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I've found the similar question and can't understand why $$ |\sin z|\>\frac{\sqrt 2}{2}\quad\forall z=x+iy\in\C:~|z|=1.\tag{1} $$ Namely, why $\sqrt{(\e^{2y}-4+\e^{-2y})+(4-2\cos(2x))}\>\sqrt 2$? Lagrange multipliers method for function $g(x,y)=(\e^{2y}-4+\e^{-2y})+(4-2\cos(2x))$ gives the system which is really hard to solve.
Found another approach here but we need ($f(z)\ne 0$ if $|z|=1$) in order to use Argument Principle.
Note that in order to apply Rouche and conclude that there are precisely $2019$ zeroes (and using that $|z^{2019}|=1$), one needs to prove that $|\sin w| >1/10, |w|=1$ since then for $|z|=1$ one has $$|10\sin z^{2019}| > |10\sin z^{2019}-(10\sin z^{2019}-z)|=1$$
But if $w=x+iy, x^2+y^2=1$ one has $|\sin w|=(\sin^2 x+\sinh^2 y)^{1/2} \ge \max (|\sin x|, |\sinh y|)$ and using that for $|x| \le \pi/2, |\sin x| \ge 2|x|/\pi >4|x|/7$, the inequality follows for $|x|>7/40$; on the other hand $2|\sinh y| =e^{|y|}-e^{-|y|}\ge 1+|y|-1=|y|$ so $|\sinh y| \ge |y|/2$ hence the inequality holds for $|y|>1/5$. Since $x^2+y^2=1$ implies at least one the above inequalities $|x|>7/40, |y|>1/5$ holds we are done!