Question

how do you write the quotient (9+i) over (9-i) in standard form? I need the answer but more importantly, I need to learn how to solve these kinds of problems.

Answer 1

Since they are the conjugates of each (9-a and 9+a are conjugates, or with any number replacing 9) , we multiply it by (9+i)/(9+i) to get (9+i)^2/(9-i^2)=(81+2i+i^2)/(9-(-1))=(81-1+2i)/(10)=80/10+2i/10=8+i/5. Conjugates work because they make i into i^2 (-1) without any of the sneaky i's in the middle!