First, calculate the result of the multiplication, as shown below
[tex]\begin{gathered} w=(-4+3i)(\sqrt{2}+2i)=-4\sqrt{2}+i3\sqrt{2}-8i+6(-1) \\ =-4\sqrt{2}-6+i(3\sqrt{2}-8) \end{gathered}[/tex]In general, given a complex number z=a+ib, its magnitude is given by the formula below
[tex]\begin{gathered} z=a+ib \\ \Rightarrow||z||=\sqrt{a^2+b^2} \end{gathered}[/tex]Then, in our case,
[tex]\begin{gathered} \Rightarrow||w||=(-4\sqrt{2}-6)^2+(3\sqrt{2}-8)^2 \\ \Rightarrow||w||=4(17+12\sqrt{2})+2(41-24\sqrt{2}) \\ \Rightarrow||w||=5\sqrt{6} \end{gathered}[/tex]Furthermore,
[tex]\begin{gathered} w=r(cos\theta+isin\theta) \\ and \\ r=5\sqrt{6} \end{gathered}[/tex]Then, finding theta,
[tex]\begin{gathered} \Rightarrow cos\theta=\frac{-4\sqrt{2}-6}{5\sqrt{6}} \\ \Rightarrow\theta\approx-2.8289... \\ \Rightarrow\theta\approx-2.83 \end{gathered}[/tex]Hence, the answer is
[tex]w=5\sqrt{6}(cos(-2.83)+isin(-2.83))[/tex]