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Homework Help: Another complex question

  1. Mar 28, 2010 #1
    1. The problem statement, all variables and given/known data
    f(z)=u(r,theta)+iy(r,theta)... where x=rcos(theta) and y=rsin(theta), use chain rule to show that [tex]\partial[/tex]u/[tex]\partial[/tex]r=1/r([tex]\partial[/tex]v/[tex]\partial[/tex][tex]\theta[/tex]) and [tex]\partial[/tex]v/[tex]\partial[/tex]r=-1/r([tex]\partial[/tex]u/[tex]\partial[/tex][tex]\theta[/tex]) are equivelent to the cauchy riemann equations.

    2. Relevant equations
    CR equations: [tex]\partial[/tex]u/[tex]\partial[/tex]x=[tex]\partial[/tex]v/[tex]\partial[/tex]y and [tex]\partial[/tex]u/[tex]\partial[/tex]y=-[tex]\partial[/tex]v/[tex]\partial[/tex]x

    3. The attempt at a solution
    Ok the im unsure by how i am meant to use the chain rule here? and instead of typing out the dirvative im goin to just write i.e d/dx..

    i did, dz/dr=dz/dx*dx/dr=1(cos(theta) and dz/dtheta=dz/dy*dy/dtheta=rcos(theta)... but that doesnt make sense... its the same as the provided equations without the 1/r.. but if i do the CR equations i get, du/dx=1 and du/dy=0????
  2. jcsd
  3. Mar 28, 2010 #2
    Your use of the chain rule is incorrect here.

    Since we're talking about functions of several variables, the chain rule must consider all the possible derivatives. So for example:

    [tex]\frac{\partial u}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial u}{\partial r}+\frac{\partial \theta}{\partial x}\frac{\partial u}{\partial \theta}[/tex]
  4. Mar 28, 2010 #3
    ok then, but can i ask how do u do dr/dx and dtheta/dx???? since r and theta are part of x?
  5. Mar 28, 2010 #4
    do i rearrange?? coz r=x/cos(theta)?? then diffrentiate with respect to x?
  6. Mar 28, 2010 #5
    ok so if i do the chain rule that elib gave... i get (1/cos(theta))*cos(theta)+-1/(r*sqrt(1-x^2/r^2)*-rsin(theta).... but when i do dv/dy i kind of get similar answer, but instead of the x^2 its y^2, and the last term is cos(theta) and not sin(theta)
  7. Mar 28, 2010 #6
    ok for the x^2 and y^2 values, i subbed in there respecful values x=rcos(theta) y=rsin(theta) then i get...
    1+(sin([tex]\theta[/tex])/sqrt(1-cos^2([tex]\theta[/tex])/r^2))=1+(cos([tex]\theta[/tex])/sqrt(1-sin^2([tex]\theta[/tex])/r^2)), which clearly does no equal???
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