We have, $z_1$ = 2 - i , $z_2 = -2 + i \quad (i)$z_1z_2$$= (2 - i) (- 2 + i) = - 4 + 2i + 2i -$i^2$= - 4 + 1 + 4i = - 3 + 4i \quad \text{Also},$z_1↖{-}$= 2 + i \quad \therefore${z_1z_2}/z_1↖{-}$=${-3+4i}/{2+i}$=${-3+4i}/{2+i}$\times${2-i}/{2-i}$\quad =${-6+3i+8i+4}/{4+1}$=${-2+11i}/5$=$-2/5 + {11}/5 i$=$({z_1z_2}/z_1↖{-})$\quad \therefore \ \text{Re}$2/5$= -$z_1z_1↖{-}$\quad (ii)$1/{z_1z_1↖{-}}$= (2 - i) (2 + i) = 4 + 1 = 5 \quad \text{Now},$1/5$=$1/5$=$(1/{z_1z_1↖{-}})$ = 0