How can I find $\Im(\log(1-e^{inx}))$

Question

I'm aware that $$\Im(\log(1-e^{inx}))=\arg(1-e^{inx})=\arctan\left(\frac{\Im(1-e^{inx})}{\Re(1-e^{inx})}\right)=-\arctan \left(\frac{\sin(nx)}{1-\cos(nx)}\right)$$

but how do I proceed from here?

Answer 1

Assuming your steps are correct, note that $$\dfrac{\sin(nx)}{1-\cos(nx)}=\dfrac{2\sin(\frac{nx}{2})\cos(\frac{nx}{2})}{2\sin^2({\frac{nx}{2}})} = \cot\left(\frac{nx}{2}\right)=\tan\left(\frac{\pi}{2}-\frac{nx}{2}\right)$$