Is it true that $|\arcsin z | \le |\frac {\pi z} {2} |$?

Question

Decide whether the following ie true or false $$\lvert\arcsin z \rvert \le \left\lvert \frac {\pi z} {2} \right\rvert $$ whenever $z\in\Bbb C$ .

$\arcsin z =-i \text{Log } (\sqrt{1-z^2}+iz)$,

$\text{Log }z=\log|z|+i\arg z,\arg z\in(-\pi,\pi] $

The problem is related to the series $\sum_{n=1}^{\infty}\arcsin(n^{-2}z) $ converges normally in the whole complex plane.

Answer 1

Proof: We split into four cases:

1) $z = x \in (1, +\infty)$: From 4.23.20 in [1], we have $$|\arcsin x| = \sqrt{\frac{1}{4}\pi^2 + \ln^2 (\sqrt{x^2-1} + x)}.\tag{1}$$ It suffices to prove that $$\tfrac{1}{2}\pi\sqrt{x^2-1} - \ln (\sqrt{x^2-1} + x) \ge 0.\tag{2}$$ With the substitution $x = \frac{1+u^2}{2u}$ for $u > 1$, the inequality above becomes $$f(u) = \tfrac{1}{2}\pi \frac{u^2-1}{2u} - \ln u \ge 0, \quad \forall u > 1.\tag{3}$$ We have $f'(u) = \frac{\pi (1 + u^2) - 4u}{4u^2} \ge \frac{\pi \cdot 2u - 4u}{4u^2} > 0$. Also, $f(1) = 0$. Thus, $f(u) \ge 0$ for $u > 1$. The inequality is true.

2) $z = x\in (-\infty, -1)$: From 4.23.21 in [1], we have $$|\arcsin x| = \sqrt{\frac{1}{4}\pi^2 + \ln^2 (\sqrt{x^2-1} - x)}.\tag{4}$$ From Case 1), the inequality is true.

3) $z = x \in [-1, 1]$: It suffices to prove that $$g(u) = \tfrac{1}{2}\pi u - \arcsin u \ge 0, \quad \forall u \in [0, 1].$$ We have $g'(u) = \tfrac{1}{2}\pi - \frac{1}{\sqrt{1-u^2}}$. Denote $u_0 = \frac{\sqrt{\pi^2 -4}}{\pi}$. We know that $g(u)$ is strictly increasing on $[0, u_0)$, and strictly decreasing on $(u_0, 1]$. Also, $g(0) = g(1) = 0$. Thus, $g(u) \ge 0$ on $[0, 1]$. The inequality is true.

4) $z = x + y\mathrm{i}$ with $y \ne 0$: From 4.23.34 in [1], we have $$|\arcsin z| = \sqrt{\arcsin^2 \beta + \ln^2 (\sqrt{\alpha^2 - 1} + \alpha) }\tag{5}$$ where \begin{align} \alpha &= \tfrac{1}{2}\sqrt{(x+1)^2 + y^2} + \tfrac{1}{2}\sqrt{(x-1)^2 + y^2}, \tag{6}\\ \beta &= \tfrac{1}{2}\sqrt{(x+1)^2 + y^2} - \tfrac{1}{2}\sqrt{(x-1)^2 + y^2}. \tag{7} \end{align} It suffices to prove that $$\tfrac{1}{4}\pi^2 (x^2 + y^2) \ge \arcsin^2 \beta + \ln^2 (\sqrt{\alpha^2 - 1} + \alpha). \tag{8}$$

Clearly, we only need to prove the case when $x\ge 0$ and $y > 0$. We have $$x\ge 0, \ y > 0 \quad \Longleftrightarrow \quad \alpha > 1, \ 0 \le \beta < 1. \tag{9}$$ Proof: “$\Longrightarrow$” part is easy. “$\Longleftarrow$” part: Indeed, from (6), (7) and $\alpha > 1, \ 0 \le \beta < 1$, we uniquely obtain $x = \alpha \beta$ and $y = \sqrt{(\alpha^2 - 1) (1-\beta^2)}$.

Also, it is easy to prove that $x^2+y^2 = \alpha^2 + \beta^2 - 1$. Thus, it suffices to prove that for $\alpha > 1$ and $0\le \beta < 1$, $$\tfrac{1}{4}\pi^2 (\alpha^2 + \beta^2 - 1) \ge \arcsin^2 \beta + \ln^2 (\sqrt{\alpha^2 - 1} + \alpha).\tag{10}$$ With the substitutions $\alpha = \frac{1+u^2}{2u}$ and $\beta = \sin v$ for $u > 1$ and $v \in [0, \frac{1}{2}\pi)$, the inequality above becomes $$\tfrac{1}{4}\pi^2 \Big(\frac{(1+u^2)^2}{4u^2} + \sin^2 v - 1\Big) \ge v^2 + \ln^2 u, \quad \forall u > 1, \ v \in [0, \tfrac{1}{2}\pi).\tag{11}$$ It is easy to prove that $\frac{\pi}{2} \sin v \ge v$ for $v \in [0, \frac{1}{2}\pi)$. Thus, it suffices to prove that $$\tfrac{1}{4}\pi^2 \Big(\frac{(1+u^2)^2}{4u^2} - 1\Big) \ge \ln^2 u, \quad \forall u > 1, \tag{12}$$ or $$\frac{\pi(u^2-1)}{4u} \ge \ln u, \quad \forall u > 1. \tag{13}$$ This has been proved in Case 1) (see (3)).

We are done.

Reference

[1] https://dlmf.nist.gov/4.23

Answer 2

Note $\arcsin(z) = \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{ z^{2n+1}}{2n+1}$. So we have that $\lvert\arcsin z \rvert \le \arcsin\lvert z \rvert$.

Further, we have that $\arcsin\lvert z \rvert$ is convex. Since $\arcsin\lvert z \rvert$ is declared on $0 \le \lvert z \rvert \le 1 $ and $\arcsin 0 = 0$, we have that $\arcsin\lvert z \rvert \le \arcsin(1) \cdot \lvert z \rvert = \frac{\pi}{2}\lvert z \rvert$, which proves the claim.