If one root of the quadratic equation $$a{x^2} + bx + c = 0$$ is equal to the $$n$$-th power of the other, then show that $$${\left( {a{c^n}} \right)^{{1 \over {n + 1}}}} + {\left( {{a^n}c} \right)^{{1 \over {n + 1}}}} + b = 0$$$

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