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Complex Analysis question

  1. Jan 21, 2014 #1
    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. jcsd
  3. Jan 22, 2014 #2
    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]?
     
  4. Jan 22, 2014 #3
    Is it asking for the Wirtinger derivative? If so, you're actually looking to compute $$\frac{\partial w}{\partial z}=\frac{1}{2}\left(\frac{\partial w}{\partial x}-i\frac{\partial w}{\partial y}\right).$$
     
  5. Jan 22, 2014 #4
    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.
     
  6. Jan 22, 2014 #5
    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.
     
  7. Jan 23, 2014 #6
    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?
     
  8. Jan 25, 2014 #7
    Yeah, I figured it out last time, but thanks for asking. Appreciate it. :)
     
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