[tex]|z|\leq 1,\quad m\ \textless \ n \\ z = a+bi \\ \\ (z+\overline{z})^m+(z+\overline{z})^{m+1}+...+(z+\overline{z})^n = 2016\\ \\ (a+bi+a-bi)^m+(a+bi+a-bi)^{m+1}+...+(a+bi+a-bi)^n =$ $ \\ =2016 \\ \\ (2a)^m+(2a)^{m+1}+...+(2a)^{n} = 2016 \\ \\ $Observam ca in paranteze e un numar real oricare ar fi \\ partea imaginara.$ \\ $Daca z = \dfrac{1}{2} \pm pi$ avem: \\ \\ 1^m+1^{m+1}+...+1^{n} = 2016 \\ \\ $Iar $ m=1,n = 2016: \\ \\ 1^1+1^2+...+1^{2016} = 2016 \\ \\[/tex]
[tex] 2016 = 2016 \quad (A) \\ \\ $ Pentru z = \dfrac{1}{2} \pm pi. \\ \\ $ Avem solutie cand $ (m,n) = (k,k+2015),\quad k\in \mathbb_{N} $ \\ \\ \Rightarrow (z,m,n) = \Big(\dfrac{1}{2}\pm pi,k,k+2015\Big), \quad p \in \Big[-\dfrac{\sqrt3}{2}, \dfrac{\sqrt3}{2}\Big], k\in \mathbb_{N}$ $[/tex]
