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I am studying for my prelims exam. I stumbled upon the following question.

Show that there are infinitely many zeroes of $\sin(z)-z^2$ in the complex plane.

Had it just been $f(z)=\sin(z)-z$, one can observe that $f(z+2\pi)=f(z)-2\pi$. One can, therefore, see that if $f$ takes zeros only finitely often, then $f$ must also take the value $-2\pi$ only finitely often. Picard’s theorem, therefore, tells us that $f$ is a polynomial, which is absurd.

This direct approach does not seem to work for my question. So, I tried using Rouche’s theorem. For that I take $g(z)=\sin(z).$ I know the zeroes of $\sin(z)$, my idea was to find a region containing $n$ zeroes of $sin(z)$ and show that on the boundary we have $$|f-g|<|f|+|g|.$$

I could not choose the suitable region such that the above relation holds on the boundary. I am not sure if it will work or not. Of course, if it works it will prove something stronger, namely, we will in a way have a handle on the location of zeroes.

Any hint would be appreciated. Moreover, my guess is that if $p(z)$ is any polynomial then $f(z)=\sin(z)-p(z)$ will have infinitely many zeroes in the complex plane. I would like to see an argument for this case. I am trying to use Rouche’s theorem but I am not able to make any progress. Also, if there is an alternate approach (which avoids Rouche’s theorem), it will also be much appreciated.

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  • $\begingroup$ @Thomas Shelby Thanks for the edit. $\endgroup$ – WhoKnowsWho Jun 14 '19 at 6:16
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Questions like this can typically be answered using the Hadamard factorization theorem. The function $f(z)=\sin z-z^2$ has finite order, so if it has finitely many zeroes then its Hadamard factorization has the form $P(z)e^{Q(z)}$ for some polynomials $P$ and $Q$. So, we would have the equation $$P(z)e^{Q(z)}=\sin z-z^2$$ for all $z\in\mathbb{C}$. Differentiating three times we get $$R(z)e^{Q(z)}=-\cos z$$ for some other polynomial $R$. But this is impossible, since the left side has finitely many zeroes and the right side has infinitely many zeroes.

(The same argument applies with $z^2$ replaced by ay polynomial; you just have to differentiate enough times to make it away.)

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  • $\begingroup$ Thank you very much. I didn’t even think of Hadamard Factorizarion theorem. Do you think an argument like the one I have described can be pushed through? $\endgroup$ – WhoKnowsWho Jun 14 '19 at 14:24
  • $\begingroup$ I don't see a way to make it work. The problem is that $\sin z$ is small near the real axis but exponentially big far away, whereas $z^2$ is intermediate in size everywhere as you move away from the origin. $\endgroup$ – Eric Wofsey Jun 14 '19 at 14:45

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