Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Analytic Complex Functions

  1. Sep 27, 2010 #1


    User Avatar
    Science Advisor
    Gold Member

    1. The problem statement, all variables and given/known data

    The following are pieces of analytic functions of the form f(z)=u(x,y)+iv(x,y). Find the missing pieces.

    a) [tex]u(x,y)=x^2-y^2+x+2[/tex]

    Find v, and f.

    b) [tex]v(x,y)=\sqrt{\frac{1}{2}(\sqrt{x^2+y^2}-x)}[/tex]

    Find u, and f.

    c) [tex]f(z)=tan^{-1}(z)[/tex]

    Find u, v.

    2. Relevant equations

    3. The attempt at a solution

    Ok, I have no idea what to even start with...I don't understand why if u is given, then v is somehow constricted... (or vice versa). What's to say that v can't be any random function? I get that if f is given, then obviously u and v are given (although even the third part completely eludes me).

    Someone help me start this problem please?

    BTW, the professor has posted solutions online, but I don't understand them at all, so I was hoping you guys could help.
  2. jcsd
  3. Sep 27, 2010 #2
    Do you know what a harmonic conjugate is ? You need to know what it is to solve your problem.

    A function is analytic on a domain iff u is the harmonic conjugate of v.
  4. Sep 27, 2010 #3


    User Avatar
    Science Advisor
    Gold Member

    Uh...I know what a complex conjugate is...I don't think I've ever heard of a harmonic conjugate...
  5. Sep 27, 2010 #4
    Well open up your book or look it up online. If you still need help after that then post again.
  6. Sep 27, 2010 #5


    User Avatar
    Science Advisor
    Gold Member

    The professor said that the books were only recommended and not mandatory...so I thought that his lecture notes would cover everything...apparently they do not? >__> Ok wikipedia here I come...
  7. Sep 27, 2010 #6
    Go read Wikipedia on Complex Analisys.
    In the first 15 rows or so there is a strong hint for you, which will bring you to another page. There you have the solution.
  8. Sep 27, 2010 #7
    Well you could also use the cauchy-riemann equations to find v when given u.
  9. Sep 27, 2010 #8

    What if you read a book that states that "book are mandatory and professors are not ?"

    Find a good book, or google everything, please.
  10. Sep 27, 2010 #9


    User Avatar
    Science Advisor
    Gold Member

    Ok, no need to be hostile...o_O

    I didn't get a book and it's now too late to buy a book for this problem haha. I do own a book on this stuff, but it's inaccessible to me at the moment. I will try to get a book soon.
  11. Sep 27, 2010 #10
    No intention to be hostile.... :)
  12. Sep 27, 2010 #11
    Haha...I like you and your humor :)

    If you know the cauchy riemann equations, simply equate the derievatives and integrate to solve for u or v.
  13. Sep 27, 2010 #12


    User Avatar
    Science Advisor
    Gold Member

    Ok, I get how they will help me find v from u or u from v. But what about the last one? Any hints on that? I can't simply split off the real and imaginary parts of that function in any method that I know of...
  14. Sep 27, 2010 #13
    For the last one artan (z) is defined in terms of the logarithm and you may be able to split the fuction into real and imaginary parts but I am not sure if it will work out.

    You could always differential arctan(z) and write down u and v for the derievative and then integrate back but there might be some issues with points where the functions are not defined.
  15. Sep 28, 2010 #14
    I still don't see any solution... :)


    [tex]u_x= 2x+1[/tex]
    [tex]u_y= -2y[/tex]

    [tex]v_y = u_x = 2x+1[/tex]
    [tex]v_x = -u_y= 2y[/tex]

    mmmmmm.... feels like [tex]v(x,y) = 2xy+y[/tex]
  16. Sep 28, 2010 #15


    User Avatar
    Science Advisor
    Gold Member

    Yea I got part 1 and 2...(sort of cheated on 2...looked at the solutions to see the hint of transforming to polar coordinates haha)

    Thanks for the help. =)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook