Question

3/ 3-i

Answer 1

oops
\[\frac{3}{3-i}\times \frac{3+i}{3+i}=\frac{3(3+i)}{3^2+1^2}=\frac{9+3i}{10}=\frac{9}{10}+\frac{3i}{10}\]

Answer 2

method is to multiply top and bottom by the conjugate of the denominator because
\[(a+bi)(a-bi)=a^2+b^2\] a real number

Answer 3

I would have just left it as (9+3i)/10

Answer 4

For me, leaving it as one fraction is the simplest expression

Answer 5

"simple" it may be, but "standard form" is
\[a+bi\]
otherwise might as well leave it as
\[\frac{3}{3-i}\]

Answer 6

3/(3-i) is not the same form as (9+3i)/10, that's why you ended up conjugating it in the first place. The original problem suggests to conjugate and simplify.

Answer 7

Good job Satellite! :)