We have,
$\frac{-1-3i}{2+i}=\frac{-1-3i}{2+i}\times\frac{2-i}{2-i}=\frac{-2-5i+3i^{2}}{4-i^{2}}$
$= -1 - i$
Now, let us put -1 = r cos$\theta, -1=r sin \theta$
Squaring and adding, r$^{2}$ = 2 i.e., r =$\sqrt{2}$
So that cos $\theta= \frac{-1}{\sqrt{2}} , sin \theta =\frac{-1}{\sqrt{2}}$
Therefore,$ \theta =225°$
Thus, argument is 225°.