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.
Best Answer
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 any polynomial; you just have to differentiate enough times to make it away.)