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I want to describe the image of the strip {${z \in \mathbb C |-1/2 \leq x \leq 1/2}$ and $y \geq 1$} under the map $f(z)=e^{2\pi iz}$.

My attempt, $e^{2\pi iz}=e^{-2\pi y}(cos2\pi x+isin2\pi x)$. Since $y \geq 1$ so the modulus of this number lies in $(0,e^{-2 \pi}]$ and since $|x|\leq1$ so we have $-1 \leq cos2\pi x \leq1$ and $-1 \leq sin2\pi x \leq1$. So I get the description of real and imaginary parts. What is this thing geometrically? Please help

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The main thing is that the argument of $e^{2\pi iz}$ as you calculated, is $2\pi x$, and that the point $\cos2\pi x+i\sin 2\pi x$ is on the unit circle, at exactly $2\pi x$ angle (from the right wing of $x$-axis).

Now $x=-1/2..1/2$ hence the argument (the angle closed with the right wing of $x$-axis) is going from $-\pi$ to $\pi$, thus describing the whole circle. And the radius of the circle varies from $0$ to $e^{-2\pi}$. So, it's a disk around the origo, without the origo itself.

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  • $\begingroup$ without the origin? and contains the boundary? $\endgroup$ – Mathematician Feb 25 '13 at 2:12
  • $\begingroup$ Well, $y\in [1,\infty)$, this is mapped to the radius $e^{-2\pi y}$, which is then indeed in $(0,e^{-2\pi}]$. So, radius $0$ is excluded, but yes, contains the boundary, exactly. $\endgroup$ – Berci Feb 25 '13 at 2:15
  • $\begingroup$ "Origo" is Latin for "origin"? $\endgroup$ – Jonas Meyer Feb 25 '13 at 2:22
  • $\begingroup$ yes, i didn't know that but it was obvious from the context $\endgroup$ – Mathematician Feb 25 '13 at 2:24

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