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Question regarding maximum on a unit disc

  1. Jun 8, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the maximum of |ez| on the closed unit disc.

    2. Relevant equations

    |ez| is the modulus of ez
    z belongs to complex plane
    Maximum Madulus Theorem - Let G be a bounded open set in complex plane and suppose f is a continuous function on G closure which is analytic in G. Then max{|f(z)|: z in G closure} = max{|f(z)|: z in boundary of G}

    3. The attempt at a solution

    At first glance I was thinking about using Maximum Modulus Theorem by setting f(z) = ez and let G be a open unit disc. Clearly f(z) is analytic in G and is continuous on G closure (the closed unit disc), so I should be able to apply MMT to conclude that the modulus assumes maximum on the boundary.
    However I got confused when determining how do I describe |f(z)| on the boundary, in another word, how do I determine what value of z to use?

    Please let me know if my approach is right and perhaps give me some advice or hint.
    I appreciate all the helps you provide.
     
  2. jcsd
  3. Jun 8, 2010 #2

    lanedance

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    as on the boundary |z| = 1, how about starting by writing it in the form z = 1. e^(i theta)
     
  4. Jun 8, 2010 #3
    Since z lies on the unit circle, we can write it as: max{|f(z)|: z in boundary of G} = max{|f(z)|: z = exp([tex]i\theta[/tex]), [tex]\theta \in [-\pi,\pi)[/tex]}

    However, what do you know about |[tex]e^z[/tex]| when [tex] z = e^{i\theta}[/tex]?
     
  5. Jun 8, 2010 #4
    I'm not sure where this is going, in this case we will just have exp(exp([tex]i\theta[/tex]))?

    I also know that e^(i[tex]\theta[/tex]) = cos([tex]\theta[/tex]) + i*sin([tex]\theta[/tex]), but I don't know if this will help my solution or not.
     
  6. Jun 8, 2010 #5
    It does. Calculate explicitly the modulus of [tex]e^z[/tex] using the identity you provided above for [tex]e^{i\theta}[/tex].
     
  7. Jun 8, 2010 #6
    So now I have |ecos([tex]\theta[/tex])+isin([tex]\theta[/tex])|

    |ecos([tex]\theta[/tex])*eisin([tex]\theta[/tex])|
    = |ecos([tex]\theta[/tex])*(cos(sin([tex]\theta[/tex]))+isin(sin([tex]\theta[/tex])))|

    Now I don't know why I got stuck here....
     
  8. Jun 8, 2010 #7

    lanedance

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    not too sure what you did there in the last step... but you need to remember the complex conjugate in the magnitude... try that before you simplify
    [tex]|e^{ cos(\theta)}e^{i sin(\theta)}|^2 = (e^{ cos(\theta)}e^{i sin(\theta)})(e^{ cos(\theta)}e^{i sin(\theta)})^*[/tex]
     
  9. Jun 8, 2010 #8

    thanks for the tip.

    Well, I know that (ez)* = e[tex]z^*[/tex]

    So I can simply [tex]|e^{ cos(\theta)}e^{i sin(\theta)}|^2 = (e^{ cos(\theta)}e^{i sin(\theta)})(e^{ cos(\theta)}e^{i sin(\theta)})^*[/tex] to:

    [tex](e^{ cos(\theta)}e^{i sin(\theta)})(e^{ cos(\theta)}e^{-i sin(\theta)})[/tex]

    Then by combing the exponential functions with the same exponent, then I obtain:

    [tex](e^{ 2cos(\theta)})[/tex]

    So after taking the square root of the whole thing,

    [tex]|e^z| = e^ { cos(\theta)}[/tex]

    The max value of [tex]cos(\theta)[/tex] = 1

    So the maximum of [tex]|e^z| = e[/tex]?
     
  10. Jun 8, 2010 #9

    lanedance

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    that looks good to me
     
  11. Jun 8, 2010 #10
    Thanks.
     
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