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Let $D=\{(x,y) : - \pi < x < \pi, y > 0 \}$. Find a subset $S$ of the boundary of $D$ such that $D \cup S$ is a fundamental domain for $\sin(z)$.

What if instead the domain was $D' = \{(x,y): - \pi/2 < x < \pi/2 \}$?

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Where is the problem from? –  Jonas Meyer Feb 14 '13 at 1:39
    
A question my prof asked that isn't in the textbook –  DJ_ Feb 14 '13 at 2:25
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By the way, what are you using as the definition of "fundamental domain"? –  Gerry Myerson Feb 20 '13 at 1:58
    
I "think" it is the largest set with f(z) is one-to-one. That is exactly how the question was presented to me, I wish I could clarify it more but that's all I have to work with. –  DJ_ Feb 20 '13 at 4:37
    
Your prof asked you the question. Are you not allowed to ask your prof what the question means? –  Jonas Meyer Feb 20 '13 at 5:04
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2 Answers 2

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Call two points $z$, $z'\in\Bbb C$ equivalent if $\sin z=\sin z'$. A fundamental domain ${\cal F}$ for this equivalence relation is a set of representatives which has a nice geometric description. In order to make ${\cal F}$ a closed (sometimes even compact) set one usually allows multiple representation along the boundary $\partial{\cal F}$. Such an ${\cal F}$ is by no means uniquely determined, as the versions $D$ and $D'$ proposed by the OP show.

Given that $\sin z={e^{iz}-e^{-iz}\over 2i}$ we introduce the auxiliary variables $e^{iz}=:w\ne0$, $\ e^{iz'}=:w'\ne0$. Then $\sin z=\sin z'$ amounts to $w-{1\over w}=w'-{1\over w'}$ or $$w-w'={1\over w}-{1\over w'}={w'-w\over w\>w'}\ .$$ The last equation is fulfilled iff at least one of the following holds:

(a) $\ w=w'$, which means $z'=z+2k\pi$ for some $k\in\Bbb Z$,

(b) $\ w\>w'=-1$, which means $z+z'=\pi+ 2k\pi$ for some $k\in\Bbb Z$.

From (a) it could seem that the strip $S:\ |{\rm Re}(z)|\leq\pi$ could serve as fundamental domain. But according to (b) the value of $\sin$ does not change under a reflection of $z$ in either of the points $-{\pi\over2}$ or ${\pi\over2}$. Drawing a figure we see that we may take as ${\cal F}$ the part $D$ of $S$ lying above the real axis, or the infinite strip $D':\ |{\rm Re}(z)|\leq{\pi\over2}$.

Now it seems that the OP wishes that his fundamental domain represents each equivalence class exactly once. This means that we have to throw away part of the boundary of $D$, resp. $D'$. In the case of $D$ I propose to retain the interval $\bigl[-{\pi\over2},{\pi\over2}\bigr]$ on the real axis and the half-line $x=\pi$, $\>y> 0$. I leave the corresponding surgery on $D'$ to the OP.

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where does $e^(iz)$ = w come from? –  DJ_ Feb 23 '13 at 23:49
    
I thought sinz = e^(iz) - e^(-iz) / 2i ?? –  DJ_ Feb 24 '13 at 1:11
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The function $z \mapsto \sin(z)$ maps the strip $S = \{z \in \mathbb{C} \mid -\frac{\pi}{2} < \operatorname{Re}(z) < \frac{\pi}{2} \textrm{ and } \operatorname{Im}(z) > 0 \}$ onto the upper half plane. Using the Schwarz reflection principle one can deduce how $\sin$ behaves on both of your domains.

In the first case, reflect the left half $\{z \in S \mid \operatorname{Re}(z) < 0 \}$ in the line $\operatorname{Re}(z) = -\frac{\pi}{2}$ and the right half $\{z \in S \mid \operatorname{Re}(z) > 0 \}$ in $\operatorname{Re}(z) = \frac{\pi}{2}$. This shows that $D$ is mapped onto $\{ z \in \mathbb{C} \mid z \not \in [0,1] \textrm{ and } i\, z \not \in [0, \infty) \}$. (Omitting a T-shaped set.)

In the second case reflect $S$ in the segment $[0,1]$ to show that $D'$ is mapped onto $\{ z \in \mathbb{C} \mid z \not \in (-\infty, -1] \textrm{ and } z \not \in [1, \infty) \}$.

Now study the behavior of $\sin$ on the boundary of both domains to see which segments you can add to fill up the missing parts in both cases (without hitting a point twice).

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What is the Schwarz reflection principle? Also when you mean it maps onto upper half plane do you entire upper plane of w=sinz? So there the function is discontinuous on entire upper imaginary axis? –  DJ_ Feb 20 '13 at 7:36
    
@DJ_ If you're not familiar with that you could use the facts that $\sin(z + \pi) = -\sin(z)$ and $\sin(\overline{z}) = \overline{\sin(z)}$ instead. I'm just used to the more geometric interpretation that the reflection principle provides. –  WimC Feb 20 '13 at 9:26
    
This explanation is too advanced for me, my textbook doesnt cover sine mapping. Is there anyway you can dumb this down. what does the set {z∈ℂ∣z∉[0,1] and iz∉[0,∞)} mean??? –  DJ_ Feb 24 '13 at 1:10
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