Let \(f:\mathbb C\to\mathbb C\) be given in polar coordinates by \[ f(r,\theta)=(r^\alpha,n\theta), \] where \(\alpha>0\) and \(n\in\mathbb Z\). Compute \(\partial f/\partial z\) and \(\partial f/\partial\overline z\).
Solution
For \(z=re^{i\theta}\neq0\), \[ \boxed{ \frac{\partial f}{\partial z} =\frac{\alpha+n}{2}r^{\alpha-1}e^{i(n-1)\theta}, \qquad \frac{\partial f}{\partial\overline z} =\frac{\alpha-n}{2}r^{\alpha-1}e^{i(n+1)\theta}.} \] At the origin, where the Wirtinger derivatives are taken to be defined only when \(f\) is \(C^1\) there, they are as follows: \[ \begin{array}{c|c|c} \text{case}&f_z(0)&f_{\overline z}(0)\\ \hline \alpha>1&0&0\\ \alpha=1,\ n=1&1&0\\ \alpha=1,\ n=-1&0&1. \end{array} \] In all remaining cases, \(f\) is not \(C^1\) at \(0\), so the Wirtinger derivatives there are not defined in that convention.
1. For \(z=re^{i\theta}\neq0\), the Wirtinger operators are\[ \frac{\partial}{\partial z} =\frac{e^{-i\theta}}2 \left(\frac{\partial}{\partial r} -\frac{i}{r}\frac{\partial}{\partial\theta}\right), \qquad \frac{\partial}{\partial\overline z} =\frac{e^{i\theta}}2 \left(\frac{\partial}{\partial r} +\frac{i}{r}\frac{\partial}{\partial\theta}\right). \]
From \[ x=r\cos\theta, \qquad y=r\sin\theta, \] one obtains \[ \partial_x =\cos\theta\,\partial_r -\frac{\sin\theta}{r}\,\partial_\theta, \qquad \partial_y =\sin\theta\,\partial_r +\frac{\cos\theta}{r}\,\partial_\theta. \] Substituting these into \[ \partial_z=\frac12(\partial_x-i\partial_y), \qquad \partial_{\overline z}=\frac12(\partial_x+i\partial_y) \] gives the two formulas.
2. For \(z\neq0\), the derivatives have the boxed values above.
The polar description means \[ f(re^{i\theta})=r^\alpha e^{in\theta}. \] Therefore \[ f_r=\alpha r^{\alpha-1}e^{in\theta}, \qquad f_\theta=inr^\alpha e^{in\theta}. \] Applying 1 gives \[ \begin{aligned} f_z &=\frac{e^{-i\theta}}2 \left(\alpha r^{\alpha-1}e^{in\theta} -\frac{i}{r}inr^\alpha e^{in\theta}\right)\\ &=\frac{\alpha+n}{2}r^{\alpha-1}e^{i(n-1)\theta}, \end{aligned} \] and similarly \[ \begin{aligned} f_{\overline z} &=\frac{e^{i\theta}}2 \left(\alpha r^{\alpha-1}e^{in\theta} +\frac{i}{r}inr^\alpha e^{in\theta}\right)\\ &=\frac{\alpha-n}{2}r^{\alpha-1}e^{i(n+1)\theta}. \end{aligned} \]
3. If \(\alpha>1\), then \(f\) is \(C^1\) at \(0\) with real differential\(Df_0=0\), so \(f_z(0)=f_{\overline z}(0)=0\).
Since \(|f(z)|=|z|^\alpha\), \[ \frac{|f(z)-f(0)|}{|z|}=|z|^{\alpha-1}\longrightarrow0, \] so \(f\) is real differentiable at \(0\) with derivative zero. The formulas in 2 satisfy \[ |f_z(z)|\le \frac{|\alpha+n|}{2}|z|^{\alpha-1}, \qquad |f_{\overline z}(z)|\le \frac{|\alpha-n|}{2}|z|^{\alpha-1}, \] and both tend to zero. Hence the first derivatives extend continuously to \(0\), so \(f\) is \(C^1\) there and both Wirtinger derivatives vanish.
4. If \(\alpha=1\), then \(f\) is real differentiable at \(0\) exactly for\(n=1\) or \(n=-1\).
For \(z=re^{i\theta}\), \[ f(z)=r e^{in\theta}. \] If \(f\) were real differentiable at \(0\), its differential would be a real linear map \(L:\mathbb C\to\mathbb C\), hence \[ L(z)=az+b\overline z \] for some \(a,b\in\mathbb C\). Dividing \[ f(re^{i\theta})=L(re^{i\theta})+o(r) \] by \(r\) and letting \(r\to0\) gives \[ e^{in\theta}=ae^{i\theta}+be^{-i\theta} \] for every \(\theta\). The Fourier modes on the two sides can agree only when \(n=1\) or \(n=-1\). In those cases \(f(z)=z\) and \(f(z)=\overline z\), respectively, yielding the stated Wirtinger derivatives.
5. If \(0<\alpha<1\), then \(f\) is not real differentiable at \(0\).
Real differentiability would imply \(|f(z)|=O(|z|)\) as \(z\to0\). But \[ \frac{|f(z)|}{|z|}=|z|^{\alpha-1}\longrightarrow\infty, \] which is impossible. Thus \(f\) is not \(C^1\) at the origin, and its Wirtinger derivatives are not defined there in this case.