- #1

MelissaJL

- 50

- 0

**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 z

_{1}and z

_{2}. The location of the vortices are then given by their respective coordinates x

_{1},y

_{1}and x

_{2}, y

_{2}. 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 z

_{1}(t)=x

_{1}(t)+iy

_{1}(t) and z

_{2}(t)=x

_{2}(t)+iy

_{2}(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) z

_{1}+z

_{2}, (b)z

_{1}-z

_{2}, (c)z

_{1}

^{2}+z

_{2}

^{2}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) z

_{1}+z

_{2}

Now I'm not sure if I should be showing whether

[itex]\frac{d}{dt}[/itex][z

_{1}+z

_{2}]=0

Or if

[itex]\frac{d}{dt}[/itex][z

_{1}]+[itex]\frac{d}{dt}[/itex][z

_{2}]=0

Either way for this question I get...

[itex]\frac{d}{dt}[/itex][z

_{1}]+[itex]\frac{d}{dt}[/itex][z

_{2}]=[[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) z

_{1}-z

_{2}

This is where I start to get lost. I apply a similar method:

[itex]\frac{d}{dt}[/itex][z

_{1}]-[itex]\frac{d}{dt}[/itex][z

_{2}]=[[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 :)