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

- anonymous

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

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- anonymous

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. :)

- anonymous

btw, the standard form is a+bi

- anonymous

by 6-9i ?

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- anonymous

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

- anonymous

i mean 6+9i ?

- anonymous

yes

- anonymous

yes , sorry

- anonymous

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.

- anonymous

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

- anonymous

yes

- anonymous

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

- anonymous

Just making sure

- anonymous

54 + 45i ?
36 + 81i ?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

why not 81i ?

- anonymous

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

- anonymous

agh sorry about not having it in one line

- anonymous

its okay (: , im just trying to understand it

- anonymous

do you understand how i^2 is -1?

- anonymous

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

- anonymous

that it will result to the opposite ?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

see this:

##### 1 Attachment

- anonymous

to multiply aswell ?

- anonymous

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.

- anonymous

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.

- anonymous

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

- anonymous

see, the answer's just the same.

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