Question

(5 + 2i) / (3+ 6i)

Answer 1

3/5 - 8/15i

Answer 2

3/5-(8 i)/15

Answer 3

gimmick is to multiply by the conjugate of the denominator

Answer 4

what do you mean by conjugate?

Answer 5

because
\[(a+bi)(a-bi)=a^2+b^2\] a real number

Answer 6

the conjugate of
\[3+6i\] i s
\[3-6i\]

Answer 7

When you multiply the denominator, you also have to multiply the numerator by the same thing, correct? Then do I use the FOIL method?

Answer 8

so you start as
\[\frac{5+2i}{3+6i}\times \frac{3-6i}{3-6i}\] then the denominator is
\[3^2+6^2=9+36=45\]

Answer 9

the numerator is what ever you get when you multiply out. in this case it is
\[27-24i\]

Answer 10

yes if you like that method use it

Answer 11

and yes you multiply both top and bottom by the conjugate of the bottom

Answer 12

I'm confused in the numerator, shouldn't it be -6^2?

Answer 13

just don't fret about the i in the denominator

Answer 14

(5 + 2i) / (3+ 6i) * [3-6i / 3-6i] =[27-24i]/45=

Answer 15

ignore it.
\[(a+bi)(a-bi)=a^2+b^2\] that is why you use it

Answer 16

for example
\[(5+3i)(5-3i)=25+9\] and
\[(1-2i)(1+2i)=1+4\] it is that easy

Answer 17

So when all is said and done, my FINAL answer is:

9-8i/5

is that correct?

Answer 18

what did you get for the numerator?

Answer 19

45

Answer 20

27 - 24i/ 45 reduced to 9 - 8i/5

Answer 21

that is the denominator, and it is right. how bout for the top?

Answer 22

\[\frac{27}{45}-\frac{24}{45}i\] then reduce

Answer 23

Thanks!

Answer 24

27 and 45 have a common factor of 9, but it does not reduce to 9

Answer 25

it reduces to
\[\frac{3}{5}\]

Answer 26

aha! I see that now!

Answer 27

so that one was a mistake.

Answer 28

3/5 - 8i/15

Answer 29

24 and 45 have a common factor of 3 so you can reduce that one as well and get
\[\frac{8}{15}\]

Answer 30

I see that now

Answer 31

got it!
standard form is
\[\frac{3}{5}-\frac{8}{15}i\]

Answer 32

Thanks!

Answer 33

yw