Author Topic: FE-P2  (Read 3722 times)

Victor Ivrii

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FE-P2
« on: December 18, 2018, 06:13:19 AM »
(a) Check that circles $\{z\colon |z|=r\}$ (with $0<r<1$) are mapped onto confocal ellipses
$\{w=u+iv\colon \frac{u^2}{a^2}+\frac{v^2}{b^2}=1\}$ with $a^2-b^2=1$ and find $a=a(r)$ and $b=b(r)$.

(b) Check that segments $\{z\colon z= e^{i\theta}r,\ r\in (-1,1)\}$  are mapped onto confocal hyperbolas
$\{w=u+iv\colon \frac{u^2}{A^2}-\frac{v^2}{B^2}=1\}$ with $A^2+B^2=1$ and find $A=A(\theta)$ and $B=B(\theta)$.

(c) Find to what domain this  function  maps the unit disk  $\mathbb{D}=\{z\colon |z|<1\}$.

(d) Draw both domains.

(e) Check if the correspondence is one-to-one.
« Last Edit: December 18, 2018, 06:15:16 AM by Victor Ivrii »

Wanying Zhang

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Re: FE-P2
« Reply #1 on: December 18, 2018, 11:32:06 AM »
Here's the solution.

Siying Li

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Re: FE-P2
« Reply #2 on: December 18, 2018, 01:00:53 PM »
Another way to do question (a)

(a)
$\frac{{\rm 1}}{{\rm 2}}\left(z+\frac{1}{z}\right)=\frac{1}{2}\left(e^{logz}+e^{-logz}\right)$
Let ${\rm a=}{\log \left(z\right)\ }$

Then
$\frac{{\rm 1}}{{\rm 2}}\left(z+\frac{1}{z}\right)=\frac{1}{2}\left(e^a+e^{-a}\right)={\rm cosh}(a)={\sin \left(\frac{\pi}{2}+ia\right)\ }={\sin \left(\frac{\pi}{2}\right)\ }{\cosh \left(a\right)\ }+icos\left(\frac{\pi}{2}\right){\rm sinh}?(a)$
Let ${\rm u=}{\sin \left(\frac{\pi}{2}\right)\ }{\cosh \left(a\right)\ }$, ${\rm v=}cos\left(\frac{\pi}{2}\right){\rm sinh}?(a)$

Then ${\sin \left(\frac{\pi}{2}\right)\ }{\rm =}\frac{u}{{\rm cosh}?(a)}\ ,{\cos \left(\frac{\pi}{2}\right)\ }=\frac{v}{{\rm sinh}?(a)}$ , and ${{\sin }^{{\rm 2}} \left(\frac{\pi}{2}\right)\ }+{{\cos }^{{\rm 2}} \left(\frac{\pi}{2}\right)\ }={\left(\frac{u}{{\cosh \left(a\right)\ }}\right)}^2+{\left(\frac{v}{{\sinh \left(a\right)\ }}\right)}^2 =1$

Then ${z:\left|z\right|=r}$ maps to
${w=u+iv:{\left(\frac{u}{{\cosh \left(a\right)\ }}\right)}^2+{\left(\frac{v}{{\sinh \left(a\right)\ }}\right)}^2 =1}$

Where ${\left({\cosh \left({\rm a}\right)\ }\right)}^{{\rm 2}}-{\left({\sinh \left(a\right)\ }\right)}^2=1$

Xianli Yu

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Re: FE-P2
« Reply #3 on: December 18, 2018, 02:30:21 PM »
Another way to solve a) through z=x+yi.
« Last Edit: December 18, 2018, 03:14:53 PM by Xianli Yu »

Yifei Wang

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Re: FE-P2
« Reply #4 on: December 18, 2018, 02:31:18 PM »
let $z = {re^{i\theta}}$
$f(Z) = f({re^{i\theta}}) = \frac{1}{2}({re^{i\theta}}+\frac{1}{r}{re^{-i\theta}})$
= $\frac{1}{2}{re^{i\theta}} + \frac{1}{2r}{re^{-i\theta}})$
= $\frac{r}{2}({cos\theta}+i{sin\theta})$ + $\frac{1}{2r}({cos(-\theta)}+i{sin(-\theta)})$

By odd and even function

= $(\frac{1}{2r}+\frac{r}{2}){cos\theta}$ +${isin\theta}(\frac{1}{2r}+\frac{r}{2})$
= $\frac{1+r^2}{2r}{cos\theta}$ + $i\frac{r^2-1}{2r}{sin\theta}$ = $U+iV$
${cos\theta} = \frac{U}{\frac{1+r^2}{2r}}$
${sin\theta} = \frac{U}{\frac{r^2-1}{2r}}$

for a:
$\frac{U}{\frac{1+r^2}{2r}}^2$ + $\frac{U}{\frac{r^2-1}{2r}}^2 = 1$
${\frac{1+r^2}{2r}}^2 - {\frac{r^2-1}{2r}}^2 = 1$
$a^2 = {\frac{1+r^2}{2r}}^2$
$b^2 ={\frac{r^2-1}{2r}}^2$

$---------------------------$
for b:
Similar to part a

= $\frac{1+r^2}{2r}{cos\theta}$ + $i\frac{r^2-1}{2r}{sin\theta}$ = $U+iV$
${\frac{1+r^2}{2r}} = \frac{U}{cos\theta}$
${\frac{r^2-1}{2r}} = \frac{V}{sin\theta}$

$\frac{U}{cos\theta}^2 - \frac{V}{sin\theta}^2 = 1$
$a^2 = {cos\theta}^2$
$b^2 = {sin\theta}^2$
$a^2 + b^2 = 1$

Victor Ivrii

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FE-P2 official
« Reply #5 on: December 20, 2018, 04:49:22 AM »
(a)--(b) If $|z|=r$ and $\arg{w}=t$, then $z=r e^{it}$ and
\begin{align*}
w= \frac{1}{2}\bigl(re^{it} +r^{-1}e^{-it}\bigr)&= \cosh(s)\cos(t)+i\sinh(s)\sin (t)
\end{align*}
with $s=\ln (r)$ (and $r=e^s$).

Then $w=u+iv$ with $\frac{u^2}{a^2}+\frac{v^2}{b^2}=1$, $a=\cos(s)$, $b=\sinh(s)$ and also
$\frac{u^2}{A^2}-\frac{v^2}{B^2}=1$,
with  $A=\cos (\theta)$, $B=\sin(\theta)$, $A^2+B^2=1$.

(c) Consider, what the circle $\{z\colon |z|=1\}$ is mapped to: $z=e^{it}$, then
$w =\frac{1}{2}(e^{it}+e^{-it})=\cos(t)$ and it runs a line segment $[-1,1]\subset\mathbb{R}$. So $f$ maps the unit disk onto compliment of it $\mathbb{C}^*\setminus [-1,1]$ (and $0$ is mapped to $\infty$).

(d) See attachment

(e) $w=\frac{1}{2}(z+z^{-1}) \implies z^2 -2wz +1=0$; it has two roots $z_1$ and $z_2$ s.t.
$z_1z_2=1$ and $z_1+z_2=2w$; exactly one of them for $w\notin [-1,1]$ is in unit disk, and the second one is in $\{z\colon |z|>1\}$.
« Last Edit: December 20, 2018, 04:51:23 AM by Victor Ivrii »