Ta có:

a) $2i(3 + i)(2 + 4i)$

= $2i(2 + 14i)$

= $-28 + 4i$

b) $\frac{(1+i)^{2}(2i)^{3}}{-2+i}$

= $\frac{2i.(-8i)}{-2+i}$

= $\frac{16(-2-i)}{5}$

= $-\frac{32}{5}-\frac{16}{5}i$

c) $3 + 2i + (6 + i)(5 + i)$

= $3 + 2i + 29 + 11i$

= $32 + 13i$

d) $4-3i+\frac{5+4i}{3+6i}$

= $4-3i+\frac{(5+4i)(3-6i)}{(3+6i)(3-6i)}$

= $4-3i+(\frac{(39}{45}-\frac{18}{45}i)$

= $\frac{73}{15}-\frac{51}{15}i$