If ` Z_(1)=2 + 3i and z_(2) = -1 + 2i` then find
(i) ` z_(1) +z_(2)`
(ii) ` z_(1) -z_(2)`
(iii) `z_(1).z_(2)`
(iv) ` z_(1)/z_(2)`

Explanation

we have ` z _(1) 2 + 3i and z_(2) = -1 + 2i`
(i)` z_(1)" + z_(2) = (2 + 3i) + ( -1 + 2i)`
= 1 + 5i
`(ii) z_(1) -z_(2) = ( 2 +3 i) - ( -1 + 2i)`
` = ( 2+1) + ( 3-2) i`
= 3 +1
(iii) ` z_(1)z_(2) = (2 + 3i) ( -1 +2i) `
` = ( -2 -6) + ( 4-3) i`
-8 + i
(iv) ` z_(1)/(z_(2)) = ( 2 + 3i)/( -1 + 2i) = ( 2 + 3i) . 1/(-1 + 2i) `
` = ( 2 + 3i) 1/(-1 + 2i) . ( -1 -2i)/(-1 -2i)`
` ( ( 2+ 3i)( -1 -2i))/((-1)^(2) - ( 2i))^(2)`
` = 1/5 [ ( -2 + 6) + ( -4-3)i]`
` 1/5 ( 4-7i)`