1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cauchy-Riemann equations

  1. Nov 6, 2007 #1
    1. The problem statement, all variables and given/known data

    z + 1/z
    is it analytic?

    2. Relevant equations

    du/dx = dv/dy and du/dy = -dv/dx

    where f = u+iv

    3. The attempt at a solution

    I'm pretty sure I did this correctly, but I ran into an unexpected answer in a more complex problem using the same method and thought I'd ask about the easier question to make sure I'm doing this correctly.

    So my attempted solution was this:

    I expanded z to be x+iy, so z + 1/z = (x+iy) + (x+iy)^-1
    I found du/dx = 1 + -(x+iy)^-2
    dv/dy = i + -i(x+iy)^-2
    du/dy = 0
    -dv/dx = 0

    So, according to my solution, the function is not analytic. What I think could be wrong is my answer to dv/dy. The derivative of iy in terms of y should just be i is what I think but for some reason I'm second guessing that.

    Thanks.
     
  2. jcsd
  3. Nov 6, 2007 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    To do Cauchy-Riemann you have to express u and v as real functions. Take z+1/z with z=x+iy and try to express the real and imaginary parts in terms of x and y. No dangling i's are allowed. I would multiply 1/z by (x-iy)/(x-iy).
     
  4. Nov 6, 2007 #3
    Hm, so what you're saying is that you basically just look at the real and imaginary parts..

    So when I'm looking at u or v I should drop the i's out? I think I just re-phrased what you said..

    How about a slightly more complicated example just to clarify things a bit? Using the same given information:
    (3z-1)/(3-z).

    I think expanding it into it's x's and y's would be rather silly. So would multiplying it by (3+z)/(3+z) be the best first step? Then expanding?

    Thanks Dick.
     
  5. Nov 6, 2007 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You have to split it into x and y's if you want to use CR. I don't think a more complicated example will clarify things. Stick with the first one. z+1/z=(x+iy)+(x-iy)/((x+iy)*(x-iy)). See, I just followed my own advice. Now can you split that into a purely real function plus another purely real function times i? Those are your u and v. Hint: it IS analytic.
     
  6. Nov 6, 2007 #5
    When you do that, you get (x+iy) + (x-iy)/(x^2 + y^2)

    I'm not sure I understand how that helps me? I apologize if I seem slow here, but I'm just not quit seeing this.
     
  7. Nov 6, 2007 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    S'ok. That's a good reason to stick with the simple example. That's a good start. Can you split that up into a purely real function plus a purely real function times i? The denominator of the second part is real, so just split the numerator up.
     
  8. Nov 6, 2007 #7
    x + i(y) + x/(x^2 + y^2) - i(y/(x^2+y^2)) ?

    If that's correct, then what you're saying is you need to isolate i in these equations in order to successfully do the C-R equations?

    I think I just didn't quite understand the C-R equations, is the problem I was having.
     
  9. Nov 6, 2007 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Now you do. u=x+x/(x^2+y^2) and v=y-y/(x^2+y^2). Since then z+1/z=u(x,y)+v(x,y)*i. If you actually do the work of finding du/dx, du/dy, dv/dx and dv/dy (partial derivatives, all). You will find C-R works. Now you should be asking me how I know it works without doing all of the work that you just did.
     
  10. Nov 6, 2007 #9
    Alright, how do you know C-R works without doing all the work?

    EDIT: I also worked out the equations, and you were correct, the function is analytic. Not that I doubted you, but just saying I seemed to have come to an understanding. I really appreciate the help Dick.
     
    Last edited: Nov 7, 2007
  11. Nov 7, 2007 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You should do all of the work just to convince yourself it works. But I know z satisfies C-R and I know 1/z satisfies C-R. The former is easy, and the latter just a little harder. And C-R are linear equations. So I then know the sum will satisfy them. If f and g are analytic then f+g is analytic and 1/g and f*g and f/g are analytic. Aren't you glad we didn't start with (3z-1)/(3-z)? If you really HAVE to verify C-R, then you have to. Otherwise, there are different shortcuts to verifying a function is analytic.
     
  12. Nov 7, 2007 #11

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Good job again. I think at least once, everybody has work through something in messy detail that seems hard to believe and then realize it's true. I did.
     
  13. Nov 7, 2007 #12
    Haha, in terms of explaining something relatively simple in the terms of a more complex problem, yes I am happy. However, that is the problem I'm on now.

    What's getting me hung up is trying to isolate the real and imaginary parts. the 3-z turns into a trinomial if you substitute in x+yi. I tried multiplying by -x+yi, and I still have an i term in the result. I also tried 3+z, and that doesn't work either. I think I'm just missing a more fundamental concept of conjugation than anything.

    EDIT: Oops, I think I got it... hold on :D
    EDIT2: Yes, after reading the wiki article on conjugation I realized that I needed to include the 3. So multiplying by 3-x+yi resulted in zero i's afterwards :D
     
    Last edited: Nov 7, 2007
  14. Nov 7, 2007 #13

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Ha, great. If you have to use C-R, then you have to. Multiply numerator and denominator by the complex conjugate of 3+z=3+(x+iy) which is (and you knew this, right?) 3+(x-iy). All imaginaries will cancel in the denominator. Check this.
     
  15. Nov 7, 2007 #14
    Yes, that is what I found. I should have been less vague in my edits. This problem is heaps of fun so far. I think my professor is a sadist.
     
  16. Nov 7, 2007 #15
    So,

    Here is one that I think applies to your linearity short-cut you mentioned earlier. Because z+1/z is analytic, then 1/[(z+1/z)^2] must be as well. I'm going to attempt to work it out though.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Cauchy-Riemann equations
Loading...