$\begin{aligned} & z=\frac{-2(1+2 i)}{3+i} \\ &=\frac{-2-4 i}{3+i}=\frac{-2-4 i}{3+i} \times \frac{3-i}{3-i} \\ &=\frac{-6+2 i-12 i+4 i^{2}}{10} \\ &=\frac{-6-10 i-4}{10}=\frac{-10-10 \mathrm{i}}{10}=-1-i \\ & z=-1-i=r(\cos \theta+i \sin \theta) \end{aligned}$
On comparing real and imaginary part on both sides, we get $r \cos \theta=-1$
$r \sin \theta=-1$
On dividing eq. (ii) by (i), we get
$\frac{\mathrm{r} \sin \theta}{\mathrm{r} \cos \theta}=\frac{-1}{-1}$
$\tan \theta=1=\tan \frac{\pi}{4}$
$\Rightarrow \quad \theta=\frac{\pi}{4}$
$\therefore \quad \theta=\frac{\pi}{4}$
$\therefore$ Option (b) is correct.