Question

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

Answer 1

please can you write your expression using the editor?

Answer 2

\[(\sqrt{-9})\div(3-2i) + (1+5i)\]

Answer 3

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

Answer 4

@jordynhazan1

Answer 5

so that simplifies to \[\frac{ 9i+6i }{ 13 } +1 +5i\]

Answer 6

right?

Answer 7

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

Answer 8

am I right?

Answer 9

please remember that:
\[\Large {i^2} = - 1\]

Answer 10

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

Answer 11

i still have to solve for a number but don't know how to

Answer 12

how do i add the 1 and the 5i to the equation

Answer 13

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

Answer 14

please simplify

Answer 15

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

Answer 16

then it becomes \[\frac{ 74i + 7}{ 13 }\]

Answer 17

ok! that's right!

Answer 18

but that isn't near the answers given

Answer 19

Please wait I'm checking my computation

Answer 20

sorry your starting expression, at initial post, is different

Answer 21

no problem we restart to compute the right expression

Answer 22

here is the first step:
\[\Large \frac{{\sqrt { - 9} }}{{3 - 2i + 1 + 5i}} = \frac{{3i}}{{4 + 3i}}\]

Answer 23

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

Answer 24

the bottom simplifies to 25

Answer 25

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

Answer 26

the answer is C!!

Answer 27

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

Answer 28

\[\frac{ 9 + 12i }{ }\]

Answer 29

that's right! It is option C

Answer 30

over 25

Answer 31

thank you so much

Answer 32

:)