Question

(2+9i)/(-3-i)

Answer 1

Are these "i" terms like the imaginary number "i" ?

Answer 2

yes

Answer 3

write the expression in the form a+bi

Answer 4

so you need to get rid of the denominator somehow?

Answer 5

remember: complex conjugate.

multiply the top and bottom by -3+i

Answer 6

You know from "normal" algebra that (a + b) (a - b) = a^2 - b^2 right? With no "ab" terms in the answer...

So (-3 - i) (-3 + i) = 9 +3i - 3i -i^2 = 9 +1 = 10

Answer 7

right

Answer 8

yes, also multiply the top by the same (-3 + i), but that's why it works...

Answer 9

(6+25i-9(-1))/10

Answer 10

-(15+25i)/10

Answer 11

-3/2-5/2i