1. Limited time only! Sign up for a free 30min personal 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!

Showing that the complex functions are constant in t. Please Help

  1. Oct 5, 2013 #1
    Showing that the complex functions are constant in t. Please Help!! :)

    1. The problem statement, all variables and given/known data
    The complex position vectors of two parallel interacting equal fluid vortices moving with their axes of rotation always perpendicular to the z-plane are z1 and z2. The location of the vortices are then given by their respective coordinates x1,y 1 and x2, y2. It turns out to be useful to think of these coordinates as specifying a point in the complex plane, in which case the locations of the two fluids at a given time t are given by two complex numbers z1(t)=x1(t)+iy1(t) and z2(t)=x2(t)+iy2(t). The equations governing their motions are:
    [itex]\frac{dz*_{1}}{dt}[/itex]=[itex]\frac{-i}{z_{1}-z_{2}}[/itex]
    [itex]\frac{dz*_{2}}{dt}[/itex]=[itex]\frac{-i}{z_{2}-z_{1}}[/itex]

    (1) Deduce that (a) z1+z2, (b)z1-z2, (c)z12+z22 are all constant in time,(2) and hence descibe the motion geometrically.

    2. Relevant equations
    General z:
    z=x+iy
    Complex Conjugate:
    z*=[itex]\overline{z}[/itex]=x-iy
    Modules of Norm of z:
    |z|=[itex]\sqrt{(z*)z}[/itex]=[itex]\sqrt{x^{2}+y^{2}}[/itex]

    A constant function is define by:
    [itex]\frac{df(t)}{dt}[/itex]=0 OR f'(t)=0

    3. The attempt at a solution
    Wow. That was a "wordy" question. So I think I got 1.a but I'm requesting help on the rest of the question.
    (a) z1+z2
    Now I'm not sure if I should be showing whether
    [itex]\frac{d}{dt}[/itex][z1+z2]=0
    Or if
    [itex]\frac{d}{dt}[/itex][z1]+[itex]\frac{d}{dt}[/itex][z2]=0
    Either way for this question I get...
    [itex]\frac{d}{dt}[/itex][z1]+[itex]\frac{d}{dt}[/itex][z2]=[[itex]\frac{d}{dt}[/itex][z*1]+[itex]\frac{d}{dt}[/itex][z*2]]*=[[itex]\frac{-i}{z_{1}-z_{2}}[/itex]+[itex]\frac{-i}{z_{2}-z_{1}}[/itex]]*=[[itex]\frac{-i}{z_{1}-z_{2}}[/itex]+[itex]\frac{-i}{-(z_{1}-z_{2})}[/itex]]*=[[itex]\frac{-i}{z_{1}-z_{2}}[/itex]+[itex]\frac{i}{z_{1}-z_{2}}[/itex]]*=[[itex]\frac{-i+i}{z_{1}-z_{2}}[/itex]]*=[0]*=0

    (b) z1-z2
    This is where I start to get lost. I apply a similar method:
    [itex]\frac{d}{dt}[/itex][z1]-[itex]\frac{d}{dt}[/itex][z2]=[[itex]\frac{d}{dt}[/itex][z*1]-[itex]\frac{d}{dt}[/itex][z*2]]*=[[itex]\frac{-i}{z_{1}-z_{2}}[/itex]-[itex]\frac{-i}{z_2-z_1}[/itex]]*=[[itex]\frac{-i}{z_{1}-z_{2}}[/itex]+[itex]\frac{i}{z_2-z_1}[/itex]]*=[[itex]\frac{-i}{z_{1}-z_{2}}[/itex]+[itex]\frac{i}{-(z_1-z_2)}[/itex]]*=[[itex]\frac{-2i}{z_{1}-z_{2}}[/itex]]*
    And I don't know what to do to make it zero.

    (c) I have not attempted yet.

    (2) I don't know how I would represent them geometrically on an x,y plane.



    Thank you for the help, I really appreciate it and the work you guys do on here :)
     
  2. jcsd
  3. Oct 5, 2013 #2
    First: yes ##\frac {d(z_1 +z_2)}{dt} = \frac {z_1}{dt} + \frac {z_2}{dt}.## d/dt is always a linear operator.
    For part b you are practically there. You didn't apply your conjugate in the last step. (It had to be a minus sign somewhere).

    For the ##z^2## situation, you need to use the chain rule. df/dt = df/dz##\cdot## dz/dt.

    Re geometric representation, unfortunately I am geometry-blind -- I have no idea how things look; but I'm sure someone else will help.
     
    Last edited: Oct 5, 2013
  4. Oct 5, 2013 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    brmath clearly meant [itex]\frac{dz_1}{dt}+ \frac{dz_2}{dt}[/itex].

    For the "geometric representation", if [itex]z_1- z_2[/itex] and [itex]z_1+ z_2[/itex] is constant, then so is [itex](z_1- z_2)(z_1+ z_2)= z_1^2- z_2^2[/itex]. If, in addition, [itex]z_1^2+ z_2^2[/itex] is constant, so is [itex](z_1^2+ z_2^2)+ (z_1^2- z_2^2)= z_1^2[/itex] which, of course, means that [itex]z_2^2[/itex] is also always constant. This set will consist of four points: (a, b), (-a, b), (a, -b), and (-a, -b).
     
  5. Oct 5, 2013 #4
     
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: Showing that the complex functions are constant in t. Please Help
Loading...