Question

write quotient in standard form 9+5i / 6-9i

Answer 1

multiply both the top and the bottom by the conjugate of the denominator. The conjugate of a+bi = a-bi. Do that and see what happens to the denominator. :)

Answer 2

btw, the standard form is a+bi

Answer 3

by 6-9i ?

Answer 4

6+9i. b in this case would be -9 so the conjugate is 6-(-9)i which is equal to 6+9i

Answer 5

i mean 6+9i ?

Answer 6

yes

Answer 7

yes , sorry

Answer 8

whenever you want to get rid of the imaginary part of a complex denominator, multiply the top and bottom by the conjugate of the denominator.

Answer 9

9+5i * 6+9i ?
6-9i * 6+9i ?

Answer 10

yes

Answer 11

I assume that the first line is the numerator and the second line is the denominator?

Answer 12

Just making sure

Answer 13

54 + 45i ?
36 + 81i ?

Answer 14

yes first line is the numerator and the second line is the denominator

Answer 15

Ok, you got the denominator right but (9+5i)*(6+9i) = 54 + 81i + 30i + 45

Answer 16

wait, sorry. the denominator is wrong too. it is 36+81

Answer 17

why not 81i ?

Answer 18

i^2 = -1
because i = \[\sqrt{-1}\]
so i^2 = \[\sqrt{-1}\] * \[\sqrt{-1}\] = -1

Answer 19

agh sorry about not having it in one line

Answer 20

its okay (: , im just trying to understand it

Answer 21

do you understand how i^2 is -1?

Answer 22

so i would have to add the top and it'll be 99 + 111i ?

Answer 23

that it will result to the opposite ?

Answer 24

\[(9+5i) / (6-9i) = [(9+5i)(6+9i)] / [(6-9i)(6+9i)]\]
Then multiply it out. Remember that (a+b)(c+d) = ac + ad + bc + bd

Answer 25

So
\[(9+5i)(6+9i) = 9*6 + 9*9i + 6*5i + 5*9*i^2\]

Answer 26

5*9*i^2 = -1*5*9. Just do the same thing for the denominator, too.

Answer 27

so its not (9+5i)*(6+9i) = 54 + 81i + 30i + 45 ?

Answer 28

suppose x+yi(x，y∈R)，satisfies (9+5i)÷(6-9i)=a+bi ,
then (x+yi)*(6-9i)=9+5i.
we must remember the logarithm of imaginaries as: (a+bi)(c+di)=(ac－bd)+(bc+ad)i.
then 6x+9y=9, 6y-9x=5. so x=1/13, y=37/39.
so 9+5i / 6-9i = 1/13 + 37/39 i

Answer 29

That's right kathy. So in standard form it is (54+45) + (81+30)i = 99+111i
Now try doing the denominator. Something interesting should happen.

Answer 30

my method is a traditional one. Johnfreeman uses the idea of rationalization.

Answer 31

yes thats the answer they give here as an option :) so to solve problems with i i would have to use logarithm of imaginaries?

Answer 32

see this:

Answer 33

to multiply aswell ?

Answer 34

I think the easiest way to think of it at this point is to multiply the top and bottom by the conjugate because then you get a rational denominator.

Answer 35

It should at least be good practice to try multiplying the denominator by its conjugate so you know what happens when you do that. It is probably how they want you to solve this anyway.

Answer 36

clear, you can use tthe data in the problem to directly work it out:
9+5i / 6-9i=[9*6+5*(-9)]/[6^2+(-9)^2] + [5*6-9*(-9)]/[6^2+(-9)^2]*i
= 1/13 + 37/39 i

Answer 37

see, the answer's just the same.