Uniform phase-multiplier bound for the square-root dispersion law

Given: Reals $k_0>0$, $0<\delta <1$, $z\in \mathbb{R}$, and $s\in [0,{\delta }^2]$.

The real numbers form a complete ordered field, and each nonnegative real has a unique nonnegative square root; a positive radicand has a positive root. The reals are complete Square roots exist: a unique $\sqrt{a}\ge 0$ with $(\sqrt{a}{)}^2=a$; the positives are $\{x^2:x\mathrm{\ne }0\}$

For real $t$, $e^{it}=\cos t+i\sin t$ and $e^{i\pi }=-1$. The parity and Pythagorean identities give $\cos (-t)=\cos t$, $\sin (-t)=-\sin t$, and ${\cos }^{2}t+{\sin }^{2}t=1$; with the complex-modulus definition this gives $\mathrm{\mid }{e}^{it}\mathrm{\mid }=1$ and $\mathrm{\mid }\cos t\mathrm{\mid },\mathrm{\mid }\sin t\mathrm{\mid }\le 1$. For $w=a+ib$, conjugation is $\bar{w}=a-ib$ and $\mathrm{\mid }w\mathrm{\mid }=\sqrt{a^2+b^2}$. $\exp (x+iy)=e^x(\cos y+i\sin y)$, $\mathrm{\mid }\exp (x+iy)\mathrm{\mid }=e^x$, and $e^{i\pi }+1=0$ Parity and the Pythagorean identity for sine and cosine Real and imaginary parts, complex conjugation, and modulus

The complex exponential satisfies $e^{a+b}=e^a{e}^b$ for complex $a,b$. The complex exponential by its power series $\exp (z+w)=\exp z\exp w$, and the complex exponential extends the real exponential

The functions $\sin$ and $\cos$ are differentiable on $\mathbb{R}$, with derivatives $\cos$ and $-\sin$, and $\sin 0=0$. Every real $t$ is a limit point of $\mathbb{R}$: for $\epsilon >0$, $t+\epsilon \mathrm{/}2$ is a distinct point within distance $\epsilon$. Thus both functions are continuous on $\mathbb{R}$. The derivatives of sine and cosine are cosine and minus sine A function differentiable at $c$ is continuous at $c$ Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ The reals are complete

If a real function is continuous on a closed nondegenerate interval and differentiable on its interior, its endpoint difference equals its derivative at some interior point times the interval length. The mean value theorem, as the case $g(x)=x$ of Cauchy's: for $f$ continuous on $[a,b]$ with $a<b$ and differentiable on $(a,b)$ there is $c\in (a,b)$ with $f(b)-f(a)=f^{\mathrm{\prime }}(c)(b-a)$

Complex conjugation is an involution, $\overline{zw}=\bar{z}\bar{w}$, and the complex modulus is multiplicative: $\mathrm{\mid }zw\mathrm{\mid }=\mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }w\mathrm{\mid }$. Conjugation is an involutive real-field automorphism, $z\bar{z}=\mathrm{\mid }z{\mathrm{\mid }}^2$, and modulus is definite, multiplicative, and subadditive