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 Exercise [10.12] b 
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Post Exercise [10.12] b
Here's my solution:
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Last edited by vasco on 29 May 2010, 06:44, edited 1 time in total.

09 May 2010, 12:55
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Post Re: Exercise [10.12] b
Here's an updated file to correct a typo. In the second part it should have been \Delta z=i\Delta y and NOT \Delta z=\Delta y.
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10 May 2010, 05:40

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Post Re: Exercise [10.12] b
I am going to be picky.

I think the point of this exercise is to take the definition of the derivative as stated in the second equation in your proof, and then take definition of a holomorphic function as explained on page 140 and 141 to reach the result.

Page 140,1 says a holomorphic function is conformal and non-reflective, that is, small shapes are preserved in the mapping by the function, and they are not flipped overturned over. Small squares are mapped to approximately small squares, the approximation getting better as the squares get smaller. In the general case a holomorphic function combines a rotation with a uniform expansion or contraction of the small square. That scaling and turning is in other words a multiplication by a complex number. The delta x and delta y that approach zero in your proof are the sides of a small square that shrinks. Both are multiplied by the same complex number when they undergo a holomorphic transformation. Therefore the partial derivative of the function with respect to x and y are equal. From there the Cauchy-Riemann equations follow.


23 Nov 2010, 11:44
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Post Re: Exercise [10.12] b
I see that a proof is possible using the conformality argument as you say, but for me Penrose is asking for a proof based solely on the traditional definition of a derivative, which is equation 1 in my proof.
Exercise 10.12 says:
Quote:
Give a more direct derivation of the Cauchy-Riemann equations, from the definition of a derivative


24 Nov 2010, 08:09

Joined: 07 May 2009, 16:45
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Post Re: Exercise [10.12] b
My problem is not that you use this equation as the definition of the derivative, but in the sentence that follows

"Remember from the definition of derivative that lim delta z goes to 0 must be the same regardless of the manner in which delta z goes to 0."

This constraint is not really in the definition of the derivative is it? I would say it is the definition of a holomorphic function, the result of the function being conformal and non-reflective.

On page 194 Penrose says the partial derivative of a holomorphic function with respect to the complex conjugate of z equals zero. In this case the condition in your sentence is not true, but he is saying the derivative is zero, which is not the same as saying it is not defined.

Is Penrose cheating here?


26 Nov 2010, 02:47
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Post Re: Exercise [10.12] b
Quote:
Remember from the definition of derivative that lim delta z goes to 0 must be the same regardless of the manner in which delta z goes to 0.

Quote:
This constraint is not really in the definition of the derivative is it?
Oh yes it is!!

Look at any classical textbook on the calculus and you will find that the limit not only has to exist, but must be the same regardless of the direction in which you come at it.

This is fundamental.

In the theory of differentiation of functions of a real variable if you are differentiating a function f(x) at x=0, then the limit coming towards x=0 through negative values must be the same as that obtained by coming through positive values.

That is why y=|x| is not differentiable at x=0 because the limits mentioned above are 1 and -1.

This same idea is carried over into complex functions, so that if the limit is approached through real values or imaginary values or any other way, the same limit must result in all cases.

On page 194 Penrose is saying that if the function phi is to be holomorphic then it cannot be a function of z bar, but only of z, and so he sets the derivate of phi with respect to z bar to zero and obtains the CR equations.

My proof shows the same thing but only using the basic definition of derivative carried over from functions of a real variable.


26 Nov 2010, 08:18
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Post Re: Exercise [10.12] b
I remember you saying in one of your posts that you were reading Visual Complex Analysis by Needham.
If you look at Chapter 4 Part IV paragraph 2, you will see the same definition there in terms of the amplitwist.


27 Nov 2010, 11:26

Joined: 07 May 2009, 16:45
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Post Re: Exercise [10.12] b
I concede the point about the definition of a derivative. The part that causes a bit of confusion is the partial derivative.

Any function phi of z has to be a function of bar z, maybe not an analytic function, but it is a function.

Penrose is making phi a function of z and bar z, as opposed to one or the other alone, then treating the two as independent variables, which is confusing because they are not, and then taking the partial with respect to bar z.

So if phi is a holomorphic function of z, the derivative of phi ( z ) with respect to bar z is not defined, but taking the partial derivative of phi ( z , bar z ) with respect to bar z is defined and is 0.


29 Nov 2010, 21:00
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Post Re: Exercise [10.12] b
So, if I understand you correctly, you are not quibbling with my document any more but with what Penrose says in section 10.5 of RTR. Is that right?


01 Dec 2010, 07:47
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Post Re: Exercise [10.12] b
Dimbulb
I think I understand what Penrose is going on about with this function of z and z bar.
I will post a document as soon as I can with what I think it's all about.
Vasco


01 Dec 2010, 10:37
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Post Re: Exercise [10.12] b
Quote:
Any function phi of z has to be a function of bar z, maybe not an analytic function, but it is a function.

This is not a correct statement. If we choose for example \Phi=f(z)=z, then this is not a function of z bar.
You could of course say \Phi=f(z)=z=x+iy=(z+z bar)/2+(z-z bar)/2, and at first glance this looks like a function of z and z bar. However if you simplify it algebraically then it reduces to just z as you would expect, and if you calculate partial df by dz bar then it is identically zero.
You might just as well claim that y=x^2-xy+xy is a function of y.


11 Dec 2010, 14:39
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