How is it true that $\vert e^{iw(r)} - 1 \vert \leq \min\{\vert w(r) \vert, 2\}$?

Question

I'm reading a thesis where the author uses this inequality (page 33) to prove a lemma:

I know $\vert e^{iw(r)} - 1 \vert = \sqrt{2-2\cos(w(r))}$, which is at least $0$ and at most $2$, but how can I conclude that $\vert e^{iw(r)} - 1 \vert \leq \min\{\vert w(r) \vert, 2\}$? I've included the link to the thesis in case there is context missing. I've been stuck with this for a while now.

It is a mean value theorem https://math.stackexchange.com/questions/208609/mean-value-theorem-for-complex-functions — HK Lee, May 24 '18 at 15:34

score 2 · Accepted Answer · answered May 24 '18 at 15:43

2

Draw a picture. $|e^{iw}-1|$ is a length of side $[e^{iw}e^{i0}]$ of a triangle $\Delta e^{iw} e^{i0} o$ and $w$ is length of arc. If $w>2$, then $|e^{iw}-1| \leq 2$.

answered May 24 '18 at 15:43

HK Lee

19,964

Perfect! Thanks. – Matheus Andrade May 24 '18 at 16:31

How is it true that $\vert e^{iw(r)} - 1 \vert \leq \min\{\vert w(r) \vert, 2\}$?

1 Answers1