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

Complex analysis - maximum/modulus principle

  1. May 3, 2006 #1
    Suppose that f is analytic on a domain D, which contains a simple closed curve lambda and the inside of lambda. If |f| is constant on lambda, then either f is constant or f has a zero inside lambda ...
    i am supposed to use maximum/modulus principle to prove it ...

    here is my take:
    if f is constant, i dont see a reason why |f| wouldnt be constant :)

    if f is not constant, then the max/min principle applies ...
    meaning |f| can not have any local max/min on D
    now i am lost at this point ...

    i would also like a little clarification on what f is constant on lambda means, because the way i see it .. .if lambda is a closed loop (say a circle), then how can f be always increasing /decreasing ...? maybe i am misinterpreting the problem
  2. jcsd
  3. May 3, 2006 #2


    User Avatar
    Science Advisor
    Homework Helper

    if your curve lambda is a(t), t from 0 to 1 then |f| constant on lambda means |f(a(t))|=K for some constant K and all t. Normally, |f(a(t))| can wobble all over the place as it winds around the loop, it can't be increasing on all of [0,1].

    if f is a constant on some set, then any function of f is a constant, specifically it's modulus.

    What happens if you have no zero? Consider 1/f.
  4. May 3, 2006 #3
    thats what i thought ....

    but i still have no insight about f having any zero inside lambda ....

    like i kinda see it visually ... like i know it makes sense ... but i got no clue how to "prove" it
  5. May 3, 2006 #4


    User Avatar
    Science Advisor
    Homework Helper

    If there's no zero of f, you can apply maximum modulus principle to both f and 1/f. What happens?
  6. May 3, 2006 #5
    if i apply max/min modulus to both f and 1/f, then it means both f and 1/f dont have a local max/min ... so what does it tell me?
    i'm sorry ... i dunno if it's lack of confidence or what .. but i still dont see it
  7. May 3, 2006 #6


    User Avatar
    Science Advisor
    Homework Helper

    If you have minimum modulus principle as well, then you can forget about 1/f.

    Apply min/max modulus to f then. What does this tell you about |f| on the interior compared to it's values on the boudnary? (remember |f| is a continuous function, lambda+interior is a closed and boundeed set)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook