### Author Topic: FE Sample--Problem 2  (Read 3149 times)

#### Victor Ivrii ##### FE Sample--Problem 2
« on: November 27, 2018, 03:51:47 AM »
$\renewcommand{\Re}{\operatorname{Re}} \renewcommand{\Im}{\operatorname{Im}}$
(a) Consider map
$$z\mapsto w=f(z):=\cos(z).$$
(b) Check that lines $\{z\colon \Im z =q \}$  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 $a=a(q)$ and $b=b(q)$.

(c) Check that lines $\{z\colon \Re z =p \}$ 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 $A=A(p)$ and $B=B(p)$.

(d) Find to what domain this  function  maps the strip  $\mathbb{D}=\{z\colon 0<\Re p < \pi\}$.

(e) Draw both domains.

(f) Check if the correspondence is one-to-one.
« Last Edit: November 27, 2018, 08:17:45 AM by Victor Ivrii »

#### hanyu Qi

• Jr. Member
•  • Posts: 12
• Karma: 4 ##### Re: FE Sample--Problem 2
« Reply #1 on: December 01, 2018, 09:17:43 PM »
(a)

$\cos z = (\cos x)(\cosh y) - i(\sin x)(\sinh y)$

$u(x,y) = (\cos x)(\cosh y) v(x,y) = -(\sin x)(\sinh y)$

$when y =q$

$u(x,q) = (\cos x)(\cosh q) = a(\cos x)$

$v(x,q) = -(\sin x)(\sinh q) = b(\sin x)$

$\frac{u^2}{a^2} = (cosx)^2$         $\frac{v^2}{b^2} = (sinx)^2$

$\frac{u^2}{a^2} + \frac{v^2}{b^2} = (cosx)^2 + (sinx)^2 = 1$

So, lines {z: Imz = q} are mapped onto confocal ellipses {w=u+iv: $\frac{u^2}{a^2} + \frac{v^2}{b^2} = 1$} with $a^2 - b^2 = 1$ since $(coshq)^2 - (-sinhq)^2 = 1$

$a = \cosh q$
$b = -\sinh q$
« Last Edit: December 01, 2018, 09:24:32 PM by Alex Qi »

#### hanyu Qi

• Jr. Member
•  • Posts: 12
• Karma: 4 ##### Re: FE Sample--Problem 2
« Reply #2 on: December 01, 2018, 09:33:23 PM »
(b)

when $x = p$

$u(b,y) = \cos b \cosh y = A \cosh y$

$v(b,y) = -\sin b \sinh y = B \sinh y$

$\frac{u}{A} = \cosh y$

$\frac{v}{B} = \sinh y$

$\frac{u^2}{A^2} - \frac{v^2}{B^2} = (\cosh y)^2 - (\sinh y)^2 = 1$ with $A^2 + B^2 = (\cos b)^2 + (\sin b)^2 = 1$

$A = \cos b$

$B = -\sin b$