If `z_1=2-i ,\ +2=-2+i ,` find : `R e((z_1z_2)/(z_1))`

Answer

N/a

Explanation

`{:(underline(z_(1))=underline(2-i,z_(2))=-2+i),(z_(1)=(2-i)=2+i):}`
(i) Now `(z_( 1)z_(2))/ overline(z_(1))=((2-i )(-2+i))/(2+i)=(-(4+i^(2)-4i))/(2+i)`
`" "=(-(3-4i))/(2+ i)xx(2-i)/(2-i)`
`" "=(-(6- 3i-8i+4i^(2)))/(4-i^(2))=(-(2-11i))/(5 )`
`" " =-(2)/(5)+(11)/(5)i`
`thereforeRe((z_(1)z_(2))/(bar(z_(1))))=-(2)/(5)`
(ii) `(1)/(z_(1)bar(z_(1)))=(1)/((2-i)(2+i))=(1)/(4-i^(2))=(1)/(5)`
`" "=(1)/(5)+0*i`
`thereforeIm((1)/(z_(1) barz_(1)))=0`