Question

the question will be below. Not sure how to do this

Answer 1

Simplify the expression \[\frac{ \sqrt{-16} }{ (3-3i)+(1-2i) }\]

Answer 2

@Jaynator495

Answer 3

@zepdrix I did some things my lesson showed me but it makes no sense....

Answer 4

Notice that there is nothing special about the brackets in the denominator.
There is nothing being applied to the outside of them.
So we can simply drop the brackets,\[\large\rm \frac{\sqrt{-16}}{3-3i+1-2i}\]And then proceed to combine like-terms in the denominator.

Answer 5

so then it becomes \[\frac{ \sqrt{-16} }{ 4-5i }\]

Answer 6

Ok great.
Let's turn the numerator into something else,\[\large\rm \sqrt{-16}\quad=\sqrt{-1\cdot16}\quad=\sqrt{-1}\cdot\sqrt{16}\quad=?\]

Answer 7

these are my possible answers and its confusing me because none of the numbers match with what I had gotten, and would that equal to 4?

Answer 8

because you cannot square a negative right? or am I wrong.. I could be

Answer 9

\[\large\rm \sqrt{-16}\quad=\sqrt{-1\cdot16}\quad=\sqrt{-1}\cdot\color{orangered}{\sqrt{16}}\quad=\sqrt{-1}\cdot \color{orangered}{4}\]
Well the positive 16 part of it is turning into a 4, ya?
But the sqrt(-1) is how we define our imaginary unit i.

Answer 10

\[\large\rm \sqrt{-16}\quad=\sqrt{-1\cdot16}\quad=\sqrt{-1}\cdot\color{orangered}{\sqrt{16}}\quad=\color{orangered}{4}i\]So we replace sqrt(-1) with i.

Answer 11

ok so then the top becomes 4i? \[\frac{ 4i }{ 4-5i }\]

Answer 12

k looks good.

Answer 13

ok, what do I need to do to the i that is on the top and bottom? I think this is the next step?..

Answer 14

Next we want to multiply the top and bottom by the `conjugate of the denominator`.
Remember what a conjugate looks like?

It looks the same as the original, but with opposite sign on the imaginary part.
So the conjugate of \(\large\rm 4-5i\) will be \(\large\rm 4+5i\).

Answer 15

the top and bottom get multiplied by 4+5i? \[\frac{ 4i }{ 4-5i }\times \frac{ 4+5i }{ 4+5i }\]

it would then look like this?

Answer 16

k good!

Answer 17

so now do I multiply the like terms?

Answer 18

You'll have to distribute.
When you multiply like that, imagine brackets around each thing being multiplied.

In the numerator: \(\large\rm 4i(4+5i)\)
In the denominator: \(\large\rm (4-5i)(4+5i)\)

Answer 19

so factoring? ok give me a minute

Answer 20

\[\frac{ 16i+20i }{ 16+20i-20i-25i }\] with it the like terms combined on the bottom \[\frac{ 16i+20i }{ 16-25i }\]

Answer 21

Messed up a couple of terms.\[\large\rm -5i(5i)\ne 25i\]\[\large\rm 4i(5i)\ne 20i\]

Answer 22

\[\large\rm 4i(5i)=20i^2\]Remember what happens when you square i?

Answer 23

when I square i it becomes \[\sqrt{-1}\]

Answer 24

\[\large\rm i=\sqrt{-1}\]When you square it, it becomes,\[\large\rm i^2\quad=\sqrt{-1}\cdot\sqrt{-1}\quad=-1\]

Answer 25

so the places where a \[i^{2}\] should be it really just gets multiplied by -1?

Answer 26

Yes.
So our expansion should have looked like this:\[\large\rm \frac{ 16i+\color{orangered}{20i^2} }{ 16+20i-20i\color{orangered}{-25i^2} }\]and then we'll replace the i^2 with -1 in each place.

Answer 27

ok so then we have \[\frac{ 16i-20 }{ 16+25 }\] then add the bottom and get \[\frac{ 16i-20 }{ 41 }\]

Answer 28

Correct!
Yay good job!

Answer 29

thanks man :D

Answer 30

np