anonymous
  • anonymous
write quotient in standard form 9+5i / 6-9i
Mathematics
chestercat
  • chestercat
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anonymous
  • 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
  • anonymous
btw, the standard form is a+bi
anonymous
  • anonymous
by 6-9i ?

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anonymous
  • anonymous
6+9i. b in this case would be -9 so the conjugate is 6-(-9)i which is equal to 6+9i
anonymous
  • anonymous
i mean 6+9i ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
yes , sorry
anonymous
  • 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
  • anonymous
9+5i * 6+9i ? 6-9i * 6+9i ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
I assume that the first line is the numerator and the second line is the denominator?
anonymous
  • anonymous
Just making sure
anonymous
  • anonymous
54 + 45i ? 36 + 81i ?
anonymous
  • anonymous
yes first line is the numerator and the second line is the denominator
anonymous
  • anonymous
Ok, you got the denominator right but (9+5i)*(6+9i) = 54 + 81i + 30i + 45
anonymous
  • anonymous
wait, sorry. the denominator is wrong too. it is 36+81
anonymous
  • anonymous
why not 81i ?
anonymous
  • anonymous
i^2 = -1 because i = \[\sqrt{-1}\] so i^2 = \[\sqrt{-1}\] * \[\sqrt{-1}\] = -1
anonymous
  • anonymous
agh sorry about not having it in one line
anonymous
  • anonymous
its okay (: , im just trying to understand it
anonymous
  • anonymous
do you understand how i^2 is -1?
anonymous
  • anonymous
so i would have to add the top and it'll be 99 + 111i ?
anonymous
  • anonymous
that it will result to the opposite ?
anonymous
  • 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
  • anonymous
So \[(9+5i)(6+9i) = 9*6 + 9*9i + 6*5i + 5*9*i^2\]
anonymous
  • anonymous
5*9*i^2 = -1*5*9. Just do the same thing for the denominator, too.
anonymous
  • anonymous
so its not (9+5i)*(6+9i) = 54 + 81i + 30i + 45 ?
anonymous
  • 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
  • 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
  • anonymous
my method is a traditional one. Johnfreeman uses the idea of rationalization.
anonymous
  • 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
  • anonymous
see this:
anonymous
  • anonymous
to multiply aswell ?
anonymous
  • 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
  • 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
  • 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
  • anonymous
see, the answer's just the same.

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