FE Sample--Problem 2

[Victor Ivrii]

$\renewcommand{\Re}{\operatorname{Re}}
  \renewcommand{\Im}{\operatorname{Im}}$
(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.

[hanyu Qi]

(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 $

[hanyu Qi]

(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 $