Prove that there exists no complex number z such that $$\left| z \right| < {1 \over 3}\,and\,\sum\limits_{r = 1}^n {{a_r}{z^r}} = 1$$ where $$\left| {{a_r}} \right| < 2$$.

If $${z_1}$$ and $${z_2}$$ are two complex numbers such that $$\,\left| {{z_1}} \right| < 1 < \left| {{z_2}} \right|\,$$ then prove that $$\,\left| {{{1 - {z_1}\overline {{z_2}} } \over {{z_1} - {z_2}}}} \right| < 1$$.

Let a complex number $$\alpha ,\,\alpha \ne 1$$, be a root of the equation $${z^{p + q}} - {z^p} - {z^q} + 1 = 0$$, where p, q are distinct primes. Show that either $$1 + \alpha + {\alpha ^2} + .... + {\alpha ^{p - 1}} = 0\,or\,1 + \alpha + {\alpha ^2} + .... + {\alpha ^{q - 1}} = 0$$, but not both together.

For complex numbers z and w, prove that $${\left| z \right|^2}w - {\left| w \right|^2}z = z - w$$ if and only if $$ z = w\,or\,z\overline {\,w} = 1$$.