## 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

(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 :)

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