anonymous
  • anonymous
Simplify the expression The root of negative nine over the quantity of three minus two i plus the quantity of one plus five i. .
Mathematics
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anonymous
  • anonymous
Simplify the expression The root of negative nine over the quantity of three minus two i plus the quantity of one plus five i. .
Mathematics
chestercat
  • chestercat
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Michele_Laino
  • Michele_Laino
please can you write your expression using the editor?
anonymous
  • anonymous
\[(\sqrt{-9})\div(3-2i) + (1+5i)\]
Michele_Laino
  • Michele_Laino
here we can rewrite your expression as follows: \[\Large \frac{{\sqrt { - 9} }}{{3 - 2i}} + 1 + 5i = \frac{{3i\left( {3 + 2i} \right)}}{{9 + 4}} + 1 + 5i\]

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Michele_Laino
  • Michele_Laino
anonymous
  • anonymous
so that simplifies to \[\frac{ 9i+6i }{ 13 } +1 +5i\]
anonymous
  • anonymous
right?
Michele_Laino
  • Michele_Laino
no, the simplified expression is: \[\Large \begin{gathered} \frac{{\sqrt { - 9} }}{{3 - 2i}} + 1 + 5i = \frac{{3i\left( {3 + 2i} \right)}}{{9 + 4}} + 1 + 5i = \hfill \\ \hfill \\ = \frac{{9i - 6}}{{13}} + 1 + 5i \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
am I right?
Michele_Laino
  • Michele_Laino
please remember that: \[\Large {i^2} = - 1\]
Michele_Laino
  • Michele_Laino
furthermore, I have used this identity: \[\Large \frac{1}{z} = \frac{{{z^*}}}{{{{\left| z \right|}^2}}}\] where z* is the complex conjugate of z
anonymous
  • anonymous
i still have to solve for a number but don't know how to
anonymous
  • anonymous
how do i add the 1 and the 5i to the equation
Michele_Laino
  • Michele_Laino
we have to compute the least common multiple, which is 13, so we can write: \[\Large \begin{gathered} \frac{{\sqrt { - 9} }}{{3 - 2i}} + 1 + 5i = \frac{{3i\left( {3 + 2i} \right)}}{{9 + 4}} + 1 + 5i = \hfill \\ \hfill \\ = \frac{{9i - 6}}{{13}} + 1 + 5i = \frac{{9i - 6 + 13\left( {1 + 5i} \right)}}{{13}} = ... \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
please simplify
Michele_Laino
  • Michele_Laino
hint: \[\Large \begin{gathered} \frac{{\sqrt { - 9} }}{{3 - 2i}} + 1 + 5i = \frac{{3i\left( {3 + 2i} \right)}}{{9 + 4}} + 1 + 5i = \hfill \\ \hfill \\ = \frac{{9i - 6}}{{13}} + 1 + 5i = \frac{{9i - 6 + 13\left( {1 + 5i} \right)}}{{13}} = \hfill \\ \hfill \\ = \frac{{9i - 6 + 13 + 65i}}{{13}} = ... \hfill \\ \end{gathered} \]
anonymous
  • anonymous
then it becomes \[\frac{ 74i + 7}{ 13 }\]
Michele_Laino
  • Michele_Laino
ok! that's right!
anonymous
  • anonymous
but that isn't near the answers given
anonymous
  • anonymous
Michele_Laino
  • Michele_Laino
Please wait I'm checking my computation
Michele_Laino
  • Michele_Laino
sorry your starting expression, at initial post, is different
Michele_Laino
  • Michele_Laino
no problem we restart to compute the right expression
Michele_Laino
  • Michele_Laino
here is the first step: \[\Large \frac{{\sqrt { - 9} }}{{3 - 2i + 1 + 5i}} = \frac{{3i}}{{4 + 3i}}\]
Michele_Laino
  • Michele_Laino
now we can write this: \[\Large \frac{1}{{4 + 3i}} = \frac{{4 - 3i}}{{16 + 9}}\] since I have used this identity: \[\Large \frac{1}{z} = \frac{{{z^*}}}{{{{\left| z \right|}^2}}}\]
anonymous
  • anonymous
the bottom simplifies to 25
Michele_Laino
  • Michele_Laino
ok! substituting into the above expression, we get: \[\Large \begin{gathered} \frac{{\sqrt { - 9} }}{{3 - 2i + 1 + 5i}} = \frac{{3i}}{{4 + 3i}} = \hfill \\ \hfill \\ = \frac{{3i\left( {4 - 3i} \right)}}{{16 + 9}} \hfill \\ \end{gathered} \]
anonymous
  • anonymous
the answer is C!!
Michele_Laino
  • Michele_Laino
or: \[\Large \begin{gathered} \frac{{\sqrt { - 9} }}{{3 - 2i + 1 + 5i}} = \frac{{3i}}{{4 + 3i}} = \hfill \\ \hfill \\ = \frac{{3i\left( {4 - 3i} \right)}}{{16 + 9}} = \frac{{3i\left( {4 - 3i} \right)}}{{25}} = ... \hfill \\ \end{gathered} \]
anonymous
  • anonymous
\[\frac{ 9 + 12i }{ }\]
Michele_Laino
  • Michele_Laino
that's right! It is option C
anonymous
  • anonymous
over 25
anonymous
  • anonymous
thank you so much
Michele_Laino
  • Michele_Laino
:)

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