Suppose \(f\) is analytic. Compute \[ \frac{\partial}{\partial z}\frac{\partial}{\partial\overline z}|f(z)|^2. \]
Solution
\[ \boxed{ \frac{\partial}{\partial z} \frac{\partial}{\partial\overline z}|f(z)|^2 =|f'(z)|^2.} \]
1. Since \(f\) is analytic,\[ \frac{\partial f}{\partial\overline z}=0, \qquad \frac{\partial\overline f}{\partial\overline z} =\overline{f'(z)}. \]
The first identity is the Cauchy–Riemann equation in Wirtinger notation. Taking complex conjugates of \[ \frac{\partial f}{\partial z}=f'(z) \] gives the second identity.
2. Therefore\[ \frac{\partial}{\partial\overline z}|f|^2 =f\,\overline{f'}. \]
Because \(|f|^2=f\overline f\), the product rule and 1 give \[ \frac{\partial}{\partial\overline z}(f\overline f) =\frac{\partial f}{\partial\overline z}\overline f +f\frac{\partial\overline f}{\partial\overline z} =f\,\overline{f'}. \]
3. Differentiating once more with respect to \(z\) yields\[ \frac{\partial}{\partial z} \frac{\partial}{\partial\overline z}|f|^2 =|f'|^2. \]
The function \(f'\) is analytic, so \(\overline{f'}\) is antiholomorphic and \[ \frac{\partial}{\partial z}\overline{f'}=0. \] Thus, by 2, \[ \frac{\partial}{\partial z}\bigl(f\,\overline{f'}\bigr) =f'\overline{f'} +f\frac{\partial\overline{f'}}{\partial z} =|f'|^2. \]