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

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

Homework Statement


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.

Homework 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

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 :)
 

Answers and Replies

  • #2
329
34
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:
  • #3
HallsofIvy
Science Advisor
Homework Helper
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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).
 
  • #4
329
34
brmath clearly meant [itex]\frac{dz_1}{dt}+ \frac{dz_2}{dt}[/itex].

yes, that is indeed what brmath mean
 

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