Prove that the complex numbers $${{z_1}}$$, $${{z_2}}$$ and the origin form an equilateral triangle only if $$z_1^2 + z_2^2 - {z_1}\,{z_2} = 0$$.

If $$z = x + iy$$ and $$\omega = \left( {1 - iz} \right)/\left( {z - i} \right),$$ then $$\,\left| \omega \right| = 1$$ implies that, in the complex plane,

Show that $$$16\cos \left( {{{2\pi } \over {15}}} \right)\cos \left( {{{4\pi } \over {15}}} \right)\cos \left( {{{8\pi } \over {15}}} \right)\cos \left( {{{16\pi } \over {15}}} \right) = 1$$$

If $$\tan \,A = \left( {1 - \cos B} \right)/\sin B,$$ then $$tan2A = tan\,B$$.