = t$. Since time-changing by the same clock preserves the covariation structure and $\langle u(W), v(W) \rangle = 0$, we have $\langle \tilde{U}, \tilde{V} \rangle_t = 0$. The pair $(\tilde{U}, \tilde{V})$ therefore consists of continuous local martingales starting at $(u(W_0), v(W_0))$ with $\langle \tilde{U} \rangle_t = \langle \tilde{V} \rangle_t = t$ and $\langle \tilde{U}, \tilde{V} \rangle_t = 0$. After centering (subtracting the initial values), the hypotheses of [Levy's Characterization of Brownian Motion](/theorems/2100) are satisfied with $d = 2$, $X^1 = \tilde{U} - u(W_0)$, $X^2 = \tilde{V} - v(W_0)$: - $X^1_0 = X^2_0 = 0$. - $\langle X^i, X^j \rangle_t = \delta_{ij} t$ for $i, j \in \{1, 2\}$. By [Levy's Characterization](/theorems/2100), $(X^1, X^2)$ is a standard two-dimensional Brownian motion. Equivalently, \begin{align*} f(W_{\sigma^{-1}(t)}) = u(W_{\sigma^{-1}(t)}) + i v(W_{\sigma^{-1}(t)}) = f(W_0) + (X^1_t + i X^2_t) \end{align*} is $f(W_0)$ plus a standard two-dimensional Brownian motion. Since $W_0 = 0$ in the standard setup, $f(W_{\sigma^{-1}(t)})$ is a standard two-dimensional Brownian motion started at $f(0)$, which after the usual translation gives the claimed result.[/step]

[guided]The time change $t \mapsto \sigma^{-1}(t)$ "straightens out" the quadratic variation. Before the time change, $u(W)$ is a continuous local martingale whose quadratic variation $\langle u(W) \rangle_t = \sigma(t) = \int_0^t |f'(W_s)|^2 \, ds$ accumulates at a rate proportional to $|f'(W_s)|^2$. Where the holomorphic map $f$ stretches more (larger $|f'|$), the quadratic variation grows faster, and the time change compensates by slowing down. After the time change, the quadratic variation becomes $\langle \tilde{U} \rangle_t = t$, which is exactly the quadratic variation of a standard Brownian motion. Combined with the zero covariation $\langle \tilde{U}, \tilde{V} \rangle = 0$ and the local martingale property, Levy's characterization theorem (proved in the preceding theorem) identifies $(\tilde{U}, \tilde{V})$ as a two-dimensional Brownian motion. The geometric meaning is that a holomorphic map is locally a rotation and scaling (this is the conformal property), and Brownian motion is invariant under rotations. The scaling is absorbed by the time change, and the rotation is invisible to Brownian motion by its rotational symmetry. This is why conformal invariance is a two-dimensional phenomenon: in higher dimensions, holomorphic maps do not exist (in the same sense), and the rotational symmetry argument breaks down.[/guided]