How to $ (3-2i)(2+3i)/(1+2i)(2-i)$

Question

I got my admit card today noon and my exam (Lateral Entry for B.Tech) is tomorrow 11:15A.M. LoL...that means I have to go through my guide of > 500 pages in less than 18 hours (I know its not possible). So, I'm trying to read my guide as a crash cource :p My question is $$\frac{(3-2i)(2+3i)}{(1+2i)(2-i)}$$

Answer 1

$$\frac{(3 - 2i)(2 + 3i)}{(1 + 2i)(2 - i)} = \frac{12 + 5i}{4 + 3i} = \frac{(12 + 5i)(4 - 3i)}{(4 + 3i)(4 - 3i)} = \frac{63 - 16i}{25} = \frac{63}{25} - \frac{16}{25}i$$