\(z_!=2-i,z_@=-2+i\)
\((i)z_1z_2=(2-i)(-2+i)=-4+2i+2i-i^2=-4+4i-(-1)=-3+4i\)
\(\bar{z_1}=2+1\)
\(∴\frac{z_1z_2}{z_1}=\frac{-3+4i}{2+i}\)
On multiplying numerator and denominator by (2i), we obtain
\(\frac{z_1z_2}{z_1}=\frac{(-3+4i)(2-i)}{(2+i)(2-i)}=\frac{-6+3i+8i-4i^2}{2^2+1^2}=\frac{-6+1li-4(-1)}{2^2+1^2}\)
\(=\frac{-2+1\,li}{5}=\frac{-2}{5}+\frac{11}{5}i\)
On comparing real parts, we obtain
\(Re(\frac{z_1z_2}{\bar{z_1}})=\frac{-2}{5}\)
(ii) \(\frac{1}{z_1\bar{z_1}}=\frac{1}{(2-i)(2+i)}=\frac{1}{(2)^2+(1)^2}=\frac{1}{5}\)
On comparing imaginary parts, we obtain
\(Im(\frac{1}{z_1\bar{z_1}})=0\)