1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    BiGyElLoWhAt

    User Avatar
    Gold Member

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

    HallsofIvy

    User Avatar
    Staff Emeritus
    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

    pasmith

    User Avatar
    Homework Helper

    fixed.

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Complex analysis quick problem
  1. Quick Analysis Problem (Replies: 4)

Loading...