# Calculating $Df$ for $f(z)=\frac{z^3}{\overline{z}}$

Please verify my attempted solution. How would one calculate $${D}{f}$$ for $${f{{\left({z}\right)}}}=\frac{{{z}^{{3}}}}{{\overline{{{z}}}}}$$? I am aware that $$\overline{{{z}}}$$ is a nowhere analytic function. If we loosen the problem to ask for $${D}{f}$$ instead of $${f}'{\left({z}\right)}$$, we need only find a $${D}{f{{\left({h}\right)}}}$$ s.t. the following holds.

$${f{{\left({a}+{h}\right)}}}={f{{\left({a}\right)}}}+{D}{f{{\left({h}\right)}}}+{\left|{h}\right|}{\epsilon}{\left({h}\right)}$$

In the above, $$\lim_{{{h}\to{0}}}{\epsilon}{\left({h}\right)}={0}$$ and $$h$$ lies in a neighborhood of sufficiently small modulus on the complex plane.

$${f{{\left({z}\right)}}}=\frac{{{z}^{{3}}}}{{\overline{{{z}}}}}=\frac{{{\left({x}+{i}{y}\right)}^{{3}}}}{{{\left({x}-{i}{y}\right)}}}$$

Writing $${z}={x}+{i}{y}$$ and using the fact that $${\mathbb{{{R}}}}^{{{2}}}$$ is isomorphic to $${\mathbb{{{C}}}}$$, we may define $${D}{f}=\frac{{\partial{f}}}{{\partial{x}}}{\left({a}\right)}{\left.{d}{x}\right.}+\frac{{\partial{f}}}{{\partial{y}}}{\left({a}\right)}{\left.{d}{y}\right.}$$.

$$\frac{{\partial{f}}}{{\partial{x}}}=\frac{{{3}{\left({x}-{i}{y}\right)}{\left({x}+{i}{y}\right)}^{{2}}-{\left({x}+{i}{y}\right)}^{{3}}}}{{{\left({x}-{i}{y}\right)}^{{2}}}}=\frac{{{3}{\left|{z}\right|}^{{2}}{z}-{z}^{{3}}}}{{{\left(\overline{{{z}}}\right)}^{{2}}}}$$ $$\frac{{\partial{f}}}{{\partial{y}}}=\frac{{{3}{i}{\left({x}-{i}{y}\right)}{\left({x}+{i}{y}^{{2}}\right)}+{i}{\left({x}+{i}{y}\right)}^{{3}}}}{{{\left({x}-{i}{y}\right)}^{{2}}}}=\frac{{{3}{i}{\left|{z}\right|}^{{2}}{z}+{i}{z}^{{3}}}}{{{\left(\overline{{{z}}}\right)}^{{2}}}}$$

$${D}{f}=\frac{{\partial{f}}}{{\partial{x}}}{\left.{d}{x}\right.}+\frac{{\partial{f}}}{{\partial{y}}}{\left.{d}{y}\right.}=\frac{{{3}{\left|{z}\right|}^{{2}}{z}{\left({\left.{d}{x}\right.}+{i}{\left.{d}{y}\right.}\right)}+{z}^{{3}}{\left({i}{\left.{d}{y}\right.}-{\left.{d}{x}\right.}\right)}}}{{{\left(\overline{{{z}}}\right)}^{{2}}}}=\frac{{{3}{\left|{z}\right|}^{{2}}{z}{\left.{d}{z}\right.}-{z}^{{3}}{d}\overline{{{z}}}}}{{{\left(\overline{{{z}}}\right)}^{{2}}}}$$

I find my solution to be inelegant and doubt that it is correct. Although I can't take the true complex derivative, would my reasoning be the most appropriate appeal to MVC?

Furthermore, if I were to try and determine when $${D}{f}\in\mathscr{L}_{{{\mathbb{{{C}}}}}}{\left({\mathbb{{{C}}}}\right)}$$, would checking the holomorphicity of $${D}{f}$$ suffice? I had read that a linear transform $${L}={P}{\left.{d}{x}\right.}+{Q}{\left.{d}{y}\right.}$$ satisfies $${L}\in\mathscr{L}_{{{\mathbb{{{C}}}}}}{\left({\mathbb{{{C}}}}\right)}\Leftrightarrow{Q}={i}{P}$$, which is truly just the Cauchy-Riemann equations. The primary reason that I believe I am mistaken is the fact that my definition for $${D}{f}$$ does not appear to be holomorphic on $${\mathbb{{{C}}}}$$ or $${\mathbb{{{C}}}}^{{\cdot}}={\mathbb{{{C}}}}\setminus{\left\lbrace{0}\right\rbrace}$$.

Your result is correct. Note that it can be simplified to $$Df = \frac{3z^2}{\overline z} dz - \frac{z^3}{\overline z^2} d\overline z$$ which makes it apparent that it is the same result as obtained by $$Df = \frac{\partial}{\partial z}(a) dz + \frac{\partial}{\partial \overline z}(a) d\overline z \, .$$
• If I hoped to find the $z\in\mathbb{C}$ s.t. $Df \in \mathscr{L}_{\mathbb{C}}(\mathbb{C})$, would I need to show that $Df=\frac{3z^2}{\overline{z}}dz-\frac{z^3}{\overline{z}^2}d\overline{z}$ satisfies the Cauchy-Riemann equations locally in a subset $\Omega \subset \mathbb{C}$? I had been under the impression that $Df$ would be nowhere holomorphic given that it includes a $\overline{z}$ term, so the task of finding a region in which the Cauchy-Riemann equations are satisfied appears impossible. Is there an alternative method to prove $Df \in \mathscr{L}_{\mathbb{C}}(\mathbb{C})$? Jun 22, 2021 at 19:18
• @JPwin: What does $\mathscr{L}_{\mathbb{C}}(\mathbb{C})$ denote? Jun 22, 2021 at 19:29
• $\mathscr{L}_{\mathbb{C}}(\mathbb{C})$ is the set of all complex linear transforms. A transform $L=Pdx + Qdy$ is $\mathbb{C}$-linear ($L\in\mathscr{L}_{\mathbb{C}}(\mathbb{C})$) iff $P=-iQ$, which is equivalent to the Cauchy-Riemann equations. If $Df$ were to be $\mathbb{C}$-linear at a point, I would presume that it would have to satisfy the CR-equations; however, my definition of $Df$ does not appear to be holomorphic on a subset $\Omega \subset \mathbb{C}$. Thank you. Jun 22, 2021 at 21:38
• @JPwin: The CR equations are also equivalent to $\frac{\partial f}{\partial \overline z} = 0$ and that is nowhere the case. $f$ is nowhere complex differentiable, and $Df$ is nowhere a complex linear transform. Jun 22, 2021 at 21:45
• Thank you, I had a hunch that $DF \not\in \mathscr{L}_{\mathbb{C}}(\mathbb{C})$; your comment confirmed it. Thanks again. Jun 22, 2021 at 22:50