[proofplan] We expand the Wirtinger derivatives using $f=u+iv$ and separate real and imaginary parts. The equation $\partial_{\bar z}f(z_0)=0$ is then equivalent to the two scalar equations $\partial_x u(z_0)=\partial_y v(z_0)$ and $\partial_y u(z_0)=-\partial_x v(z_0)$, which are exactly the Cauchy-Riemann equations at $z_0$. Finally, under those equations, the expression for $\partial_z f(z_0)$ simplifies to the asserted formula. [/proofplan] [step:Expand $\partial_{\bar z}f(z_0)$ into real and imaginary parts] Write $z_0=x_0+iy_0$ with $x_0,y_0\in\mathbb R$. Since $f=u+iv$ and $u,v$ have first partial derivatives at $z_0$, the partial derivatives of the complex-valued function $f$ at $z_0$ are computed componentwise: \begin{align*} \partial_x f(z_0) &= \partial_x u(z_0)+i\partial_x v(z_0), \\ \partial_y f(z_0) &= \partial_y u(z_0)+i\partial_y v(z_0). \end{align*} Using the definition $\partial_{\bar z}=\frac{1}{2}(\partial_x+i\partial_y)$ gives \begin{align*} \partial_{\bar z}f(z_0) &= \frac{1}{2}\left(\partial_x f(z_0)+i\partial_y f(z_0)\right) \\ &= \frac{1}{2}\left(\partial_x u(z_0)+i\partial_x v(z_0)+i\partial_y u(z_0)+i^2\partial_y v(z_0)\right) \\ &= \frac{1}{2}\left(\partial_x u(z_0)-\partial_y v(z_0)\right) +\frac{i}{2}\left(\partial_x v(z_0)+\partial_y u(z_0)\right). \end{align*} [guided] We first translate the Wirtinger derivative into ordinary real partial derivatives. Write $z_0=x_0+iy_0$ with $x_0,y_0\in\mathbb R$. The function $f$ is complex-valued, and its real and imaginary parts are the real-valued functions \begin{align*} u: \Omega &\to \mathbb R, & v: \Omega &\to \mathbb R. \end{align*} Because $f=u+iv$ and $u,v$ have first partial derivatives at $z_0$, differentiation with respect to the real variables $x$ and $y$ is performed componentwise: \begin{align*} \partial_x f(z_0) &= \partial_x u(z_0)+i\partial_x v(z_0), \\ \partial_y f(z_0) &= \partial_y u(z_0)+i\partial_y v(z_0). \end{align*} Now we substitute these identities into the definition of the Wirtinger operator $\partial_{\bar z}=\frac{1}{2}(\partial_x+i\partial_y)$: \begin{align*} \partial_{\bar z}f(z_0) &= \frac{1}{2}\left(\partial_x f(z_0)+i\partial_y f(z_0)\right) \\ &= \frac{1}{2}\left(\partial_x u(z_0)+i\partial_x v(z_0)+i\left(\partial_y u(z_0)+i\partial_y v(z_0)\right)\right). \end{align*} Distributing the factor $i$ over the second parenthesis gives \begin{align*} \partial_{\bar z}f(z_0) &= \frac{1}{2}\left(\partial_x u(z_0)+i\partial_x v(z_0)+i\partial_y u(z_0)+i^2\partial_y v(z_0)\right). \end{align*} Since $i^2=-1$, the real terms and imaginary terms separate as \begin{align*} \partial_{\bar z}f(z_0) &= \frac{1}{2}\left(\partial_x u(z_0)-\partial_y v(z_0)\right) +\frac{i}{2}\left(\partial_x v(z_0)+\partial_y u(z_0)\right). \end{align*} This is the key computation: the vanishing of $\partial_{\bar z}f(z_0)$ is now reduced to the vanishing of two real scalar quantities. [/guided] [/step] [step:Identify the vanishing of $\partial_{\bar z}f(z_0)$ with the Cauchy-Riemann equations] A complex number $a+ib$ with $a,b\in\mathbb R$ is zero if and only if $a=0$ and $b=0$. Applying this to the expansion from the previous step, \begin{align*} \partial_{\bar z}f(z_0)=0 \end{align*} holds if and only if \begin{align*} \partial_x u(z_0)-\partial_y v(z_0)&=0, \\ \partial_x v(z_0)+\partial_y u(z_0)&=0. \end{align*} Equivalently, \begin{align*} \partial_x u(z_0)&=\partial_y v(z_0), \\ \partial_y u(z_0)&=-\partial_x v(z_0). \end{align*} These are precisely the [Cauchy-Riemann equations](/page/Cauchy-Riemann%20Equations) for $f=u+iv$ at $z_0$. Therefore $f$ satisfies the Cauchy-Riemann equations at $z_0$ if and only if $\partial_{\bar z}f(z_0)=0$. [/step] [step:Compute $\partial_z f(z_0)$ under the Cauchy-Riemann equations] Assume now that the Cauchy-Riemann equations hold at $z_0$, so \begin{align*} \partial_x u(z_0)&=\partial_y v(z_0), \\ \partial_y u(z_0)&=-\partial_x v(z_0). \end{align*} Using the definition $\partial_z=\frac{1}{2}(\partial_x-i\partial_y)$ and differentiating $f=u+iv$ componentwise, \begin{align*} \partial_z f(z_0) &= \frac{1}{2}\left(\partial_x f(z_0)-i\partial_y f(z_0)\right) \\ &= \frac{1}{2}\left(\partial_x u(z_0)+i\partial_x v(z_0)-i\partial_y u(z_0)-i^2\partial_y v(z_0)\right) \\ &= \frac{1}{2}\left(\partial_x u(z_0)+\partial_y v(z_0)\right) +\frac{i}{2}\left(\partial_x v(z_0)-\partial_y u(z_0)\right). \end{align*} Substituting $\partial_y v(z_0)=\partial_x u(z_0)$ and $\partial_y u(z_0)=-\partial_x v(z_0)$ gives \begin{align*} \partial_z f(z_0) &= \frac{1}{2}\left(\partial_x u(z_0)+\partial_x u(z_0)\right) +\frac{i}{2}\left(\partial_x v(z_0)+\partial_x v(z_0)\right) \\ &= \partial_x u(z_0)+i\partial_x v(z_0). \end{align*} This is the asserted formula for $\partial_z f(z_0)$ when the Cauchy-Riemann equations hold. [/step]