If $z_{1}=2-i, z_{2}=-2+i$, find
(i) $\operatorname{Re}\left(\frac{z_{1} z_{2}}{z_{1}}\right)$
(ii) $\operatorname{lm}\left(\frac{1}{z_{1} \bar{z}_{1}}\right)$
(i) $z_{1}=2-i, z_{2}=-2+i, \overline{z_{1}}=2+i$
$\therefore\left(\frac{z_{1} z_{2}}{z_{1}}\right)=\left(\frac{[2-i][-2+i]}{2+i}\right)$
$=\left(\frac{-4+2 i+2 i-i^{2}}{2+i}\right)$
$=\left(\frac{-3+4 i}{2+i}\right)$
$=\left[\frac{-3+4 i}{2+i} \times\left(\frac{2-i}{2-i}\right)\right]$
$=\left(\frac{-6+3 i+8 i-4 i^{2}}{2^{2}-i^{2}}\right)$
$=\left(\frac{-2+11 i}{4-(-1)}\right)$
$=\left(\frac{-2+11 i}{5}\right)$
$\operatorname{Re}\left(\frac{z_{1} z_{2}}{z_{1}}\right)=\frac{-2}{5}$
(ii) $\left(\frac{1}{z_{1} \overline{z_{1}}}\right)=\frac{1}{(2-i)(2+i)}$
$=\frac{1}{2^{2}-i^{2}}$
$=\frac{1}{5}$
$\operatorname{Im}\left(\frac{1}{z_{1} \overline{z_{1}}}\right)=0$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.