• # question_answer Additive inverse of $1-i$is A) $0+0i$ B) $-1-i$ C) $-1+i$ D) None of these

If $z=x+iy$ is the additive inverse of $1-i$ then $(x+iy)+(1-i)=0$ Þ $x+1=0$, $y-1=0$                           Þ $x=-1$, $y=1$ $\therefore$ The additive inverse of $1-i$is $z=-1+i$ Trick: Since $(1-i)+(-1+i)=0$