Do you know how to prove $2(1-\cos x))<x^{2}$

Question

Let $z\in\mathbb{C}$ such that $z = x+iy$ for some real numbers $x$ and $y$.

I have to prove that $$\Biggl|\frac{z}{|z|}-1\Biggr|<\arg(z)$$

And now the problem reduces to prove that $2(1-\cos x))<x^{2}$

Answer 1

We have that

$$\left|\frac{z}{|z|}-1\right|=\left|\frac{x}{\sqrt{x^2+y^2}}-1+i\frac{y}{\sqrt{x^2+y^2}}\right|=\lvert \cos \theta-1+i\sin \theta\rvert=\sqrt{(\cos \theta-1)^2+\sin^2 \theta}=$$

$$=\sqrt{2-2\cos \theta}<\theta \iff1-\cos \theta <\frac{\theta^2}2 \quad\theta>0$$

then consider for $x>0$

$$f(x)=\cos x-1+\frac{x^2}2>0$$

indeed $f(0)=0$ and

$$g(x)=f'(x)=-\sin x+x\ge0$$

indeed $g(0)=0$ and

$$g'(x)=1-\cos x \ge 0$$

Answer 2

$z= |z| e^{it} = |z|(\cos t + i\sin t)\\ \arg z = t$

$|\frac {z}{|z|} - 1| \le \arg z\\ |(\cos t - 1) + i\sin t| \le t$

What is this geometrically?

The red line is shorter that the black curve. And since the line connecting two points is the shortest path between those points, we could drop it there.

I am guessing that you have: $|(\cos t - 1) + i\sin t| = \sqrt {(\cos t - 1)^2 + \sin^2 t} = \sqrt {2+2\cos t}$

and then you squared both sides, giving the question in this post.

I would rather suggest: $\sqrt {2+2\cos t}= 2|\sin \frac 12 t|$

and $|\sin x| \le x$ if $x\ge 0$

$2|\sin \frac 12 t| \le t$