Showing that the complex functions are constant in t.

Thank you for catching my typo! I have corrected it.In summary, the complex functions z1+z2, z1-z2, and z1^2+z2^2 are all constant in time. This can be shown by taking the derivative with respect to time, applying the chain rule, and using the given equations governing their motions. Geometrically, this means that the locations of the two fluids at a given time t are represented by four points on the complex plane, (a, b), (-a, b), (a, -b), and (-a, -b).
  • #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 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 :)
 
Physics news on Phys.org
  • #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:
  • #3
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
HallsofIvy said:
brmath clearly meant [itex]\frac{dz_1}{dt}+ \frac{dz_2}{dt}[/itex].

yes, that is indeed what brmath mean
 

1. What does it mean for a complex function to be constant in t?

When a complex function is constant in t, it means that its value does not change as t varies. This can also be referred to as being time-independent, as the value of the function does not depend on the variable t.

2. How can you show that a complex function is constant in t?

To show that a complex function is constant in t, you can use the definition of a constant function, which states that the function has the same value at every point in its domain. This means that by substituting different values of t into the function, the result will remain the same.

3. Why is it important to show that a complex function is constant in t?

It is important to show that a complex function is constant in t because it allows us to better understand the behavior and properties of the function. It also simplifies the analysis of the function and makes it easier to solve equations involving the function.

4. What are some common techniques used to show that a complex function is constant in t?

Some common techniques used to show that a complex function is constant in t include using the Cauchy-Riemann equations, using the definition of a constant function, and using the identity theorem for analytic functions.

5. Can a complex function be constant in t and still be a non-constant function?

Yes, a complex function can be constant in t and still be a non-constant function. This can occur if the function is constant in t but varies with respect to other variables, such as x and y. In this case, the function would be considered constant in t but not in the overall domain.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
542
  • Topology and Analysis
Replies
2
Views
623
Replies
1
Views
596
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
13
Views
1K
Replies
1
Views
783
  • Calculus and Beyond Homework Help
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
746
  • Calculus and Beyond Homework Help
Replies
6
Views
520
  • Calculus and Beyond Homework Help
Replies
1
Views
522
Back
Top