Gọi ${{z}_{1}}=a+bi,{{z}_{2}}=x+yi,\left( a,b,x,y\in \mathbb{R} \right)$.

Ta có $\left| {{z}_{1}}+{{z}_{2}} \right|=\sqrt{2}\left| {{z}_{1}} \right|$

$\begin{align}  & \Leftrightarrow \left| \left( a+x \right)+\left( b+y \right)i \right|=\sqrt{2}\left| a+bi \right| \\  & \Leftrightarrow {{\left( a+x \right)}^{2}}+{{\left( b+y \right)}^{2}}=2\left( {{a}^{2}}+{{b}^{2}} \right) \\  & \Leftrightarrow {{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}-2ax-2by \\  & \Leftrightarrow {{\left( a-x \right)}^{2}}+{{\left( b-y \right)}^{2}}=2\left( {{x}^{2}}+{{y}^{2}} \right) \\  & \Rightarrow \left| \left( a-x \right)+\left( b-y \right)i \right|=\sqrt{2}\left| x+yi \right| \\  & \Leftrightarrow \left| {{z}_{1}}-{{z}_{2}} \right|=\sqrt{2}\left| {{z}_{2}} \right| \\ \end{align}$

Ta chọn phương án B.