\[a=1+3i\]
and common ratio \[r= \frac{1+i}{3}\]
.\[\| r \| = \| \frac{1+i}{3} \| = \frac{\sqrt{2}}{3} \lt 1\]
so we can use the formula for the sum of a geometric series \[S= \frac{a}{1-r}\]
.\[\begin{equation} \begin{aligned} S &= \frac{a}{1-r} \\ &= \frac{1+3i}{1- \frac{1+i}{3}} \\ &= \frac{1+3i}{\frac{2-i}{3}} \\ &= \frac{3+9i}{2-i} \\ &= \frac{3+9i}{2-i} \times \frac{2+i}{2+i} \\ &= \frac{6+3i+18i-9}{4+2i-2i-(-1)} \\ &= \frac{-3+21i}{5} \end{aligned} \end{equation}\]