Let $\Omega \subset \Bbb C$ be an open domain and $f\colon \Omega \to \Bbb C$ be a holomorphic function. By identifying the point $z = x+iy$ with $(x,y) \in \Bbb R^2$, we may write $f(z) = u(x,y) + i v(x,y)$ where $u,v$ are real-valued functions on $\Omega$ (considered as a subset of $\Bbb R^2$). We can then define a real-valued function
$$
W(x,y) = |f(z)|^2 = u(x,y)^2 + v(x,y)^2.
$$
It is not hard to verify that $W$ is differentiable and its gradient is
$$
\nabla W = 2\begin{bmatrix}
uu_x+vv_x \\
uu_y + vv_y
\end{bmatrix}
=2\begin{bmatrix}
uu_x+vv_x \\
-uv_x + vu_x
\end{bmatrix},
$$
where we used the Cauchy-Riemann equation to substitute $u_x = v_y$ and $u_y = -v_x$. Again, by the Cauchy-Riemann equation, we know that $f'(z) = u_x(z) + iv_x(z)$ and hence
$$
f\overline{f'} = (u+iv)(u_x-iv_x) = (uu_x + vv_x) + i(-uv_x + vu_x).
$$
One would be tempted to identify the real gradient of the real-valued function $W$ with the complex number $2f\overline{f'}$ and write
$$
\nabla W(x,y) = 2 f(z)\overline{f'(z)}.
$$
Is there a neat way to derive this "identity" $\nabla W = 2 f\overline{f'}$?
Of course, this looks suspiciously like the identity $(g^2)' = 2gg'$ for a differentiable $g\colon\Bbb R\to\Bbb R$, but with a complex conjugation sign, so I think it is reasonable to expect that one should be able to derive it using a product rule or chain rule of some kind.
I tried writing $W = f\bar f$, thinking of $f,\bar f$ as $\Bbb R^2$-valued functions, and then differentiated it. Alas, I am quite bad at complex analysis and couldn't recover the desired identity. One of the problems I encountered is that the $2\times2$ Jacobian matrix $D(\bar f)$ cannot be identified with $\overline {f'}$. In fact, $D(\bar f)$ doesn't even represent any complex number (since $f$ is holomorphic). By that I meant the identification of $\Bbb C$ as isomorphic to the subalgebra of $M^{2\times 2}(\Bbb R)$
$$
\left\{ \begin{bmatrix}
a &-b \\
b &a
\end{bmatrix} : a,b \in \Bbb R \right\} \cong \left\{ z = a+ib : a,b \in \Bbb R \right\} \cong \Bbb C.
$$
Best Answer
If you identify $\nabla W$ with $\frac{\partial}{\partial x}W + i \frac{\partial}{\partial y}W $ then you can use the Wirtinger derivatives $$ {\begin{aligned}{\frac {\partial }{\partial z}}&={\frac {1}{2}}\left({\frac {\partial }{\partial x}}-i{\frac {\partial }{\partial y}}\right)\\{\frac {\partial }{\partial {\bar {z}}}}&={\frac {1}{2}}\left({\frac {\partial }{\partial x}}+i{\frac {\partial }{\partial y}}\right)\end{aligned}} $$ which satisfy a product rule like a “normal” derivative. For holomorphic functions is (because of the Cauchy-Riemann equations) $$ \frac {\partial }{\partial z} f = f' \, , \,\frac {\partial }{\partial \bar z} f = 0 \,,\\ \frac {\partial }{\partial z} \bar f = 0 \,, \,\frac {\partial }{\partial \bar z} \bar f = \overline{f'} \, . $$ Then $$ \nabla (|f|^2) = 2 \frac {\partial }{\partial \bar z} (f \bar f) = 2 \left( \bar f \frac {\partial }{\partial \bar z} f + f \frac {\partial }{\partial \bar z} \bar f \right) = 2 f \overline{f'} \, . $$