Concept:Quadrant:x > 0, y > 0 point lies in I quadrantx < 0, y > 0 point lies in II quadrantx < 0, y < 0 point lies in III quadrantx > 0, y < 0 point lies in IV quadrantCalculation:(I) let z = \( \rm \frac{1 - 3i}{2 + i}\)z = \(\rm \frac{(1 - 3i)(2 “ i)}{(2 + i)(2 “i)}\)z = \( \rm \frac{2 - 6i - i + 3i^{2}}{4 - i^{2}}\)z = \( \rm \frac{2 “ 7i - 3}{4 + 1}\)z = \(\rm \frac{-1 “ 7i }{5}\)\(\rm \overline{z} = \frac{-1 + 7i }{5}\)Since, x < 0, y > 0so, \(\rm \left ( \frac{-1}{5},\frac{7}{5} \right ) \)lies in II quadrant. (II) let z = 2 + ion reciprocal\(\rm \frac{1}{z} = \frac{1}{2 + i}\)\(\rm \frac{1}{z} = \frac{(2 “ i)}{(2 + i)(2 “ i)}\)\(\rm \frac{1}{z} = \frac{(2 “ i)}{(4 - i^{2})} = \rm \frac{1}{5} ( 2 “ i)\)x > 0, y < 0so, \(\rm \left ( \frac{2}{5},\frac{-1}{5} \right ) \)lies in IV quadrant.so, statement I and II both are correct