Question

giving medal and becoming a fan if somebody can help me with this one question

Answer 1

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 2

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

Answer 3

@whpalmer4 @mathstudent55 @Preetha

Answer 4

thats not an answer choice

Answer 5

\[\frac{ 12-9i }{ 25}\]

Answer 6

\[\frac{ 12-9i }{ 7 }\]

Answer 7

\[\frac{ 9+12i }{ 25 } \frac{ 9+12i }{ 7 }\]

Answer 8

FIrst thing to do is to simplify the denominator.

\[(3−2i)+(1+5i)=\]

Answer 9

\[(3−2i)+(1+5i) = 3-2i+1+5i = 3+1-2i+5i = 4+3i\]Agreed?

Answer 10

yes i agree

Answer 11

That gives us \[\frac{\sqrt{-9}}{4+3i}\]Now we need to rid the denominator of the complex number. We do that by multiplying both numerator and denominator by the conjugate of the denominator, which is the same number, except we change the sign on the imaginary part.

\[\frac{\sqrt{-9}}{(4+3i)}*\frac{(4-3i)}{(4-3i)} = \]Can you do that for me, please?

Answer 12

ok hold on

Answer 13

for the denominators do you use the distribute property?

Answer 14

Just multiply \[(4+3i)(4-3i)\]you can use whatever method you use for multiplying polynomials.

Answer 15

or you can use the fact that they are the factors of a difference of squares.

Answer 16

\[a^2-b^2 = (a+b)(a-b)\]

Answer 17

oh ok

Answer 18

Here it is a number of different ways:

FOIL:
\[(4+3i)(4-3i) = 4*4 -4*3i + 4*3i -3i*3i = 16 - 12i + 12i -9i^2\]\[\qquad = 16-9i^2\]

Distributive property:
\[(4+3i)(4-3i) = 4(4-3i) + 3i(4-3i) = 16-12i + 12i - 9i^2\]\[\qquad=16-9i^2\]

Difference of squares:
\[(4+3i)(4-3i) = (a+b)(a-b) = a^2-b^2 = 4^2-(3i)^2\]\[\qquad=16-9i^2\]

In all of those cases, we replace \(i^2 = -1\) to get
\[(4+3i)(4-3i) = 16-9i^2 = 16-9(-1) = 16+9\]\[\qquad = 25\]

That means our fraction is now \[\frac{\sqrt{-9}(4-3i)}{25}\]

Answer 19

We're getting close to the end :-)

\[\sqrt{-9} = \sqrt{-1*9} = \sqrt{-1}*\sqrt{9} = i*3 = 3i\]Right?

Answer 20

right

Answer 21

Okay, so now we have \[\frac{3i(4-3i)}{25} = \]

Answer 22

You should be able to finish that...

Answer 23

yes thank you so much:))

Answer 24

you're welcome.