Answer
$-\frac{1}{3}+\frac{1}{3}i$
Work Step by Step
$\frac{1+i}{-3i}\times\frac{-3i}{-3i}$
=$\frac{(1+i)\times -3i}{-3i\times -3i}$
=$\frac{-3i-3i^{2}}{9i^{2}}$
=$\frac{-3i-3(-1)}{9(-1)}$
=$\frac{3-3i}{-9}$
=$\frac{-3(-1+i)}{-9}$
=$\frac{(-1+i)}{3}$
=$-\frac{1}{3}+\frac{1}{3}i$
