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Natura
#1
Jan21-14, 05:08 PM
P: 4
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))' = [itex]\frac{∂u(x,y)}{∂x}[/itex] +i[itex]\frac{∂v(x,y)}{∂x}[/itex]

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

[itex]\frac{∂w(z)}{∂z}[/itex] = [itex]\frac{∂u(x,y)}{∂x}[/itex][itex]\frac{∂x}{∂z}[/itex] +[itex]\frac{∂u(x,y)}{∂y}[/itex][itex]\frac{∂y}{∂z}[/itex] + i([itex]\frac{∂v(x,y)}{∂x}[/itex][itex]\frac{∂x}{∂z}[/itex] + [itex]\frac{∂v(x,y)}{∂y}[/itex][itex]\frac{∂y}{∂z}[/itex])

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)
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jackmell
#2
Jan22-14, 07:32 AM
P: 1,666
Quote Quote by Natura View Post
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

[tex]\frac{dw}{dx}[/tex]

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 [itex]\frac{dx}{dz}[/itex] and [itex]\frac{dy}{dz}[/itex]?
Mandelbroth
#3
Jan22-14, 09:00 AM
Mandelbroth's Avatar
P: 615
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
#4
Jan22-14, 02:56 PM
P: 4
Complex Analysis question

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
[itex]\frac{∂x}{∂z}[/itex] = [itex]\frac{∂z}{∂z}[/itex] = 1

Similarly for y we get
[itex]\frac{∂y}{∂z}[/itex] = -i


Then using the Cauchy-Riemann relations to eliminate all of the y derivatives and substituting the above results for [itex]\frac{∂x}{∂z}[/itex] and [itex]\frac{∂y}{∂z}[/itex] I get that
[itex]\frac{∂w}{∂z}[/itex] = 2*[itex]\frac{∂w}{∂x}[/itex]


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
#5
Jan22-14, 03:26 PM
P: 4
Nevermind, I can see that my expressions for [itex]\frac{∂x}{∂z}[/itex] and [itex]\frac{∂y}{∂z}[/itex] are wrong and are off by a factor of (1/2) ... Thanks again.
jackmell
#6
Jan23-14, 11:48 AM
P: 1,666
Quote Quote by Natura View Post
Nevermind, I can see that my expressions for [itex]\frac{∂x}{∂z}[/itex] and [itex]\frac{∂y}{∂z}[/itex] are wrong and are off by a factor of (1/2) ... Thanks again.
Natura, let me make sure you understand this ok?

We have [itex]w=f(z)=u(x,y)+iv(x,y)[/itex]

and:

[tex]x=\frac{z+\overline{z}}{2}[/tex]
[tex]y=\frac{z-\overline{z}}{2i}[/tex]

so that:

[tex]\frac{dx}{dz}=1/2[/tex]
[tex]\frac{dy}{dz}=\frac{1}{2i}[/tex]

You got that right?
Natura
#7
Jan25-14, 06:24 PM
P: 4
Yeah, I figured it out last time, but thanks for asking. Appreciate it. :)


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