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. .

1. Michele_Laino

2. anonymous

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

3. Michele_Laino

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

4. Michele_Laino

@jordynhazan1

5. anonymous

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

6. anonymous

right?

7. Michele_Laino

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

8. Michele_Laino

am I right?

9. Michele_Laino

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

10. 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

11. anonymous

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

12. anonymous

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

13. Michele_Laino

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

14. Michele_Laino

15. Michele_Laino

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

16. anonymous

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

17. Michele_Laino

ok! that's right!

18. anonymous

but that isn't near the answers given

19. anonymous

20. Michele_Laino

Please wait I'm checking my computation

21. Michele_Laino

sorry your starting expression, at initial post, is different

22. Michele_Laino

no problem we restart to compute the right expression

23. Michele_Laino

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

24. Michele_Laino

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

25. anonymous

the bottom simplifies to 25

26. Michele_Laino

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

27. anonymous

28. Michele_Laino

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

29. anonymous

$\frac{ 9 + 12i }{ }$

30. Michele_Laino

that's right! It is option C

31. anonymous

over 25

32. anonymous

thank you so much

33. Michele_Laino

:)