I have this exercise:

Let $f: \mathbb{C} \to \mathbb{C}$ be defined as $f(0) = 0$ and $f(z) = \dfrac{(\bar{z})^2}{z}$ for $z \neq 0$.

Show that the Cauchy-Riemann equations are satisfied on $z = 0$, but $f'(0)$ doesn't exist.

I learned two different forms of the Cauchy-Riemann equations:

$$\dfrac{\partial u}{\partial x} = \dfrac{\partial v}{\partial y} \text{ }\text{ }\text{ }\text{ and }\text{ }\text{ }\text{ } \dfrac{\partial u}{\partial y} = -\dfrac{\partial v}{\partial x}$$

$$\dfrac{\partial f}{\partial \bar{z}} = 0$$

With the first form, I would have to write $f$ as a function of $x$ and $y$ rather than $z$ and $\bar{z}$, and then taking the derivative would lead to big expressions. I would like to avoid that, and try to use the second form, because the exercise seems to be asking for it.

My problem is, the function $f$ is piecewise defined. I can't just blindly do

$$\dfrac{\partial}{\partial \bar{z}} \dfrac{(\bar{z})^2}{z} = \dfrac{2\bar{z}}{z}$$

since I'm interested in $z = 0$.

I would like to use the limit definition of the partial derivative to try to solve this. But it seems that $\dfrac{\partial}{\partial \bar{z}}$ is not a regular partial derivative (here and here).

How can I proceed here?


When you are not absolutely sure of a notation such as $\partial f\over\partial\bar z$, then it is time to give up on "I would like to avoid that", bite the bullet, and go with what you know.

$$f(z) = \frac{\bar z^2}z = \frac{\bar z^3}{|z|^2} = \frac{x^3 - 3xy^2 -i3x^2y + iy^3}{x^2+y^2} = x\left(1 - \frac{4y^2}{x^2+y^2}\right) + iy\left(1 - \frac{4x^2}{x^2+y^2}\right)$$

So $$u = x\left(1 - \frac{4y^2}{x^2+y^2}\right)\\v = y\left(1 - \frac{4x^2}{x^2+y^2}\right)$$

  • $\begingroup$ Well, instead of giving up I usually prefer to go ahead and try to learn more. That was the purpose of my question. Thanks for the attempt though, maybe I didn't write the question well enough. Is the method on your answer the only way to solve this exercise? Or maybe is there a way to use $\frac{\partial f}{\partial \bar{z}}$ to solve it? $\endgroup$ – Pedro A Mar 27 '16 at 20:52
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    $\begingroup$ My point is, when something is unsure, you need to back up to where it is sure, then you can move forward more carefully and see how things actually do apply. $\frac{\partial f}{\partial\bar z}$ is a shorthand you can use when things are smoothly defined. But in fringe cases where smoothness is not assured, you have to look more carefully into the definitions to see if they apply. That is the case here. You need to examine carefully what is going on at $0$. To do that, shorthands are not going to be good enough. $\endgroup$ – Paul Sinclair Mar 27 '16 at 22:02
  • $\begingroup$ Oh I understand now. The thing is, we can't use the $\frac{\partial f}{\partial \bar{z}}$ in this case; this fact only got clear to me with your comment. If you don't mind, you should add your comment to the answer itself, so it gets more complete for other readers (though I already understood). Thanks for your help! $\endgroup$ – Pedro A Mar 27 '16 at 23:44

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