# Complex Functions

• Sep 18th 2006, 11:56 AM
Red_Fox
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 (:
• Aug 19th 2007, 08:03 AM
Rebesques
Quote:

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.

Quote:

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.

Quote:

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.

Quote:

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). Apply Rouche's theorem on the curve $\partial D$.

Quote:

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.