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Math Help - Complex Functions

  1. #1
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    Complex Functions

    Hi everybody!
    I need to frove the following (even some of those will help):
    1. assume f(z) is an entire function [analytical on the whole plane], that holds for every n (natural) ==> if |z|=n^2 then |f(z)|<13. prove that f(z) is a constant function.
    2. assume f(z) is a Meromorphic function [its finite singular points are only poles] on C, that holds for every z (complex) ==> f(z)=f(2f(z)). find all the functions f(z) [general form].
    3. assume f(z) is periodical and analytical on a domain D that contains the point z=infinity. prove that f(z)=const on D.
    4. assume f(z) is analytical on a bounded domain D, and continuous on D* (with the "boundary"). assume also that f is not constant on D, but |f| is constant on the "boundary". prove that f has a zero on D.
    5. assume f(z) is analytical on a pierced surroundings of z=0 (sorry about the stammerer english, its not my mother tongue language...). prove that f(z) and h(z)=f(z+z^2) have the same "kind" of singular point on z=0, and if it is a zero or a pole, then the order is also the same.

    TNX A LOT !!!
    RedFox (:
    Last edited by Red_Fox; September 18th 2006 at 02:26 PM.
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  2. #2
    Super Member Rebesques's Avatar
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    assume f(z) is an entire function [analytical on the whole plane], that holds for every n (natural) ==> if |z|=n^2 then |f(z)|<13. prove that f(z) is a constant function.

    If f is not constant, then {\rm lim}_{|z|\rightarrow\infty}|f|=+\infty. But f\bigg|_{\{z:|z|=n^2\}} remains bounded, a contradiction.


    assume f(z) is a Meromorphic function [its finite singular points are only poles] on C, that holds for every z (complex) ==> f(z)=f(2f(z)). find all the functions f(z) [general form].

    Consider the function near a pole: f(z)=(z-a)^mg(z)+h(z), where m<0 and h(a), g(a) are finite. Write f=f(2f) and show that poles change order.


    ssume f(z) is periodical and analytical on a domain D that contains the point z=infinity. prove that f(z)=const on D.
    Again, we must have {\rm lim}_{|z|\rightarrow\infty}|f|=+\infty. But f is periodical, and thus remains bounded.


    assume f(z) is analytical on a bounded domain D, and continuous on D* (with the "boundary"). assume also that f is not constant on D, but |f| is constant on the "boundary". prove that f has a zero on D.
    Suppose not. Let |f|\bigg|_{\partial D}=c. Consider a fixed a\in D^o and the function g(z)=\frac{c(z-a)}{(c+1){\rm sup}_{w\in \partial D}|w-a|}, which has a zero at z=a. Now |g(z)|\bigg|_{z\in \partial D}\leq c/(c+1)<c= |f(z)|\bigg|_{z\in \partial D}. Apply Rouche's theorem on the curve \partial D.


    prove that f(z) and h(z)=f(z+z^2) have the same "kind" of singular point on z=0, and if it is a zero or a pole, then the order is also the same.

    Near zero, let  f(z)=z^mg(z)+z^kv(z), \ k<0. Note that


    h(z)=(z+z^2)^mg(z+z^2)+(z+z^2)^kv(z+z^2)=z^mg(z+z^  2)+z^kv(z+z^2)+w(z),

    where w is analytic at 0.
    Last edited by Rebesques; August 19th 2007 at 08:05 AM. Reason: bad digestion
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