I want to solve this question: find the images of lines $y = k = \mbox{constant}$ under the mapping $w =\cos(z).$ I know that $w=\cos(z)=\cos(x)\cosh(y)-i\sin(x)\sinh(y)$ so $u=\cos(x)\cosh(y)$ and $v=-\sin(x)\sinh(y)$ but $y=k$ is only a line and how can i map it? I don't know what values should be given to x. I put k in $u=\cos(x) \cosh(k)$ and in $v=-\sin(x)\sinh(k).$ Lastly, I ask what should I do?
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1$x$ traverses $\mathbb{R}$. What sort of curve do you get from $x\mapsto a\cos x - ib\sin x$? – Daniel Fischer Dec 20 '14 at 21:20
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Related: http://math.stackexchange.com/questions/54713/complex-cosine-and-sine – Hans Lundmark Dec 20 '14 at 21:23
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i can solve it when y and x got boundaries like this $1<y<2$ and $2<x<3$ but i dont know how to solve this question – Milad Qasemi Dec 20 '14 at 21:26
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you have $\cos x = {u \cosh k}, \sin x = {-v \over \sinh k.}$ use the fact $\cos ^2 x + \sin ^2 x = 1$ that gives you bunch ellipses for level curves. – abel Dec 20 '14 at 21:52
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As Daniel Fisher suggested you should think of $\cosh(k)$ and $\sinh(k)$ as constants. For example $a$ en $b$.
The resulting function $x\mapsto a\cos x-i b\sin x$ should ring a bell.
Perhaps it's more clear if explicitly written as a coordinate.
$$w(t) = (a\cos t, -b\sin t) \qquad t\in [0,2\pi]$$
(since the function is $2\pi$ periodic)
If $a=b$ then you'd find a circle. When they are not equal you find an ellips. Perhaps you remember the cartians form. Just set $x=a\cos t$ and $y=-b\sin t$, then
$$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$$
Here's a picture of the map. Bigger values of $k$ make the ellipse more circle-like. (since $\cosh k$ would be closer to $\sinh k$).
dietervdf
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