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Show that $\sin(z) = z$ has infinitely many solutions in complex numbers. Little Picard theorem should help, but using big Picard theorem is undesirable. Thanks a lot!

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    $\begingroup$ why downvoting? I'm new here.. May be style is bad. But say it please! Why just downvoting? $\endgroup$
    – sbeliakov
    May 11 '13 at 17:32
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    $\begingroup$ Welcome to math.SE! We expect you to show your thoughts and ideas, even say "I don't know where to start" would of been good, but just stating the problem, without context or effort shown, will get you downvoted. You can edit your question using the edit button $\endgroup$
    – Belgi
    May 11 '13 at 17:38
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    $\begingroup$ thanks Belgi. I'll keep it in mind $\endgroup$
    – sbeliakov
    May 11 '13 at 18:26
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Using a non trivial result by Picard: $\sin(z) - z$ has an essential singularity at $\infty$. Therefore it attains all values infinitely many times with at most one exception. Since $$\sin(z + 2\pi) - (z+2\pi) = \sin(z) - z - 2\pi$$ one can shift the image over multiples of $2\pi$. By Picard's theorem there cannot be any value (including $0$) that is only attained a finite number of times since only a single exception is allowed.

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  • $\begingroup$ @user17762 Yes and also highly non-trivial unfortunately. $\endgroup$
    – WimC
    May 11 '13 at 18:12
  • $\begingroup$ @WimC Surely these thoughts solve the problem. Thank you a lot! But is there any solution using more trivial results? For example little Picard theorem is ok. $\endgroup$
    – sbeliakov
    May 11 '13 at 18:39
  • $\begingroup$ @maggot092 Little Picard suffices to show surjectivity but I don't see a way that shows that there must be infinitely many roots. Probably something rather pedestrian like this is possible here as well. $\endgroup$
    – WimC
    May 11 '13 at 18:48
  • $\begingroup$ Oh, Picard theorem accepted :) Thanks a lot $\endgroup$
    – sbeliakov
    Jun 2 '13 at 11:44
  • $\begingroup$ @maggot092 My second answer below might be interesting (I think it is a charming approach). It is more elementary and also gives estimates of where the fixed points are located. $\endgroup$
    – WimC
    Jun 2 '13 at 12:17
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The function $\sin z - z$ is entire and of order $1$, so if it had finitely-many zeros then the Hadamard factorization theorem would say that

$$ \sin z - z = e^{az+b} p(z), $$

where $p(z)$ is some polynomial and $a$ and $b$ are some complex numbers.

This can't be true. We know that $\sin z - z \sim e^{-iz}$ as $z\to +i\infty$ and $\sin z - z \sim e^{iz}$ as $z \to -i\infty$ for $\operatorname{Re} z = 0$, and when taken together these properties imply the absurd conclusions that, not only is $p(z)$ constant, but $\operatorname{Im} a < 0$ and $\operatorname{Im} a > 0$.

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  • $\begingroup$ I learned this technique from this answer by ybungalobill. $\endgroup$ May 12 '13 at 6:03
  • $\begingroup$ Sorry if I comment now, I don't understand why if the zeroes are finitely many, then Hadamard factorization is in the form $e^{g (z)} p (z)$. Thanks a lot $\endgroup$
    – Dorian
    Feb 25 '20 at 8:49
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    $\begingroup$ @Dorian If the product in the factorization theorem were an infinite product, the function would have infinitely many zeros. $\endgroup$ Feb 27 '20 at 14:30
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Let $k \in \mathbb{Z}$, $y \in \mathbb{R}$, $y>0$ and consider the solid rectangle $$S = \{z \in \mathbb{C} \mid 0 \leq \operatorname{Im}(z) \leq y \textrm{ and } (2k - \tfrac{1}{2})\pi \leq \operatorname{Re}(z) \leq (2k+\tfrac{1}{2})\pi \}. $$

The function $\sin(z)$ is single valued on $S$ and maps onto to the upper half of a solid ellipse with foci $\pm 1$. The lower, left and right sides of $S$ map onto the real interval and the top side onto the elliptic arc. (The following image may help to visualize this.)

image of <span class=$\sin(z)$" />

On the top side (where $\operatorname{Im}(z) = y$) the inequality $|\sin(z)| \geq \sinh(y)$ holds. Choose $y>0$ such that $$\sinh(y) \geq \max_{z\in S}|z|$$ (note that $S$ itself depends on $y$ but $\sinh$ grows fast enough). Then it follows that $$S \subset \sin(S).$$ If $\sin^{-1}\!: \sin(S) \to S$ is the inverse this can be written as $$S = \sin^{-1}(\sin(S)) \subset \sin(S).$$

By Brouwer's fixed point theorem $\sin^{-1}$ must have a fixed point in $S$, which will also be a fixed point of $\sin$. By symmetry of $\sin$ there is also a fixed point in $\overline{S}$, $-S$ and $-\overline{S}$. Varying $k$ shows that $\sin$ has infinitely many fixed points.

There are also relevant complex analytic fixed point results (see here and here for example). To name some consequences: there is at most one fixed point in the interior of $S$ and if there is one then iterating $\sin^{-1}$ on any starting point will converge to the fixed point. (There is the technical difficulty that $\sin^{-1}$ does not map $\sin(S)$ into the interior of $S$. Note for example that for $k=0$ there is no fixed point in the interior of $S$!)

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