I am told to get the absolute value of $$\frac{1}{e^{i\omega t}-1}$$
I sense that there's something ridiculously simple about this, but I tried working from the fact that if I square it, the absolute value is the square root since the square of anything is always positive unless it involves i. So: $$\left(\frac{1}{e^{i\omega t}-1}\right)^2=\frac{1}{e^{2i\omega t}+2e^{i\omega t}+1}=\frac{1}{(\cos{i\omega t}+i\sin{i\omega t}-1)^2}=\frac{1}{(\cos^2{i\omega t}+2i\cos{i\omega t}\sin{i\omega t}-\sin^2{i\omega t}-2\cos{i\omega t}-2i\sin{i\omega t}+1)}=\frac{1}{2\cos^2{i\omega t}+i\sin{2i\omega t}-2[\cos{i\omega t}-i\sin{i\omega t}])}$$
Well, looking at this I know that $\cos^2{i\omega t}$ is always going to be between 0 and 2. And $i\sin{2i\omega t}$ is going to be between -i and i. And the last term should always be between -2 and 2.
But I am still a little stuck as I feel I am missing the last step. Or I started the problem wrongly and over-thought it a bit. So I am asking if I did something wrong here.
ANy help (or a reminder of some silly principle I forgot) is appreciated.
$$\begin{align} \left|\frac{1}{e^{i\omega t} - 1}\right| &= \left|\frac{1}{e^{i\omega t/2}(e^{i\omega t/2} - e^{-i\omega t/2})}\right| \\ &= \frac{1}{|e^{i\omega t/2}|} \left|\frac{1}{2i \sin(\omega t/2)}\right|\\ &= \frac{1}{2|\sin(\omega t/2)|} \end{align}$$