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