Number of zeros of the complex function in the unit disk

Question

$ \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\Res}{Res} \DeclareMathOperator{\VP}{V.P.} \DeclareMathOperator{\e}{e} \DeclareMathOperator{\AC}{AC} \DeclareMathOperator{\BB}{B} \DeclareMathOperator{\RHO}{\rho} %\DeclareMathOperator{\CCC}{C^1} \DeclareMathOperator{\contt}{C} \DeclareMathOperator{\PCC}{PC} \DeclareMathOperator{\LL}{L} \DeclareMathOperator{\RE}{Re} \DeclareMathOperator{\IM}{Im} %\DeclareMathOperator{\dd}{d} \DeclareMathOperator{\VEC}{vec} \DeclareMathOperator{\EXP}{exp} \DeclareMathOperator{\COS}{cos} \DeclareMathOperator{\SIN}{sin} \newcommand{\ph}{\varphi} \newcommand{\inth}{\Int_{-h_k}^0} \newcommand{\teta}{\theta} \newcommand{\R}{\mathbb{R}} \renewcommand{\C}{\mathbb{C}} %\renewcommand{\L}{\mathbb{L}} \newcommand{\N}{\mathbb{N}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\<}{\leq} \renewcommand{\>}{\geq} \renewcommand{\d}{\mathrm{d}} \newcommand{\Int}{\int\limits} \newcommand{\Sum}{\sum\limits} \newcommand{\Sup}{\sup\limits} \newcommand{\eqdef}{\stackrel{\mathrm{def}}{=}} \renewcommand{\ae}{\stackrel{\mathrm{a.e.}}{=}} \newcommand{\mh}{\mathfrak{h}} \newcommand{\bh}{\mathbf h} \newcommand{\bN}{\mathbf N} \newcommand{\eps}{\varepsilon} \newcommand{\dt}{\mathrm{d}t} \newcommand{\dd}{\mathrm{d}} \newcommand{\x}{\textup{x}} \newcommand{\y}{\textup{y}} %\newcommand{\w}{\textup{w}} \newcommand{\al}{\alpha} \newcommand{\ti}{\times} \newcommand{\D}{\Delta} \newcommand{\del}{\delta} \newcommand{\F}{\mathbb{F}} \renewcommand{\O}{\Omega} \newcommand{\til}[1]{\widetilde{#1}} \newcommand{\g}{\gamma} \newcommand{\wD}{\widehat{\Delta}} \newcommand{\imp}{\implies} \newcommand{\ot}{\otimes} $ Consider the function $f(z)=10 \sin (z^{2019})-z.$ I'd like to find the number of zeros of $f$ in the unit disk $D=\left\{z\in\C:~|z|\<1\right\}.$

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.

I don't think you really need all those DeclareMathOperators and newcommands. — Arthur, May 17 '22 at 14:23
It looks true but not obvious that with real variables $x^2+y^2=1 \implies e^{2y}+e^{-2y} - 2 \cos(2x) \geq 2$. I don't see how the $\pm 4$ might help. — aschepler, May 17 '22 at 14:43
Where is the $2019$ power? Looks like it is $z^{2019}$ and then sine; is that correct? — Conrad, May 17 '22 at 15:56
@Arthur why it's bad if I write these commands but not use it? — Dmitry, May 18 '22 at 06:12
(It does make it a little more difficult to edit when you have to scroll past all those lines.) — Arthur, May 18 '22 at 06:56

score 1 · Answer 1 · answered May 18 '22 at 15:00

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!

Number of zeros of the complex function in the unit disk

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