Question

(Complex Numbers) Find each quotient: (2-3i)/(2+3i)

Answer 1

do you know what a conjugate is?

Answer 2

yes, I multiplied the numerator and denominator by: (2-3i) but I don't see how my answer is going to be -5/13-12/13i. So I don't know what I am doing wrong.

Answer 3

That's what the answer key says it's suppose to be.

Answer 4

that answer looks good to me

Answer 5

How do I get there? I'm at (4-12i+9i^2)/(4-9i^2)

Answer 6

the denominator is \(a^2+b^2\) which in your case is \(2^2+3^2=4+9=13\)

Answer 7

I^2=-1

Answer 8

oh i see, you have the right answer, just forget that \(i^2=-1\)

Answer 9

I got: (3+3i) after I did the rest, and that's still not in the fractions that the back of the book has, so how does it get to be: -5/13-12/13i ?

Answer 10

I mean (-3+3i) was what I got.

Answer 11

you wrote
\[\frac{4-12i+9i^2}{4-9i^2}\]right?

Answer 12

Yeah, and then put (-1) in place of the I^2 and got (-3+3i)

Answer 13

\[\frac{4-12i+9i^2}{4-9i^2}=\frac{4-12i+9\times (-1)}{4-9\times (-1)}\]

Answer 14

you get
\[\frac{4-9-12i}{4+9}=\frac{-5-12i}{13}\]

Answer 15

then in standard form this is
\[-\frac{5}{13}-\frac{12}{13}i\]

Answer 16

clear or no? i am not really sure how you got the \(3-3i\) from this

Answer 17

Yeah I see what I was doing wrong right there now, I was subtracting the -1 instead of multiplying it, thank you for your help!