1. Find the general form of bilinear

mapping that maps the upper halfplane $\displaystyle H^+=\{z : Imz >0\}$

to yourself.

2. Prove that for all $\displaystyle R: R>0$, $\displaystyle n \in N$ (large enough),

all zeros of polynom

$\displaystyle p_n(z)=\frac{1}{n!}z^n + ... + \frac{1}{2!}z^2 + 1$

are in $\displaystyle \{z \in \mathbb{C}: |z| > R\}$

Coment: For 2. $\displaystyle p_n -> e^z$, and... ?

New question:

$\displaystyle H^+ = \{z : Im z>0\}. $

$\displaystyle f(z)=z^5. $

Now, $\displaystyle f(H^+) = \{ z= r e^{i \theta} : r >0, 0 < \theta < 5\pi\} = \mathbb{C} \ \{x \leq 0\}$

or $\displaystyle f(H^+) = \{z= r e^{i\theta} : r >0, 0 < \theta < \pi\}$ ?

Is needed $\displaystyle mod 2\pi$ ?