Proof of an inequality involving complex analysis

Question

I'm trying to prove : $$\left|\frac{1-\exp\{2\pi i(b-a)\lambda\}}{1-\exp(2\pi i\lambda)}\right|\le \frac{1}{|\sin (\pi \lambda)|}.$$

Expanding the exponentials I'm unable to prove ! Any way to prove this inequality ?

Answer 1

Observe \begin{align} |1-\exp(2\pi i(b-a)\lambda)| =&\ 2\left|\frac{\exp(\pi i(b-a)\lambda)-\exp(-\pi i(b-a)\lambda)}{2}\right|\\ =&\ 2|\sin(\pi \lambda(b-a))| \end{align} then we see that \begin{align} \left| \frac{1-\exp(2\pi i(b-a)\lambda)}{1-\exp(2\pi i\lambda)}\right| = \frac{|\sin(\pi\lambda(b-a))|}{|\sin\pi\lambda|} \leq \frac{1}{|\sin\pi \lambda|}. \end{align}