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# Let $\;z_{1}=2-i\;,z_{2}=-2+i\;$. Find (i)$\;Re(\large\frac{z_{1}z_{2}}{\overline{z_{1}}})\;,$ (ii) $\;Im(\large\frac{1}{z_{1} \overline{z_{2}}})$

$(a)\;-\large\frac{2}{5},0\qquad(b)\;-\large\frac{3}{5},0\qquad(c)\;-\large\frac{1}{5},0\qquad(d)\;-\large\frac{7}{5},0$

Answer : $\;-\large\frac{2}{5},0$
Explanation :
$\;z_{1}=2-i\;,z_{2}=-2+i\;$
(i) $\;z_{1}z_{2}= (2-i)(-2+i)= -4+2i+2i-i^{2} = -4+4i-(-1) \qquad [i^{2}=-1]$
$= -4 +4i+1=-3+4i$
$\overline{z_{1}} = 2+i$
Therefore , $\;\large\frac{z_{1}z_{2}}{\overline{z_{1}}} = \large\frac{-3+4i}{2+i}$
On multiplying numerator and denominator bainy $\;2-i\;$ , we obtain
$\;\large\frac{z_{1}z_{2}}{\overline{z_{1}}} = \large\frac{-3+4i}{2+i} \times \large\frac{2-i}{2-i}$
$= \large\frac{-6+3i+8i-4i^{2}}{2^{2}-i^{2}}$
$= \large\frac{-6+11i+4}{4+1} = \large\frac{-2+11i}{5}$
$= -\large\frac{2}{5} + \large\frac{11}{5} i$
On comparing Real parts , we obtain
$Re(\large\frac{z_{1}z_{2}}{\overline{z_{1}}}) = -\large\frac{2}{5}$
(ii) $\; \large\frac{1}{z_{1} \overline{z_{2}}} = \large\frac{1}{(2-i)(2+i)}$
$= \large\frac{1}{2^{2}-i^{2}} = \large\frac{1}{4+1} = \large\frac{1}{5}$
On comparing imaginary parts , we obtain
$Im(\large\frac{1}{z_{1} \overline{z_{2}}}) =0$