# Zeroes of $\sin(z)-z^2$

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.

• @Thomas Shelby Thanks for the edit. – WhoKnowsWho Jun 14 '19 at 6:16

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.)
• 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. – Eric Wofsey Jun 14 '19 at 14:45