Explanation

(c)
$\left( \frac{- 1 + 3i}{2 + i} \right) = \frac{- 1 - 3i}{2 + i} \times \frac{2 - i}{2 - i}$
$= \frac{- 2 + i - 6i + 3i^{2}}{4 + 1} = - 1 - i$
$\therefore$ Argument of $\left( \frac{- 1 - 3i}{2 + i} \right) = \tan^{- 1}{\left( \frac{- 1}{- 1} \right) = 225{^\circ}}$
[Since, the given complex number lies in IIIrd quadrant]