[tex](1+2i)x^3-2(3+i)x^2+(5-4i)x+2a^2 = 0 \\ \\ x=1 \\ \\ 1+2i-2(3+i)+(5-4i)+2a^2 = 0 \\ 2a^2 =4i \\ a^2 = 2i \\ a = \pm \sqrt{2i} \\ \\ \sqrt{2i} = a+bi \\ 2i = a^2+2abi-b^2\\ \Rightarrow a^2-b^2 = 0 ~~\text{ si }~~ab = 1 \\ \Rightarrow a^2 - \dfrac{1}{a^2} = 0\Big|\cdot a^2~~(a\neq 0) \Rightarrow a^4-1 = 0 \Rightarrow a^4 = 1 \Rightarrow a = \pm 1\\ \\ (1)\quad a = 1 \Rightarrow b = 1\\ (2)\quad a = -1 \Rightarrow b = -1 \\ \\ \Rightarrow \sqrt{2i} = \Big\{-1-i;~1+i\Big\}[/tex]
[tex]a = \pm \sqrt{2i} \Rightarrow a = \pm\Big\{-1-i;~1+i\Big\} \Rightarrow a = \pm(1+i) \Rightarrow \\ \\ \Rightarrow a \in \Big\{-1-i;~1+i\Big\}\\ \\ \displaystyle \sum\limits_{a\in A} |a| = |-1-i|+|1+i| = \sqrt{(-1)^2+(-1)^2}+\sqrt{1^2+1^2}= \\ \\ = \sqrt 2+\sqrt 2=\boxed{2\sqrt 2}[/tex]
