FE-P4

[Victor Ivrii]

(a) Find the Möbius transformation (fractional-linear transformation) $f(z)$ mapping the unit disk $\{z\colon |z|<1\}$ onto itself, such that $f(0)=\frac{1}{2}$ and $f(1)=-1$.

(b) Find the fixed points of $f$ (points s.t. $f(z)=z$)

(c) Find the stretch ($|f'(z)|$) and the rotation angle ($\arg(f'(z))$) of $f$ at $z$.

[hz12]

$f\left(z\right)=\lambda \cdot \frac{a-z}{1-\overline{a}z}$
$\left|\lambda \right|=1,\ so\ \lambda =e^{it}$
f(0) = 1/2, so $\lambda \cdot \frac{a}{1}=\frac{1}{2}\ $, $\lambda a=\frac{1}{2}$ , $\left|\lambda \right|\cdot \left|a\right|=\frac{1}{2}$ , so $\left|a\right|=\frac{1}{2}\ \ $
We let a = $\frac{1}{2}\cdot e^{i\theta }$, wo can have $\lambda a=\ \frac{1}{2}\cdot e^{i(\theta +t)}$
Since         
        f(1) = $\lambda \cdot \frac{a-1}{1-\overline{a}}=-1$
$\lambda \left(a-1\right)=\overline{a}-1 $
$\lambda a-\lambda =\ \frac{1}{2}e^{-i\theta }-1$
$\frac{1}{2}-e^{it}=\frac{1}{2}e^{-i\theta }-1$
$\frac{1}{2}-\left[cost+isint\right]=\frac{1}{2}\left[cos\theta -isin\theta \right]-1$
From the above, we can have 1, cost =$\ -\frac{1}{2}cos\theta +\frac{3}{2}$   2, sint = $\frac{1}{2}sin\theta \ $, 3, $\theta +t=2k\pi $

Solving these equation, can have $t=\ \theta =0,\ so\ a=\frac{1}{2\ }\ ,\ \ \lambda =1\ $
$f\left(z\right)=\frac{\frac{1}{2}-z}{1-\frac{1}{2}z}=\frac{1-2z}{2-z}$

b, 

let $f\left(z\right)=$ $\frac{1-2z}{2-z}=z$,  we can have ${(z-2)}^2=3,\ z1=\sqrt{3}+2,\ \ z2=-\sqrt{3}+2\ $
$(f')$=$\frac{-2\left(2-z\right)-(1-2z)(-1)}{{(z-2)}^2}$= $\frac{-3}{{(z-2)}^2}$
$|f'|$=$\left|\frac{-3}{{(z-2)}^2}\right|$=$\frac{3}{{(z-2)}^2}$
                                         $arg(f')$ = arg($\frac{-3}{{(z-2)}^2}$) = arg(-1) - arg($\frac{{(z-2)}^2}{3}$)
                                                                              =$\ \ \pi -\ \mathrm{arg(}\frac{{(z-2)}^2}{3}\mathrm{)\ }$

[Siying Li]

(a) 

Let ${\rm f}\left({\rm z}\right){\rm =}\lambda \frac{a-z}{1-\overline{a}z}$ , $\left|\lambda \right|{\rm =1}$
${\rm where\ }\lambda {\rm =1}{{\rm e}}^{{\rm it}},   {\rm a=}{{\rm re}}^{{\rm i}\theta }$
Then
${\rm f}\left(0\right){\rm =}\lambda \frac{a-0}{1-\overline{a}0}=\lambda a=\frac{1}{2}\ \ \Rightarrow a=\frac{1}{2}e^{i \theta}$
${\rm f}\left({\rm 1}\right){\rm =}{{\rm e}}^{{\rm it}}\frac{\frac{1}{2}e^{i \theta}-1}{1-\frac{1}{2}e^{-i \theta}}\ \Rightarrow \frac{1}{2}-e^{it}=-1+\frac{1}{2}e^{-i \theta}\Rightarrow t= \theta =0$
Thus $\lambda {\rm =1,\ a=}\frac{{\rm 1}}{{\rm 2}}$

Then ${\rm f}\left({\rm z}\right){\rm =}\lambda \frac{a-z}{1-\overline{a}z}=\frac{\frac{1}{2}-z}{1-\frac{1}{2}z}=\frac{1-2z}{2-z}$

(b)
${\rm f}\left({\rm z}\right)=\frac{1-2z}{2-z}=z\ \Rightarrow z^2-4z+1=0\Rightarrow z=2\pm \ \sqrt{3}$
Then the fixed points are $z=2\pm \ \sqrt{3}$

(c)
${\rm f}\left({\rm z}\right){\rm =}\frac{1-2z}{2-z}$
Then
${{\rm f}}^{{\rm '}}\left({\rm z}\right){\rm =}\frac{-2\left(2-z\right)+\left(1-2z\right)}{{\left(2-z\right)}^2}=\frac{-3}{{\left(2-z\right)}^2}$
Then
${{\rm |f}}^{{\rm '}}\left({\rm z}\right)|=\left|\frac{-3}{{\left(2-z\right)}^2}\right|=\frac{3}{{\left(2-z\right)}^2}$
${\arg  \left(f\left(z\right)\right)\ }={\arg  \left(\frac{-3}{(2-z)(2-z)}\right)\ }={\arg \left(-1\right)} - {\arg \frac{{\left(2-z\right)}^2}{3}}={\pi} - {\arg \frac{{\left(2-z\right)}^2}{3}}$

[Victor Ivrii]

(a) We consider inverse transform $z=g(w)$, since we know that it maps $\frac{1}{2}$ into $0$. It must be
\begin{gather*}
g(w)=\lambda \frac{w-a}{1- w\bar{a}}\\
\text{with $|\lambda|=1$ and $a=\frac{1}{2}$:}\\
g(w)=\lambda \frac{2w-1}{2- w}.
\end{gather*}
Since $g(-1)=1$ we have $1=\lambda \frac{-2-1}{2+1}\implies \lambda=-1 $ and therefore
$$
g(w)= - \frac{2w-1}{2- w}=z \implies  (2w-1)=-z(2-w)\implies w= \frac{2z-1}{z-2}=: f(z).  $$

(b)
$$
\frac{2z-1}{z-2}=z\implies 2z-1=z^2-2z\implies z^2-4z +1=0\implies z=2\pm \sqrt{3}.$$
These are "black" points on the picture.

(c) Then $f' = \frac{2(z-2)-(2z-1)}{(z-2)^2}=-3 (z-2)^{-2}$; $f'(0)= -3/4$ and therefore stretch is $|-3/4|= 3/4$ and rotation is $\arg(-3/4)= \pi$.