Finding Derivatives of Analytic Functions: Chain Rule Confusion

[Natura]

Hello,

I'm sorry if I'm not posting this to the correct place - this is my first post on PhysicsForums.com

My question regards derivatives of analytic functions. Here it goes:

Let

w(z) = u(x,y) +iv(x,y)​

be an analytic function,
where

z = x + iy,​

for some x,y that are real numbers.

In order to find the derivative of this function, since it is analytic it does not matter from which direction I take the limit in the limiting process so I can easily derive that

(w(z))' = $\frac{∂u(x,y)}{∂x}$ +i$\frac{∂v(x,y)}{∂x}$​

So here is where my problem begins. I was doing some problems and then one of them asked me to find $\frac{∂w(z)}{∂z}$, which I believe should be exactly the same thing as the derivative above, but I tried to apply chain rule to it and thus:

$\frac{∂w(z)}{∂z}$ = $\frac{∂u(x,y)}{∂x}$$\frac{∂x}{∂z}$ +$\frac{∂u(x,y)}{∂y}$$\frac{∂y}{∂z}$ + i($\frac{∂v(x,y)}{∂x}$$\frac{∂x}{∂z}$ + $\frac{∂v(x,y)}{∂y}$$\frac{∂y}{∂z}$)​

I get this to equal twice the initially mentioned derivative for all the functions I tried it on.
It seems that differentiating only the real or only the imaginary component (the latter multiplied by i) gives the derivative. I can't explain this to myself. I would be happy if someone points out where my error is.

Thanks in advance (apologies for my poor Latex use)

 

[jackmell]

Thanks in advance (apologies for my poor Latex use)

What's poor about it? Well, that (w(z))' thing is a little unclear. Would have been more clear to say

$$\frac{dw}{dx}$$

Now, if you did the differentiating correctly, then you should get the same results. So if you don't, then you won't right?

What exactly are all those $\frac{dx}{dz}$ and $\frac{dy}{dz}$?

 

[Mandelbroth]

Is it asking for the Wirtinger derivative? If so, you're actually looking to compute $$\frac{\partial w}{\partial z}=\frac{1}{2}\left(\frac{\partial w}{\partial x}-i\frac{\partial w}{\partial y}\right).$$

 

[Natura]

Firstly, thank you for the responses.

I agree I wasn't clear enough in my initial post. I'll try to correct that now.

Since

z = x + iy​

We can rearrange to get

x = z -iy​

therefore

$\frac{∂x}{∂z}$ = $\frac{∂z}{∂z}$ = 1​

Similarly for y we get

$\frac{∂y}{∂z}$ = -i​

Then using the Cauchy-Riemann relations to eliminate all of the y derivatives and substituting the above results for $\frac{∂x}{∂z}$ and $\frac{∂y}{∂z}$ I get that

$\frac{∂w}{∂z}$ = 2*$\frac{∂w}{∂x}$​

As for the Wirtinger derivative, it makes sense the way it is defined but I would like to see how it is derived because I don't see where the factor of (1/2) comes from which is apparently what I am missing.

Thanks again.

 

[Natura]

Nevermind, I can see that my expressions for $\frac{∂x}{∂z}$ and $\frac{∂y}{∂z}$ are wrong and are off by a factor of (1/2) ... Thanks again.

 

[jackmell]

Nevermind, I can see that my expressions for $\frac{∂x}{∂z}$ and $\frac{∂y}{∂z}$ are wrong and are off by a factor of (1/2) ... Thanks again.

Natura, let me make sure you understand this ok?

We have $w=f(z)=u(x,y)+iv(x,y)$

and:

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

so that:

$$\frac{dx}{dz}=1/2$$
$$\frac{dy}{dz}=\frac{1}{2i}$$

You got that right?

 

[Natura]

Yeah, I figured it out last time, but thanks for asking. Appreciate it. :)