Results 1 to 2 of 2

Math Help - Complex analysis

  1. #1
    Member
    Joined
    May 2008
    Posts
    171

    Complex analysis

    Hey guys.

    I have this problem, I need to show that it's true and I don't have a clue.
    I tried to do like alpha = x+yi but it got me nowhere, any ideas?

    Thanks.
    Attached Thumbnails Attached Thumbnails Complex analysis-1.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by asi123 View Post
    Hey guys.

    I have this problem, I need to show that it's true and I don't have a clue.
    I tried to do like alpha = x+yi but it got me nowhere, any ideas?

    Thanks.
    For |\alpha| < 1, \alpha \in \mathbb{C} define f(z) = \frac{z-\alpha}{1 - \overline{\alpha} z}.

    It is defined for |z|\leq 1, this is because |1 - \overline{\alpha}z| \geq \bigg||1| - |\overline{\alpha}||z|\bigg| \geq 1 - |\alpha| > 0.

    Therefore, on the closed disk |z|\leq 1 the function f is a continous function that is analytic on the open disk |z|<1. By the maximum modulos theorem theorem the largest value that |f| attains happens on the boundary of the disk i.e. on |z|=1.

    We know that |f|^2 = f ~ \overline{f}, therefore, if |z|= 1 we have:

    |f(z)|^2 = \frac{z-\alpha}{1-\overline{\alpha}z} \cdot \overline{ \frac{z-\alpha}{1-\overline{\alpha} z} } = \frac{z-\alpha}{1-\overline{\alpha} z} \cdot \frac{\overline{z}-\overline{\alpha}}{1-\alpha \overline{z}} = 1.

    Therefore, |f| \leq 1 for all |z|\leq 1 and furthermore |f| = 1 if and only if |z|=1.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: October 4th 2011, 05:30 AM
  2. Replies: 6
    Last Post: September 13th 2011, 07:16 AM
  3. Replies: 1
    Last Post: October 2nd 2010, 01:54 PM
  4. Replies: 12
    Last Post: June 2nd 2010, 02:30 PM
  5. Replies: 1
    Last Post: March 3rd 2008, 07:17 AM

Search Tags


/mathhelpforum @mathhelpforum