Question

Rewrite 4+i over 3-2i in the form a+bi, where a and b are rational numbers. please show steps, reviewing for exams and have no clue how to do.

Answer 1

`Use:(a-b)(a+b)=a^2-b^2\\-------------------------\\\\(4+i)/(3-2i)=(4+i)/(3-2i)\cdot(3+2i)/(3+2i)=((4+i)(3+2i))/((3-2i)(3+2i))=(4(3)+4(2i)+i(3)+i(2i))/(3^2-(2i)^2)\\\\=(12+8i+3i+2i^2)/(9-2^2i^2)=(12+11i+2(-1))/(9-4(-1))=(12+11i-2)/(9+4)=(10+11i)/(13)\\\\=\boxed{(10)/(13)+(11)/(13)i}\\\\z=a+bi\Rightarrow a=(10)/(13)\ and\ b=(11)/(13)`

Answer 2

`(4 + i)/(3 - 2i) \\\\ = ((4 + i)(3 + 2i))/((3 - 2i)(3 + 2i)) \text{ (rationalising the denominator) } \\\\ = (12 + 3i + 8i + 2i^2)/(9 - 6i + 6i - 4i^2) \\\\ = (12 + 11i + 2i^2)/(9 - 4i^2) \\\\ = (12 + 11i + 2(-1))/(9 - 4(-1)) \\\\ = (10 + 11i)/(13) \\\\ = (10)/(13) + (11)/(13)i`