A complex number z has argument θ and modulus 1. Show that (z^n)-(z^-n)=2iSin(nθ).

This problem wants you to use De Moivre's theorem to prove a trigonometric identity.We will tackle this problem by taking the left hand side and using theorems and manipulation to show it is equivalent to the right hand side. Since we have been given the modulus and argument, and the right side of the identity involves a trig function, it is logical to rewrite z in its polar form: z=cosθ+isinθ. De Moivre's theorem tells us that z^n=cos(nθ)+isin(nθ) and that z^-n=cos(-nθ)+isin(-nθ). Since the right side of the identity has no (-n) in it we need to find away to get rid of that negative. Recall firstly that cosine is an even function, meaning that cos(-x)=cosx, and secondly that sine is an odd function meaning sin(-x)=-sinx. Applying this to z^-n gives us z^-n=cos(n θ)-isin(n θ).Plugging this back into the left side of the identity gives z^n-z^-n=(cos(n θ)+isin(n θ))-(cos(n θ)-isin(n θ))Which simplifies to give z^n-z^-n=2isin(n θ)