Answer:
[tex]4+3i[/tex]
Step-by-step explanation:
=[tex]\frac{(10+20i)}{4+2i} \\\\\frac{10+20 i}{4+2i}[/tex]x [tex]\frac{(4-2i)}{(4-2i)}[/tex]
=[tex]\frac{(10+20i)(4-2i)}{(4^{2}-2i^{2}) } \\\frac{40-20i+80i+40}{16+4}\\ \frac{80+60i}{20}\\ 4+3i[/tex](Taking 20 common from both numerator and deniminator)
