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anonymous

  • one year ago

Simplify the expression The root of negative nine over the quantity of three minus two i plus the quantity of one plus five i. .

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  1. Michele_Laino
    • one year ago
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    please can you write your expression using the editor?

  2. anonymous
    • one year ago
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    \[(\sqrt{-9})\div(3-2i) + (1+5i)\]

  3. Michele_Laino
    • one year ago
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    here we can rewrite your expression as follows: \[\Large \frac{{\sqrt { - 9} }}{{3 - 2i}} + 1 + 5i = \frac{{3i\left( {3 + 2i} \right)}}{{9 + 4}} + 1 + 5i\]

  4. Michele_Laino
    • one year ago
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    @jordynhazan1

  5. anonymous
    • one year ago
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    so that simplifies to \[\frac{ 9i+6i }{ 13 } +1 +5i\]

  6. anonymous
    • one year ago
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    right?

  7. Michele_Laino
    • one year ago
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    no, the simplified expression is: \[\Large \begin{gathered} \frac{{\sqrt { - 9} }}{{3 - 2i}} + 1 + 5i = \frac{{3i\left( {3 + 2i} \right)}}{{9 + 4}} + 1 + 5i = \hfill \\ \hfill \\ = \frac{{9i - 6}}{{13}} + 1 + 5i \hfill \\ \end{gathered} \]

  8. Michele_Laino
    • one year ago
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    am I right?

  9. Michele_Laino
    • one year ago
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    please remember that: \[\Large {i^2} = - 1\]

  10. Michele_Laino
    • one year ago
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    furthermore, I have used this identity: \[\Large \frac{1}{z} = \frac{{{z^*}}}{{{{\left| z \right|}^2}}}\] where z* is the complex conjugate of z

  11. anonymous
    • one year ago
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    i still have to solve for a number but don't know how to

  12. anonymous
    • one year ago
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    how do i add the 1 and the 5i to the equation

  13. Michele_Laino
    • one year ago
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    we have to compute the least common multiple, which is 13, so we can write: \[\Large \begin{gathered} \frac{{\sqrt { - 9} }}{{3 - 2i}} + 1 + 5i = \frac{{3i\left( {3 + 2i} \right)}}{{9 + 4}} + 1 + 5i = \hfill \\ \hfill \\ = \frac{{9i - 6}}{{13}} + 1 + 5i = \frac{{9i - 6 + 13\left( {1 + 5i} \right)}}{{13}} = ... \hfill \\ \end{gathered} \]

  14. Michele_Laino
    • one year ago
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    please simplify

  15. Michele_Laino
    • one year ago
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    hint: \[\Large \begin{gathered} \frac{{\sqrt { - 9} }}{{3 - 2i}} + 1 + 5i = \frac{{3i\left( {3 + 2i} \right)}}{{9 + 4}} + 1 + 5i = \hfill \\ \hfill \\ = \frac{{9i - 6}}{{13}} + 1 + 5i = \frac{{9i - 6 + 13\left( {1 + 5i} \right)}}{{13}} = \hfill \\ \hfill \\ = \frac{{9i - 6 + 13 + 65i}}{{13}} = ... \hfill \\ \end{gathered} \]

  16. anonymous
    • one year ago
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    then it becomes \[\frac{ 74i + 7}{ 13 }\]

  17. Michele_Laino
    • one year ago
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    ok! that's right!

  18. anonymous
    • one year ago
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    but that isn't near the answers given

  19. anonymous
    • one year ago
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  20. Michele_Laino
    • one year ago
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    Please wait I'm checking my computation

  21. Michele_Laino
    • one year ago
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    sorry your starting expression, at initial post, is different

  22. Michele_Laino
    • one year ago
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    no problem we restart to compute the right expression

  23. Michele_Laino
    • one year ago
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    here is the first step: \[\Large \frac{{\sqrt { - 9} }}{{3 - 2i + 1 + 5i}} = \frac{{3i}}{{4 + 3i}}\]

  24. Michele_Laino
    • one year ago
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    now we can write this: \[\Large \frac{1}{{4 + 3i}} = \frac{{4 - 3i}}{{16 + 9}}\] since I have used this identity: \[\Large \frac{1}{z} = \frac{{{z^*}}}{{{{\left| z \right|}^2}}}\]

  25. anonymous
    • one year ago
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    the bottom simplifies to 25

  26. Michele_Laino
    • one year ago
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    ok! substituting into the above expression, we get: \[\Large \begin{gathered} \frac{{\sqrt { - 9} }}{{3 - 2i + 1 + 5i}} = \frac{{3i}}{{4 + 3i}} = \hfill \\ \hfill \\ = \frac{{3i\left( {4 - 3i} \right)}}{{16 + 9}} \hfill \\ \end{gathered} \]

  27. anonymous
    • one year ago
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    the answer is C!!

  28. Michele_Laino
    • one year ago
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    or: \[\Large \begin{gathered} \frac{{\sqrt { - 9} }}{{3 - 2i + 1 + 5i}} = \frac{{3i}}{{4 + 3i}} = \hfill \\ \hfill \\ = \frac{{3i\left( {4 - 3i} \right)}}{{16 + 9}} = \frac{{3i\left( {4 - 3i} \right)}}{{25}} = ... \hfill \\ \end{gathered} \]

  29. anonymous
    • one year ago
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    \[\frac{ 9 + 12i }{ }\]

  30. Michele_Laino
    • one year ago
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    that's right! It is option C

  31. anonymous
    • one year ago
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    over 25

  32. anonymous
    • one year ago
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    thank you so much

  33. Michele_Laino
    • one year ago
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    :)

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