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.