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I would like to know what the mapping properties of the complex sine and cosine are. To start with one can say that $\sin(z)$ and $\cos(z)$ are conformal where their derivatives are nonzero, which means $\sin(z)$ preserves angles on $\mathbb{C}$ without $\frac{\pi}{2}+k\pi$ and $\cos(z)$ preserves angles on $\mathbb{C}$ without $\pi k$ for $\cos(z)$, with $k \in \mathbb{Z}$. What else can we say about these mappings? Thx.

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They're conformal and satisfy trigonometric identities, they're awesome, they're cool, they're practical... isn't that enough? If you're just curious, I guess it could stop there, but are you trying to do something with those and try to imagine some nasty trick to get rid of your problem? If that is so, you should post your motivations here too. – Patrick Da Silva Jul 31 '11 at 7:13
Well, if someone could post some pictures that would be great. I am trying to represent how they transform the plane. – user786 Jul 31 '11 at 7:29
@Arjang, Thx but that doesn't help. For instance what does $\cos$ do to a disc of radius R or to a strip of height $2\pi$? – user786 Jul 31 '11 at 7:50
for the disc of radius r we have $r e^{i\theta} $ shove that into $\cos x = \frac{e^{ix}+e^{-ix}}{2}$ and behold the hidocity! just use a polar quation for what ever shape you need ( line, circle, etc..) and shove it into the Cos or Sin's Euler form, but most insightingthing is to find what shaps would remain mostly the same, e.g. circles to rotated circles or lines to streteched lines etc. , When I did the course (with Trance Tao) I found this exercise a waste of time, rather than focusing on what remains invariant, we just got bunch of distortions that didn't show anything enlightening. – Arjang Jul 31 '11 at 8:18
up vote 7 down vote accepted

Lets consider the rectangle $-\pi/2<x<\pi/2$, $-3<y<3$ in the complex $z$-plane:

Original plane

Its image under the map $\zeta=\sin(z)$ will look like

Image of sin(z)

As you can see, the lines $y=\text{constant}$ are deformed to ellipses which approach circles for $|y|\to\infty$ and the lines $x=\text{constant}$ are mapped to hyperbolae. The angles between the two sets of lines are preserved. If you extend the rectangle in the $y$-direction you'll eventually cover the whole $\zeta$-plane (except for branch cuts from $\zeta=-\infty$ to $\zeta=-1$ and from $\zeta=1$ to $\zeta=\infty$).

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If you want to see pictorially how these maps work it is enough to look at the function $z\mapsto \cosh z=(e^z +e^{-z})/2$, because $\cos z=\cosh(i z)$ and $\sin z=\cos\bigl(z-{\pi\over2}\bigr)$.

Now $\cosh$ is the composition of the two maps $\exp:\ z\mapsto w:=e^z$ and $j:\ w\mapsto \zeta:={1\over2}(w+w^{-1})$. Therefore we draw figures for these maps separately.

Presumably you have a mental image of $\exp$: It is periodic with period $2\pi i$, and it maps the strip $|y|\leq\pi$ of the $z$-plane essentially 1:1 onto the $w$-plane whereby lines $y=$const. are mapped onto rays emanating from the origin and the segments $x=$const are mapped onto concentric circles of radius $e^x$.

As for the so-called Joukowski function $j$ a detailed study shows that it maps circles $|w|=$const. onto confocal ellipses with foci $\pm1$ and rays $\arg w=$const. onto arcs of hyperbolae with the same foci. Since $j(w)=j(w^{-1})$ the map $j$ is essentially 2:1: The interior and the exterior of the unit disk in the $w$-plane each map onto the full $\zeta$-plane whereby the unit circle itself is mapped back and forth onto the segment $[{-1},1]$.

The map $j$ is explained in detail and with figures, e.g., in Peter Henrici: Applied and computational complex analysis, Vol. 1, pp. 294–298.

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