Solution 3.1:2b

The two terms do not have the same denominator, so it is not possible to subtract them directly. It is perhaps simplest to calculate each quotient individually and then subtract the result.

We multiply the top and bottom of each fraction by the complex conjugate of its denominator:

\displaystyle \begin{align}\frac{3i}{4-6i}-\frac{1+i}{3+2i} &= \frac{3i(4+6i)}{(4-6i)(4+6i)}-\frac{(1+i)(3-2i)}{(3+2i)(3-2i)}\\ &= \frac{3i\cdot 4 + 3i\cdot 6i}{(4^2-(6i)^2}-\frac{1\cdot 3 - 1\cdot 2i + i \cdot 3 - i\cdot 2i}{3^2-(2i)^2}\\ &= \frac{12i + 18i^2}{16+36}-\frac{3 - 2i + 3i - 2i^2}{9+4}\\ &= \frac{-18+12i}{52}-\frac{3 +(-2 + 3)i + 2}{13}\\ &= \frac{-18+12i}{52}-\frac{5 +i}{13}\end{align}

Then, we multiply the top and bottom of the last fraction by 4, so as to give make both fractions have the same denominator, and after that we subtract the numerators:

\displaystyle \begin{align}\frac{-18+12i}{52}-\frac{(5 +i)\cdot 4}{13\cdot 4}&=\frac{-18+12i}{52}-\frac{20+4i}{52}\\ &=\frac{-18+12i-20-4i}{52}\\ &=\frac{-38+8i}{52}\\ &=-\frac{19}{26}+\frac{2}{13}i.\end{align}