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#1
Jan2114, 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) 


#2
Jan2214, 07:32 AM

P: 1,666

[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]? 


#3
Jan2214, 09:00 AM

P: 615

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



#4
Jan2214, 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 CauchyRiemann 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. 


#5
Jan2214, 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.



#6
Jan2314, 11:48 AM

P: 1,666

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? 


#7
Jan2514, 06:24 PM

P: 4

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



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