# Understanding partial derivatives related to the Cauchy-Riemann equations

So I am reading a soft introductory book on complex variables (I have complex analysis next year.) The book I am reading is called "Complex Variables Demystified". I am currently reading a chapter where the main goal is to arrive at the Cauchy-Riemann equations. I'm having problems following the deduction; specifically, given a function $f$ of a complex variable $z=x+iy$, the author starts out by expressing the partial derivatives of $f$ with respect to $z$ and $\bar z$. To do this, he first arrive at : $\frac{\partial}{\partial z}=\frac{\partial x}{\partial z}\frac{\partial}{\partial x}+\frac{\partial y}{\partial z}\frac{\partial}{\partial y}$, and similarly $\frac{\partial}{\partial \bar z}=\frac{\partial x}{\partial \bar z}\frac{\partial}{\partial x}+\frac{\partial y}{\partial \bar z}\frac{\partial}{\partial y}$.

At this point in the text i am a littlebit lost. First of all, how did he deduce these equations? (Where do they come from?) And what do they mean? Is the author expressing a general differential operator $\frac{\partial}{\partial z}$ of a complex variable $z=x+iy$ in which $x$ and $y$ again are funcions $x=x(z,\bar z)$, $y=y(z,\bar z)$ of $z$ and $\bar z$ ? And what does it mean in the two equations above when $\frac{\partial}{\partial x}$ is written without anything in the numerator? Is this again a general differential operator? I'm essentially just trying to understand how the author gets to the final expressions here:

I would be happy if anyone could explain what's going on as simple as possible :) -Thanx

In brief, the author is simply applying the chain rule to the partial differentiation operators w.r.t. $z$ and $\bar{z}$. These don't depend on the function being differentiated, hence the lack of a function in the numerator.

And, as you point out, because $z$ and $\bar{z}$ are functions of $x$ and $y$, the expressions derived from the chain rule can be simplified to what the author is describing in the page you posted.

For a more detailed description:

Let's say you have some function $f(w)$ where $w$ is some variable related to $x$ by $w = g(x)$. If I wanted to calculate how this function $f(w)$ is changing with respect to $x$, I need to use the chain rule to do so, obtaining:

$$\frac{\partial f(w)}{\partial x} = \frac{\partial f(w)}{\partial w}\frac{\partial w}{\partial x}$$

This equality is true independent of the form of $f(w)$ (as long as it's not pathological, but that's usually the case). Because of that, the author is defining $\frac{\partial}{\partial x}$ without a function in the numerator because he can manipulate the resulting expression without needing to define an explicit function $f$. You'll find that inserting some arbitrary function $f$ into the numerator of the expressions you're asking about won't change anything about the proof or the nature of the problem.

To demonstrate this, we note that $$z = x + iy$$ $$\bar{z} = x - iy$$

We can do algebra on both of these relationships to solve for $x$ and $y$ as functions of $z$ and $\bar{z}$:

$$x = \frac{z + \bar{z}}{2}$$ $$y = \frac{z - \bar{z}}{2i}$$

That means we can use the chain rule as we did above to find:

$$\frac{\partial f(z,\bar{z})}{\partial z} = \frac{\partial f(z,\bar{z})}{\partial x}\frac{\partial x}{\partial z} + \frac{\partial f(z,\bar{z})}{\partial y}\frac{\partial y}{\partial z}$$

$$\frac{\partial f(z,\bar{z})}{\partial \bar{z}} = \frac{\partial f(z,\bar{z})}{\partial x}\frac{\partial x}{\partial \bar{z}} + \frac{\partial f(z,\bar{z})}{\partial y}\frac{\partial y}{\partial \bar{z}}$$

From here, our expressions for $x$ and $y$ allow us to calculate the partial derivatives of $x$ and $y$ with respect to $z$ and $\bar{z}$, leading directly to equations 3.21 and 3.22 once you identify $f = u + iv$. Note again that these partial derivatives are totally independent of $f$, which is why the author ignores putting $f$ in the numerator. (It's just operator notation.)

Hopefully that clears things up!

is this correct? :)

and this? Thanks for your response by the way!