Why only $z$ ?

Given: ${ a }_{ 1 }{ z }^{ 1 }+{ a }_{ 2 }{ z }^{ 2 }+....+{ a }_{ n }{ z }^{ n }=1$ Now using the triangular inequality we get: $\left| { a }_{ 1 }{ z }^{ 1 }+{ a }_{ 2 }{ z }^{ 2 }+....+{ a }_{ n }{ z }^{ n } \right| \le \left| { a }_{ 1 }{ z }^{ 1 } \right| +\left| { a }_{ 2 }{ z }^{ 2 } \right| +..+\left| { a }_{ n }{ z }^{ n } \right| \\ 1\le \left| { a }_{ 1 }{ z }^{ 1 } \right| +\left| { a }_{ 2 }{ z }^{ 2 } \right| +...\left| { a }_{ n }{ z }^{ n } \right|$ Now maximum of R.H.S. occurs at $\left| { a }_{ r } \right| =2$ Therefore, $\frac { 1 }{ 2 } \le \left| { z }^{ 1 } \right| +\left| { z }^{ 2 } \right| +...+\left| { z }^{ n } \right|$ Now adding extra $\left| z \right|$ will not effect the inequality.Therefore, $\frac { 1 }{ 2 } \le { \left| z \right| }^{ 1 }+{ \left| z \right| }^{ 2 }+...+{ \left| z \right| }^{ n }+{ \left| z \right| }^{ n+1 }+....\infty$ Hence, $\frac { 1 }{ 2 } \le \frac { \left| z \right| }{ 1-\left| z \right| } \\ \left| z \right| \ge \frac { 1 }{ 3 } \Rightarrow { \left| z \right| }_{ min }=\frac { 1 }{ 3 }$