Prove Cauchy-Riemann respected with $\frac{\partial f}{\partial \bar z} = 0$

Question

I struggle a lot with complex analysis currently. I need to find the domain of analycity of $f(z) =: u(x, y) + iv(x, y)$, so the function needs to be continuous at $z$, its partial derivates must exist and the Cauchy-Riemann rules must be respected.

Here is my function:

$$f(z) = \frac{z^2}{z-3}$$

I know $z = 3$ is to be rejected. But what about Cauchy-Riemann ? In order to find the solution within this century, I need to verify:

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

So, naively, I tried:

$$\frac{\partial f}{\partial \bar z} = \frac{\partial f}{\partial z}.\frac{\partial z}{\partial \bar z} $$

with:

$$\frac{\partial z}{\partial \bar z} = \frac{\partial z}{\partial x}.\frac{1}{\frac{\partial \bar z}{\partial x}} + \frac{\partial z}{\partial y}.\frac{1}{\frac{\partial \bar z}{\partial y}} = 1+ i^2 = 0$$

A dead end, thus.

So, what's the easiest way to prove Cauchy-Riemann in this exercise ?

Answer 1

There are multiple more "sophisticated" ways by e.g. using that $z$ can be seen as a constant when applying the operator $\frac{\partial}{\partial \bar{z}}$ (see Wirtinger derivatives for more info on this), but here is a more elementary way:
The operator $\frac{\partial}{\partial \bar{z}}$ is defined as $$ \frac{\partial}{\partial \bar{z}}:=\frac{1}{2}(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}) $$ for $z=x+iy$ and your function can be see as a function of $z=x+iy$: $$ f(x+iy)=\frac{(x+iy)^3}{x+3-iy}. $$ Now you just need to compute $$ \frac{1}{2}(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y})\frac{(x+iy)^3}{x+3-iy}=\frac{1}{2}\frac{\partial}{\partial x}\frac{(x+iy)^3}{x+3-iy}+i\frac{1}{2}\frac{\partial}{\partial y}\frac{(x+iy)^3}{x+3-iy} $$ for $x\neq 3$ and it should be $0$ in the end.

Answer 2

Several ways, depending on what you know...

1. $f$ is the ratio of two polynomial maps, hence is analytic where the denominator doesn't vanish. So $f$ is analytic on $\mathbb C \setminus\{3\}$.

2. Using Cauchy-Riemann equations, which means tedious derivatives.

3. Using the Taylor series.

The first one is the simplest one...

Answer 3

From the tree $f\rightarrow x,y\rightarrow z, \overline z$, of dependence of variables, we have $$\frac{\partial f}{\partial\overline{z}}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial\overline{z}}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial\overline{z}}=f_x.\frac{1}{2}+f_y.\frac{i}{2}=\frac{1}{2}(f_x+if_y).\tag{1}$$ If $f(z)$ is analytic function, the Cauchy-Riemann equations that $u_x=v_y$ and $v_x=-u_y$ are satisfied, where $f(z)=u(x,y)+iv(x,y)$. But, then $$f_x+if_y=(u_x+iv_x)+i(u_y+iv_y)=(u_x-v_y)+i(v_x+u_y)=0.\tag{2}$$ From $(1)$ and $(2)$, $\frac{\partial f}{\partial\overline{z}}=0$ in its domain of analyticty.