Alternative proof for $\Bbb R\ni x\mapsto e^{ix}\in S^1$ Lipschitz condition with $K=1$?
I need to prove that the function defined as
$${\rm cis}:\Bbb R\to S^1,\;x\mapsto e^{ix}$$
is Lipschitz with $K=1$. I can prove it with geometric reasoning, using basic facts about the cosine function, but it is supposed I must prove it just using the analytic definition of trigonometric functions like
$$\cos x=\sum_{k=0}^\infty (-1)^k\frac{x^{2k}}{(2k)!}$$
and probably some bound as
$$|s-s_n|\le |x_{n+1}|$$
where $s$ is a convergent alternating series, $s_n$ is a partial sum and $x_n$ are the elements to add in the series.
I found that
$$\frac{|e^{ix}-e^{iy}|}{|x-y|}=\frac{|e^{iz}-1|}{|z|}=\frac{\sqrt{2(1-\cos z)}}{|z|}=\sqrt{2\sum_{k=1}^\infty(-1)^{k-1}\frac{z^{2k-1}}{(2k)!}}$$
where $z=x-y$, but I dont get something useful from here to prove the Lipschitz condition with $K=1$ for any $0\le z<2$.
I know too that ${\rm cis}$ is a (continuous and) surjective homomorphism between the groups $(\Bbb R,+)$ and $(S^1,\cdot)$, but I dont get any idea from here to prove that it is Lipschitz with $K=1$.
In this line of work I tried to think about the relation of Lipschitz condition and compactness, but anyway I need to prove that $K=1$ not only that the function is Lipschitz, what it is not clear from this approach.
Some hint or solution will be appreciated, thank you.
UPDATE:
If I show that
$$\frac{|e^{ix}-1|}{|x|}=\left|i\sum_{k=1}^\infty\frac{(ix)^{k-1}}{k!}\right|=\left|\sum_{k=0}^\infty\frac{(ix)^k}{(k+1)!}\right|\le \left|\sum_{k=0}^\infty\frac{(ix)^k}{k!}\right|=|e^{ix}|=1$$
Im done, but the last inequality is not as easy to prove from the power series as it seems.
It follows from the fundamental theorem of calculus that if $x\leq y$ then $$ e^{iy}-e^{ix}=i\int_x^ye^{it}\;dt$$ and therefore $$ |e^{iy}-e^{ix}|=\Big|\int_x^ye^{it}\;dt\Big|\leq \int_x^y\;dt=y-x$$