Proof of $|e^{i\theta} -1| \leq |\theta|$

Question

https://math.stackexchange.com/a/584507/978073
In the above answer given by Julien, They made use of the Fundamental Theorem of Calculus.

They rewrote the LHS of the inequality as $$\int_0^\theta ie^{it}\mathrm{d}t$$
and left us, the reader, to finish the thought.
I understand the Fundamental Theorem of Calculus but I somehow fail to see how that proves the inequality. With the way I understand FTC, it's that the integral is lesser than (or equal to) [ Maximum Value of Function ]*[ Difference of Limits ]

Which would be $$e^{i\theta}*[\theta-0]$$

But it doesn't help the inequality, does it?

It seems like its easy to grasp but somehow its going right over my head. Any help is appreciated.

Answer 1

There is a direct geometric way. The distance from the point $A=(\cos \theta,\sin\theta)$ to the point $B=(1,0)$ is less than the arc length between $A$ and $B,$ which is equal $\theta.$ The inequality is strict if $A\neq B.$

Answer 2

Once you apply the Fundamental Theorem of Calculus, take the absolute value of both sides and then use the triangle inequality: $$ \left| e^{i\theta} -1 \right| = \left| \int_0^\theta i e^{it} dt \right| \leq \int_0^\theta \left| i e^{it} \right| dt = \int_0^\theta 1 dt = \theta.$$