Question

Valoarea nr. (z+3)^2016, unde z€M, M={z€C/ |z|+z=2-4i} este... ?

Answer 1

$\displaystyle Fie ~z=a+bi,~unde~a,b \in \mathbb{R}. \\ \\ |z|+z=2-4i \Leftrightarrow \sqrt{a^2+b^2}+a+bi=2-4i. \\ \\ Prin~urmare: \left \{ {{ \sqrt{a^2+b^2}+a=2} \atop {b=-4}} \right. .~Substituind~b=-4~in~prima \\ \\ ramura,~obtinem:~ \sqrt{a^2+16}=2-a. \\ \\ Radicalul~este~\e{de}finit~ \forall~ a \in \mathbb{R}. \\ \\ \sqrt{a^2+16} \geq 4 \Rightarrow 2-a \geq 4 \Rightarrow a \leq -2. \\ \\ Ridicand~la~patrat,~obtinem:~a^2+16=4-4a+a^2 \Rightarrow a=-3.$

$\displaystyle (Si~putem~observa~ca~a=-3~respecta~conditia~a \leq -2.) \\ \\ Deci~ \boxed{z=-3-4i}. \\ \\ (z+3)^{2016}=(-4i)^{2016}=(-4)^{2016} \cdot i^{2016}=4^{2016} \cdot \left(i^4 \right )^{504}=4^{2016}.$