Gọi $z=a+bi\,\left( a,b\in R \right)$ và $w=x+yi\,\left( x,y\in R \right)$

               Ta có :    $\left| z-1 \right|=2\Leftrightarrow {{(a-1)}^{2}}+{{b}^{2}}=4(1)$

             Từ    $\,w=(1+i\sqrt{3})z+2\Rightarrow x+yi=\left( 1+i\sqrt{3} \right)\left( a+bi \right)+2$

                        $\Leftrightarrow \left\{ \begin{align}  & x=a-b\sqrt{3}+2 \\  & y=\sqrt{3}a+b \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align}  & x-3=a-1-b\sqrt{3} \\  & y-\sqrt{3}=\sqrt{3}(a-1)+b \\ \end{align} \right.$

               Từ đó : $\,\,{{(x-3)}^{2}}+{{(y-\sqrt{3})}^{2}}=4\left[ {{\left( a-1 \right)}^{2}}+{{b}^{2}} \right]=16.$ (do (1))

            Suy ra r = 4