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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    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. :)

  2. anonymous
    • 5 years ago
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    btw, the standard form is a+bi

  3. anonymous
    • 5 years ago
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    by 6-9i ?

  4. anonymous
    • 5 years ago
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    6+9i. b in this case would be -9 so the conjugate is 6-(-9)i which is equal to 6+9i

  5. anonymous
    • 5 years ago
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    i mean 6+9i ?

  6. anonymous
    • 5 years ago
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    yes

  7. anonymous
    • 5 years ago
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    yes , sorry

  8. anonymous
    • 5 years ago
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    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.

  9. anonymous
    • 5 years ago
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    9+5i * 6+9i ? 6-9i * 6+9i ?

  10. anonymous
    • 5 years ago
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    yes

  11. anonymous
    • 5 years ago
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    I assume that the first line is the numerator and the second line is the denominator?

  12. anonymous
    • 5 years ago
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    Just making sure

  13. anonymous
    • 5 years ago
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    54 + 45i ? 36 + 81i ?

  14. anonymous
    • 5 years ago
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    yes first line is the numerator and the second line is the denominator

  15. anonymous
    • 5 years ago
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    Ok, you got the denominator right but (9+5i)*(6+9i) = 54 + 81i + 30i + 45

  16. anonymous
    • 5 years ago
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    wait, sorry. the denominator is wrong too. it is 36+81

  17. anonymous
    • 5 years ago
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    why not 81i ?

  18. anonymous
    • 5 years ago
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    i^2 = -1 because i = \[\sqrt{-1}\] so i^2 = \[\sqrt{-1}\] * \[\sqrt{-1}\] = -1

  19. anonymous
    • 5 years ago
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    agh sorry about not having it in one line

  20. anonymous
    • 5 years ago
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    its okay (: , im just trying to understand it

  21. anonymous
    • 5 years ago
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    do you understand how i^2 is -1?

  22. anonymous
    • 5 years ago
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    so i would have to add the top and it'll be 99 + 111i ?

  23. anonymous
    • 5 years ago
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    that it will result to the opposite ?

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

  25. anonymous
    • 5 years ago
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    So \[(9+5i)(6+9i) = 9*6 + 9*9i + 6*5i + 5*9*i^2\]

  26. anonymous
    • 5 years ago
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    5*9*i^2 = -1*5*9. Just do the same thing for the denominator, too.

  27. anonymous
    • 5 years ago
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    so its not (9+5i)*(6+9i) = 54 + 81i + 30i + 45 ?

  28. anonymous
    • 5 years ago
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    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

  29. anonymous
    • 5 years ago
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    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.

  30. anonymous
    • 5 years ago
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    my method is a traditional one. Johnfreeman uses the idea of rationalization.

  31. anonymous
    • 5 years ago
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    yes thats the answer they give here as an option :) so to solve problems with i i would have to use logarithm of imaginaries?

  32. anonymous
    • 5 years ago
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    see this:

  33. anonymous
    • 5 years ago
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    to multiply aswell ?

  34. anonymous
    • 5 years ago
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    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.

  35. anonymous
    • 5 years ago
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    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.

  36. anonymous
    • 5 years ago
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    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

  37. anonymous
    • 5 years ago
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    see, the answer's just the same.

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