Ta có:

$\begin{align}  & \left( \left| z \right|+1 \right)\overline{z}=\frac{\left( 2a+4b \right)\left( 2b-4a \right)i}{\left( a+2b \right)+\left( b-2a \right)i} \\ & \Rightarrow \left| \left( \left| z \right|+1 \right)\overline{z} \right|=\left| \frac{\left( 2a+4b \right)\left( 2b-4a \right)i}{\left( a+2b \right)+\left( b-2a \right)i} \right|=\frac{\left| \left( 2a+4b \right)\left( 2b-4a \right)i \right|}{\left| \left( a+2b \right)+\left( b-2a \right)i \right|} \\ & \Rightarrow \left| \left| z \right|+1 \right|\left| \overline{z} \right|=\frac{\sqrt{{{\left( 2a+4b \right)}^{2}}+{{\left( 2b-4a \right)}^{2}}}}{\sqrt{{{\left( a+2b \right)}^{2}}+{{\left( b-2a \right)}^{2}}}} \\ & \Rightarrow \left( \left| z \right|+1 \right)\left| z \right|=\frac{\sqrt{20{{a}^{2}}+20{{b}^{2}}}}{\sqrt{5{{a}^{2}}+5{{b}^{2}}}}=2 \\ & \Rightarrow {{\left( \left| z \right| \right)}^{2}}+\left| z \right|-2=0 \\ & \Rightarrow \left( \left| z \right|-1 \right)\left( \left| z \right|+2 \right)=0 \\ & \Rightarrow \left| z \right|-1=0\Leftrightarrow \left| z \right|=1 \\ \end{align}$

Đáp án đúng là B