Modulus of exponential function with real and complex arguments

Question

Can anyone please explain why $$|e^{\frac12 \sin(2x) }|\le e^{1/2}$$ for all real $x$, while $$|e^{-i\sin(x)^{2}}|=1$$ for all real $x$?

Answer 1

If $x \in \mathbb R$, then $-1 \le \sin 2x \le 1$. Since exponentiation by a real number is a monotonically increasing function, $$|e^{(\sin 2x)/2}|$$ attains its maximum when $\sin 2x$ is maximized; i.e., when $\sin 2x = 1$. (The absolute value is irrelevant because $e^x > 0$ for all $x \in \mathbb R$. It follows that $$e^{(\sin 2x)/2} \le e^{1/2}.$$

For the second identity, we note that $$|e^{i\theta}| = |\cos \theta + i \sin \theta| = \sqrt{\cos^2 \theta + \sin^2 \theta} = 1$$ for any $\theta \in \mathbb R$. Therefore $$|e^{-i\sin x^2}| = 1$$ since $-\sin x^2$ or $-\sin^2 x$ (whichever you prefer) is a real number.