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Homework Help: Complex analysis quick problem

  1. Apr 24, 2014 #1
    1. The problem statement, all variables and given/known data

    f(z) = u(x, y) + iv(x, y)
    where z ≡ x + iy. Let the fluid velocity be V = ∇u. If f(z) is analytic, show that
    df/dz = V_x − iV_y

    2. Relevant equations

    V_x = du/dx
    V_y = idu/dy

    The CR equations du/dx = dv/dy, du/dy = -dv/dx.

    3. The attempt at a solution

    I attempted to use this form for d/dz: d/dz = d/dx - i*d/dy.
    This gave me df/dz = 2(V_x - iV_y) though.
  2. jcsd
  3. Apr 24, 2014 #2


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    Gold Member

    Just to clarify before I try to work this out, V and v are different functions, correct?
  4. Apr 24, 2014 #3


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    Science Advisor

    I have no idea how you could possibly have got that "2". Please show your work.
  5. Apr 25, 2014 #4
    V = grad(u); so V_x = du/dx and V_y = du/dy

    df/dz = (d/dx - id/dy)(u+iv) = du/dx + i dv/dx - i du/dy + dv/dy

    Using the CR equations, du/dx = dv/dy and du/dy = -dv/dx.

    So df/dz = 2(du/dx) -2i(du/dy) = 2(V_x - iV_y).
  6. Apr 25, 2014 #5
    Yes, from the question it appears they are different functions.
  7. Apr 25, 2014 #6


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    That is not an acceptable choice for [itex]d/dz[/itex].

    If [itex]f = u + iv[/itex] is analytic in a neighbourhood of [itex]z[/itex], then by definition the following limit exists, and is independent of how [itex]h \in \mathbb{C}[/itex] approaches zero: [tex]
    \frac{df}{dz} = \lim_{h \to 0} \frac{f(z + h) - f(z)}{h} = \lim_{h \to 0} \frac{f(x + \mathrm{Re}(h), y + \mathrm{Im}(h)) - f(x,y)}{h}.
    [/tex] Thus you can take [itex]h \in \mathbb{R}[/itex] so that [tex]
    \frac{df}{dz} = \frac{\partial f}{\partial x} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x},[/tex] but instead you can take [itex]h \in i\mathbb{R}[/itex] so that [tex]
    \frac{df}{dz} = -i\frac{\partial f}{\partial y} = -i\frac{\partial u}{\partial y} + \frac{\partial v}{\partial y}[/tex] and since these are equal we must have [tex]
    \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad
    \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}[/tex] which are the Cauchy-Riemann equations.

    One also has [tex]
    \frac{df}{dz} = \lim_{h \to 0} \frac{f(x + \mathrm{Re}(h), y + \mathrm{Im}(h)) - f(x,y)}{h}\\ = \lim_{h \to 0} \frac{f(x + \mathrm{Re}(h), y + \mathrm{Im}(h)) - f(x, y + \mathrm{Im}(h))}{h} + \lim_{h \to 0}\frac{f(x, y + \mathrm{Im}(h)) - f(x,y)}{h} \\
    = \lim_{h \to 0} \frac{\mathrm{Re}(h)}h \frac{\partial f}{\partial x} + \lim_{h \to 0} \frac{\mathrm{Im}(h)}h \frac{\partial f}{\partial y}.[/tex]
  8. Apr 25, 2014 #7
    I looked over my notes again, and apparently d/dz = 1/2 (d/dx - i d/dy).

    From here, that would suggest [itex] \lim_{h \to 0} \frac{\mathrm{Re}(h)}h = \frac{1}{2} [/itex] and [itex]\lim_{h \to 0} \frac{\mathrm{Im}(h)}h= \frac{-i}{2}[/itex].

    I can't think of how to get there, though.
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