In order to compute 2u and 3z, we distribute the real number into each term: 2 u = 2 ( 1 + 3 i ) = 2 + 6 i {\displaystyle 2u=2(1+3i)=2+6i} 3 z = 3 ( 2 − i ) = 6 − 3 i {\displaystyle 3z=3(2-i)=6-3i} Now, we subtract the real and imaginary parts to get 2 u − 3 z = − 4 + 9 i . {\displaystyle 2u-3z=-4+9i.} Answer: 2 u − 3 z = − 4 + 9 i {\displaystyle \color {blue}2u-3z=-4+9i}
