Toronto Math Forum
MAT334-2018F => MAT334--Tests => End of Semester Bonus--sample problem for FE => Topic started by: Victor Ivrii on November 27, 2018, 03:51:47 AM
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$\renewcommand{\Re}{\operatorname{Re}}
\renewcommand{\Im}{\operatorname{Im}}$
(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.
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(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 $
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(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 $