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Partial Derivatives for Functions f(z) of a Complex Variable.

  1. Jul 23, 2010 #1
    Hi, Everyone:

    I was never clear n this point: given that z is a single complex variable,
    how/why does it make sense to talk about z having partial derivatives.?

    I mean, if we are given, say, f(x,y); R<sup>2</sup> -->R<sup>n</sup>
    then it makes sense to talk about f<sub>x</sub> and f<sub>y</sub>, since
    x and y are different variables. But , in f(z), z is a single variable, so there are no
    additional variables to refer to, to meaningfully talk about partial derivatives.

    Is it the case that a function of a complex variable z is also a function of two complex
    variables.?. If not, is there a formal/theoretical argument to support this use.?
  2. jcsd
  3. Jul 23, 2010 #2
    If we were to use the natural identification [tex]\mathbf{R}^2 \simeq \mathbf{C}[/tex], then you can write a complex valued function on [tex]\mathbf{R}^2[/tex] via:

    [tex]f(x,y) = f\left( \frac{z+\bar z}{2}, \frac{z-\bar z}{2\mathrm{i}}\right) \equiv F(z,\bar z)[/tex]

    The notion of complex differentiability is determined by the Cauchy-Riemann equations: [tex] \partial F/\partial \bar z =0[/tex].
  4. Jul 23, 2010 #3
    Anthony's argument supports the idea that analytic functions are true functions of a complex variable as opposed to simply complex functions of two real variables.

    As to your very first question, I'm not sure what the problem is. We can write a complex function of a complex variable as f(x + iy) = u(x,y) + iv(x,y), where u and v are real functions. Partial derivatives then work exactly as they do in real analysis...
  5. Jul 23, 2010 #4
    But you do say that f(z) is a function of two real variables, through:

    f(z)=u(x,y)+iv(x,y) .?

    Only then does it make sense to take partials f_x and f_y.

    But if f is a function of z as well as a function of x,y, is there
    some functional dependence between z and x and z and y.?

    My point is that if an analytic function depends only on a single
    complex variable, then the argument z has only 1 "part" , so that
    a partial derivative would not make sense, just like the partial
    derivative of a function of a single real variable would not make sense.
    (unlike a function f(x,y) , whose argument has two "parts" )

    I thought that, re what Anthony said, that we may be using implicitly,
    the diffeomorphism between C and R^2 : x+iy -->(x,y) .

    But this last (maybe combined with my lack of sleep )seems confusing,
    since there are functions f: R^2-->R that are differentiable , while those
    same functions f are not differentiable as functions from C-->R (specifically,
    all differentiable functions that don't satisfy Cauchy-Riemann ) ; I thought
    that if manifolds M,N were diffeomorphic, that meant that every function f:M-->R
    is differentiable iff f:N-->R is also differentiable.

    Hope I am not too far of.
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