Question

divide 5-2i/3+6i. Please help!

Answer 1

multiply top and bottom by the conjugate of the denominator

Answer 2

the conjugate of \(a+bi\) is \(a-bi\) and this works because \((a+bi)(a-bi)=a^2+b^2\) a real number

Answer 3

in your case it will be
\[\frac{5-2i}{3+6i}\times \frac{3-6i}{3-6i}=\frac{(5-2i)(3-6i)}{9+36}\]

Answer 4

and you finish it up from there?

Answer 5

i got 15-36i+12i^2/9-6i

Answer 6

the denominator should have no \(i\) in it
that is what you are trying to achieve

Answer 7

i^2 is 1 right?

Answer 8

\[(3+6i)(3-6i)=3^2+6^2=9+36=45\]

Answer 9

no \(i=\sqrt{-1}\) so \(i^2=-1\) not \(1\)

Answer 10

i see

Answer 11

make sure it is clear what you are trying to do in this problem
your goal is to divide, which really means turn \[\frac{5-2i}{3+6i}\] into standard form as \(a+bi\)

Answer 12

i wrote the steps above, although you still have to compute the numerator

Answer 13

i got 1/15 - 36square root of -1/45

Answer 14

there is not square root in it

Answer 15

only i instead of square root

Answer 16

first part is right, it is \(\frac{1}{15}\) for the real part

Answer 17

36i over 45?

Answer 18

oh i guess the whole thing you wrote is correct

Answer 19

except you should write
\[\frac{1}{15}-\frac{4}{5}i\]

Answer 20

thank you so much! you are my hero now!

Answer 21

my pleasure