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Homework Help: 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

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    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

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    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

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    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

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    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

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    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

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    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

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    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.
     
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