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Let $f:\mathbb{C} \to \mathbb{C} \\ \,\ \ \ \ \ z \to (|z|^2-2)\overline{z}$

Determine the points where f is differentiable, presenting the due expression $f'(z)$.

$u(x,y)=Re((x^2+y^2-2)(x-yi))=x^3+xy^2-2x; \\ v(x,y)=Im((x^2+y^2-2)(x-yi))=-yx^2-y^3+2y $

$u_x=3x^2+y^2-2; \,u_y=2xy \\ v_x= \ \ -2xy;\qquad \ \ v_y=-x^2-3y^2+2$

The CR equality happens when: $$4(x^2+y^2)=4$$

And as all partial derivatives are continuous, f is differentiable at $\partial{D(0,1)}$

And now, for $z_{0} \in \partial{D(0,1)}$ and knowing $\textit{'a priori'}$ that the limit exists:

$$ lim_{z \to z_0; z \in \partial{D(0,1)}} \frac{f(z)-f(z_0)}{z-z_0} = lim_{z \to z_0; z \in \partial{D(0,1)}}\frac{(|z|^2-2)\overline{z}+\overline{z_0}}{z-z_0}=lim_{z \to z_0; z \in \partial{D(0,1)}}-\frac{z-z_0}{z-z_0}=-1$$ So $f'(z_0)=-1$ for $z_0 \in \partial{D(0,1)}$

Is this correct? Any further suggestions?

J P
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    Note that for $z$ on the unit circle, $\bar z = z^{-1}$ (and even more plainly, $|z|=1$), so on that circle the function is identical to $g(z) = -z^{-1}$. Surely you can differentiate that. – Torsten Schoeneberg Mar 21 '24 at 00:04

3 Answers3

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You are right about the points at which $f\require{cancel}$ is differentiable, but not about the computation of $f'$. If $x^2+y^2=1$, then\begin{align}f'(x+yi)&=\frac{\partial u}{\partial x}(x,y)+\frac{\partial v}{\partial x}(x,y)i\\&=3x^2+y^2-2-2xyi\\&=\cancel{2x^2+2y^2-2}+x^2-y^2-2xyi\\&=(x-yi)^2.\end{align}In other words, if $|z|=1$, then $f'(z)=\overline z^2$.

José Carlos Santos
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    Note that on the unit circle, $\bar z = z^{-1}$ which makes your result $f'(z) = z^{-2}$, matching what one gets from my comment under the question. – Torsten Schoeneberg Mar 21 '24 at 00:06
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Following an interesting exchange with @Torsten Schoeneberg who as pointed a fundamental misunderstanding of mine, the present state of this text is now a "side remark" that can still have an interest to show how complex differentiability of $f:\mathbb{C} \to \mathbb{C}$ is much more coercitive than $\mathbb{R^2}$ differentiability ($f$ considered as a function $\mathbb{R^2} \to \mathbb{C}$).

Here is an experimental approach using graphical tools that we have at our disposal.

Have a look at fig. 1 featuring the absolute value map i.e., the graphical representation of function

$$\varphi(x,y)=|(x^2+y^2-2)(x-iy)|$$

enter image description here

Fig. 1 : Modulus information. The black circle corresponds to the unit circle, the locus of points where $f$ is $\mathbb{C}$-differentiable.

"Visibly", function $f$ is $\mathbb{R^2}$ differentiable except :

  • along the circle with equation $x^2+y^2=2$ and

  • in the origin.

It is also interesting to understand how the "argument" (aka "phase", or even "angle" in software like Matlab) evolves (see at the bottom the short Matlab script which has generated fig. 2).

enter image description here

Fig. 2 : Phase information.

displaying "phase" values between $-\pi$ and $\pi$ (see right bar) taking a circular convention in order that the colors attributed to the two ends are the same (thanks to @AccidentalTaylorExpansion for his remark).

Phase transition $\varphi \to \varphi \pm \pi$ occurs along circle $|z|=\sqrt{2}$ ; it takes its source in the sign change of $|z|^2-2$ from negative inside the disk to positive outside it.

These graphics are by no means a proof ; they can be useful either for a first understanding of such a question or for a confirmation of already obtained results.

Matlab script for fig. 2 (see in particular the use of HSV color system in order to obtain a circular coding) :

 a=2.5;
 [X,Y]=meshgrid(-a:0.01:a);
 Z=X+i*Y;
 F=((abs(Z).^2)-2).*conj(Z);
 hmap(1:256,1) = linspace(0,1,256);
 hmap(:,[2 3]) = 0.7;
 huemap = hsv2rgb(hmap);
 imagesc(angle(F));colormap(huemap);axis equal tight;colorbar
Jean Marie
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  • Not sure what the picture is supposed to show, but to me it seems to show that $f$ is not even continuous at any point on the circle and the line segments indicated by you -- and that is false, because the function is quite obviously continuous everywhere. – Torsten Schoeneberg Mar 21 '24 at 01:17
  • @Torsten Schoeneberg You are right. I have had a bad interpretation of the colormap. I will correct it. – Jean Marie Mar 21 '24 at 06:45
  • I like to use circular colormaps (e.g. rainbow) for phase, because then every discontinuitity is a true discontinuity in phase. – AccidentalTaylorExpansion Mar 21 '24 at 09:53
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    @AccidentalTaylorExpansion I have taken your advice into consideration. – Jean Marie Mar 21 '24 at 13:43
  • Note that still, 1) the function does not have discontinuities even along that circle (@AccidentalTaylorExpansion is right that every discontinuity gives a disc. in phase, but here we see discont. in phase which are not true discontinuities because they happen where the modulus is $0$); 2) now the image together with the other answers is a good illustration that complex differentiability is much stronger than differentiability as real map: This function is real-differentiable away from $0$ and the $2$-circle. But the CR equations are satisfied only on the (different) circle $\partial D(0,1)$. – Torsten Schoeneberg Mar 21 '24 at 15:58
  • @Torsten Schoeneberg Thank you a lot ! You are completely right : I was thinking in terms of differentiability on $\mathbb{R^2}$ and had completely forgotten how CR constraints are "much stronger" as you say. I have to modify again my answer ! – Jean Marie Mar 21 '24 at 16:49
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    @TorstenSchoeneberg : I have modified my text. Besides, I was wondering if this "polar vision" (modulus vs. argument) could have been connected with the polar version of Cauchy-Riemann equations (see for example here). – Jean Marie Mar 21 '24 at 21:49
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Use complex CR equality: $$\frac{\partial f}{\partial \overline z}=0,$$ and $f(z)=(|z|^2-2)\overline{z}=z\overline z^2-2\overline z$, then $$\frac{\partial f}{\partial \overline z} =2z\overline z-2=0\iff|z|=1,$$ and $$f'(z)=\frac{\partial f}{\partial z}=\overline z^2,\quad \forall z \in \partial{D(0,1)}.$$

Riemann
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