Question

$If{z}_1=2-i,z_2=-2+i,\text{find}\left(i\right)Re\left(\frac{z_1{z}_2}{z_1}\right)\left(ii\right)Im\left(\frac{1}{z_1{z}_2}\right)$

Answer 1

$\frac{z_1{z}_2}{z_1}=\frac{z_1{z}_2}{z_1}\times \frac{z_1}{z_2}\text{(rationalising the denominator)}=\frac{(z_1{)}^2{z}_2}{z_1{z}_2}=\frac{(2-i{)}^2(-2+i)}{|z_1{|}^2}(\because z\overline{z}|z{|}^2)=\frac{(2^2+i^2-2\times 2\times i)(-2+i)}{|2-i{|}^2}=\frac{(4-1-4i)(-2+i)}{2^2+(-1{)}^2}=\frac{(3-4i)(-2+i)}{4+i}=3(-2+i)-4i(-2+i)=\frac{-6+3i+8i+4}{5}=\frac{-2+11i}{5}\therefore Re\left(\frac{z_1{z}_2}{z_1}\right)=Re(\frac{-2}{5}+\frac{11}{5}i)=\frac{-2}{5}$

$\left(ii\right)\frac{1}{z_1\overline{z_1}}=\frac{1}{|z_1{|}^2}=\frac{1}{|2-i{|}^2}=\frac{1}{2^2+(-1{)}^2}=\frac{1}{4+1}=\frac{1}{5},\text{which is purely real}\therefore I_m\left(\frac{1}{z_1{z}_1}\right)=0$