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.


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.


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

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


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 ${\mathbb{{{C}}}}^{{\cdot}}={\mathbb{{{C}}}}\setminus{\left\lbrace{0}\right\rbrace}$.


1 Answer 1


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 $$ Df = \frac{\partial}{\partial z}(a) dz + \frac{\partial}{\partial \overline z}(a) d\overline z \, . $$

  • $\begingroup$ 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 $Df \in \mathscr{L}_{\mathbb{C}}(\mathbb{C})$? $\endgroup$
    – Talmsmen
    Jun 22, 2021 at 19:18
  • $\begingroup$ @JPwin: What does $\mathscr{L}_{\mathbb{C}}(\mathbb{C})$ denote? $\endgroup$
    – Martin R
    Jun 22, 2021 at 19:29
  • $\begingroup$ $\mathscr{L}_{\mathbb{C}}(\mathbb{C})$ 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. $\endgroup$
    – Talmsmen
    Jun 22, 2021 at 21:38
  • 1
    $\begingroup$ @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. $\endgroup$
    – Martin R
    Jun 22, 2021 at 21:45
  • $\begingroup$ Thank you, I had a hunch that $DF \not\in \mathscr{L}_{\mathbb{C}}(\mathbb{C})$; your comment confirmed it. Thanks again. $\endgroup$
    – Talmsmen
    Jun 22, 2021 at 22:50

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