Showing that the complex functions are constant in t.

[MelissaJL]

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:
$\frac{dz*_{1}}{dt}$=$\frac{-i}{z_{1}-z_{2}}$
$\frac{dz*_{2}}{dt}$=$\frac{-i}{z_{2}-z_{1}}$

(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*=$\overline{z}$=x-iy
Modules of Norm of z:
|z|=$\sqrt{(z*)z}$=$\sqrt{x^{2}+y^{2}}$

A constant function is define by:
$\frac{df(t)}{dt}$=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
$\frac{d}{dt}$[z₁+z₂]=0
Or if
$\frac{d}{dt}$[z₁]+$\frac{d}{dt}$[z₂]=0
Either way for this question I get...
$\frac{d}{dt}$[z₁]+$\frac{d}{dt}$[z₂]=[$\frac{d}{dt}$[z*₁]+$\frac{d}{dt}$[z*₂]]*=[$\frac{-i}{z_{1}-z_{2}}$+$\frac{-i}{z_{2}-z_{1}}$]*=[$\frac{-i}{z_{1}-z_{2}}$+$\frac{-i}{-(z_{1}-z_{2})}$]*=[$\frac{-i}{z_{1}-z_{2}}$+$\frac{i}{z_{1}-z_{2}}$]*=[$\frac{-i+i}{z_{1}-z_{2}}$]*=[0]*=0

(b) z₁-z₂
This is where I start to get lost. I apply a similar method:
$\frac{d}{dt}$[z₁]-$\frac{d}{dt}$[z₂]=[$\frac{d}{dt}$[z*₁]-$\frac{d}{dt}$[z*₂]]*=[$\frac{-i}{z_{1}-z_{2}}$-$\frac{-i}{z_2-z_1}$]*=[$\frac{-i}{z_{1}-z_{2}}$+$\frac{i}{z_2-z_1}$]*=[$\frac{-i}{z_{1}-z_{2}}$+$\frac{i}{-(z_1-z_2)}$]*=[$\frac{-2i}{z_{1}-z_{2}}$]*
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 :)

 

[brmath]

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.

 

[HallsofIvy]

brmath clearly meant $\frac{dz_1}{dt}+ \frac{dz_2}{dt}$.

For the "geometric representation", if $z_1- z_2$ and $z_1+ z_2$ is constant, then so is $(z_1- z_2)(z_1+ z_2)= z_1^2- z_2^2$. If, in addition, $z_1^2+ z_2^2$ is constant, so is $(z_1^2+ z_2^2)+ (z_1^2- z_2^2)= z_1^2$ which, of course, means that $z_2^2$ is also always constant. This set will consist of four points: (a, b), (-a, b), (a, -b), and (-a, -b).

 

[brmath]

brmath clearly meant $\frac{dz_1}{dt}+ \frac{dz_2}{dt}$.

yes, that is indeed what brmath mean