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How would you find the image of ${ x+yi : -\pi < x < \pi, y > 0 }$ under $\sin(z)$? I see that $\text{Re } \sin(x+yi) = \sin x \cosh y$ and $\text{Im } \sin(x+yi) = \cos x \sinh y$, but I'm not sure what to do next since $\sinh$ and $\cosh$ are unbounded.

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2 Answers 2

up vote 4 down vote accepted

What shape do you get when you fix y and consider all values of x? Then see how the shape depends on y. Of course, you could try the reverse as well (fix x and vary y first), but my way, you don't have to deal with the "y > 0" condition until the end.

In fact, draw a picture showing the images of the lines x=constant and y=constant. You get two families of curves that are orthogonal to one another (since sin(z) is holomorphic and thus generally preserves angles). I promise that this will be very enlightening.

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The image is the whole complex plane minus the interval $[-1,1]$ (on the real axis) and the negative imaginary axis.

Here is a figure illustrating the mapping. (The color-coding is explained at the top of that page.)

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Hans, those pics look awesome. May I know what you used to plot these? –  J. M. Sep 29 '10 at 0:42
    
@J. M.: For that page I used GIMP with the MathMap plugin. See mai.liu.se/~halun/complex, especially the section "Technical stuff". –  Hans Lundmark Sep 29 '10 at 6:34

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