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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. .
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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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2please can you write your expression using the editor?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(\sqrt{9})\div(32i) + (1+5i)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2here we can rewrite your expression as follows: \[\Large \frac{{\sqrt {  9} }}{{3  2i}} + 1 + 5i = \frac{{3i\left( {3 + 2i} \right)}}{{9 + 4}} + 1 + 5i\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so that simplifies to \[\frac{ 9i+6i }{ 13 } +1 +5i\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2no, 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
 one year ago
Best ResponseYou've already chosen the best response.2please remember that: \[\Large {i^2} =  1\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2furthermore, I have used this identity: \[\Large \frac{1}{z} = \frac{{{z^*}}}{{{{\left z \right}^2}}}\] where z* is the complex conjugate of z

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i still have to solve for a number but don't know how to

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how do i add the 1 and the 5i to the equation

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2we 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
 one year ago
Best ResponseYou've already chosen the best response.2hint: \[\Large \begin{gathered} \frac{{\sqrt {  9} }}{{3  2i}} + 1 + 5i = \frac{{3i\left( {3 + 2i} \right)}}{{9 + 4}} + 1 + 5i = \hfill \\ \hfill \\ = \frac{{9i  6}}{{13}} + 1 + 5i = \frac{{9i  6 + 13\left( {1 + 5i} \right)}}{{13}} = \hfill \\ \hfill \\ = \frac{{9i  6 + 13 + 65i}}{{13}} = ... \hfill \\ \end{gathered} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then it becomes \[\frac{ 74i + 7}{ 13 }\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2ok! that's right!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but that isn't near the answers given

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2Please wait I'm checking my computation

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2sorry your starting expression, at initial post, is different

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2no problem we restart to compute the right expression

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2here is the first step: \[\Large \frac{{\sqrt {  9} }}{{3  2i + 1 + 5i}} = \frac{{3i}}{{4 + 3i}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2now 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
 one year ago
Best ResponseYou've already chosen the best response.0the bottom simplifies to 25

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2ok! substituting into the above expression, we get: \[\Large \begin{gathered} \frac{{\sqrt {  9} }}{{3  2i + 1 + 5i}} = \frac{{3i}}{{4 + 3i}} = \hfill \\ \hfill \\ = \frac{{3i\left( {4  3i} \right)}}{{16 + 9}} \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2or: \[\Large \begin{gathered} \frac{{\sqrt {  9} }}{{3  2i + 1 + 5i}} = \frac{{3i}}{{4 + 3i}} = \hfill \\ \hfill \\ = \frac{{3i\left( {4  3i} \right)}}{{16 + 9}} = \frac{{3i\left( {4  3i} \right)}}{{25}} = ... \hfill \\ \end{gathered} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 9 + 12i }{ }\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2that's right! It is option C
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