Question

Can somebody PLEASE help me?

Answer 1

\[\frac{3-3i}{3+4i}\]

To rationalize the denominator, multiply both numerator and denominator by the conjugate of the denominator. The conjugate of the denominator is the same as the denominator with the sign of the term containing \(i\) changed to the opposite sign.

In other words, multiply your fraction by \(\large\frac{3-4i}{3-4i}\) and simplify the result. Don't forget that \(i^2 = -1\)

Answer 2

I'm very confused....

Answer 3

Wait...
Is it.. A?

Answer 4

did you work the problem, or are you guessing?

Answer 5

I'm wondering if we need to work on your guessing skills, or your algebra skills :-)

Answer 6

I feel that it's A

Answer 7

Algebra skills, I'm completely freaking lost /:

Answer 8

Okay, no problem.

First, it's not A :-)

We need to get rid of the \(i\) in the denominator. To do that, we'll make use of the fact that \(i^2 = -1\) and the difference of squares factoring.

\[(a+bi)(a-bi) = a(a-bi) + bi(a-bi) = a*a - abi + abi -b^2i^2\]\[\qquad=a^2-b^2i^2\]\[\qquad=a^2-b^2(-1)\]\[\qquad = a^2+b^2\]
Do you have any questions about that?

Answer 9

I believe I understand that..

Answer 10

okay, good. In this problem, we have \(3+4i\) as the denominator. That means \(a = 3, b = 4\). To use the difference of squares thing to get rid of that pesky \(i\), we'll need to multiply the denominator by \(3-4i\). Agreed?

Answer 11

Yes.

Answer 12

Okay, could you multiply \[(3+4i)(3-4i)\]for me and tell me what you get?

Answer 13

9-12i+12i-12i?

Answer 14

So 9-12i

Answer 15

regrettably, no.

do you use that awful FOIL crutch, or do you use the distributive property to do your multiplication?

Answer 16

I thought I had to use the FOIL.... The lesson tells me to..
That's why I'm so confused/:

Answer 17

You did look at the picture right? Don't I need the i in the problem?

Answer 18

Well, okay, minor rant time: I dislike people being taught FOIL because it only teaches them how to multiply something like \[(a+b)(c+d)\]and they are totally up the creek without a paddle as soon as they have to do something like \[(a+b+c)(d+e)\]

Whereas if you learn multiplication with the distributive property (which you already know!), you can handle any number of terms (subject to your ability to keep track of your place, of course!)

In any case, you've made a mistake in your multiplication.

\[(3+4i)(3-4i) = 3*3 + 3(-4i) + 4i(3) +4i(-4i)\]yes?

Answer 19

I think I did that in FOIL order...

Answer 20

Okay yes.

Answer 21

Okay, could you simplify that for me, please?

Answer 22

9-12i+12i-12i....? I don't get it.

Answer 23

what is \(4i*(-4i)=\)

Answer 24

-12i^2?

Answer 25

what is 4*4?

Answer 26

16 sorry... I so knew that. This is just really making me mad is all. I can't understand it...... so 16i^2

Answer 27

yes, so we have \[9-12i + 12i -16i^2 = 9\cancel{-12i}\cancel{+12i}-16i^2 = 9-16i^2\]right?

Answer 28

Okay yes..

Answer 29

and \(i^2=-1\) right?
\[i = \sqrt{-1}\]\[i*i = \sqrt{-1}*\sqrt{-1} = -1\]

Answer 30

so \[9-16i^2 = \]

Answer 31

Yes.

Answer 32

9+16?

Answer 33

which equals...

Answer 34

25

Answer 35

very good. so 25 is the denominator of our rationalized fraction. for the numerator, we have to multiply the existing numerator by the same thing we multiplied the denominator by:

\[(3-3i)(3-4i) =\]

Answer 36

Okay, so ... I'm lost again...

Answer 37

are you lost because you don't know how to do that multiplication, or you don't understand why we need to it, or <insert your question here>

Answer 38

Idk how to do that multiplication..

Answer 39

do you know how to multiply \[(3-3x)(3-4x)\]?

Answer 40

With the FOIL method /:

Answer 41

well, that's exactly what you do here, too. Except when you are done, if you have an \(i^2\) in there somewhere, you can replace it with -1

Answer 42

just treat \(i\) like you would any other variable, it's just that you happen to know its value and can use that to simplify some more in some cases

Answer 43

Okay so... 9-12x-9x+12x^2
which simplifies to 9-21x+12x^2..

Answer 44

yes, except we're using \(i\) instead of \(x\)

Answer 45

now, do you have a term which has \(i^2\) in it?

Answer 46

Okay so 9-21i+12i^2
Yes so that changes too 9-21x-12
Which turns into -3-21x?

Answer 47

So the answer is C!

Answer 48

???

Answer 49

no, we don't have any x's in our work, please...

\[9-21i-12 = -3-21i\]

so the whole problem is \[\frac{3-3i}{3+4i} = \frac{(3-3i)(3-4i)}{(3+4i)(3-4i)} = \frac{9-12i-9i+12i^2}{3^2+4^2}=\frac{9-21i-12}{25}\]\[\qquad=\frac{-3-21i}{25}\]

Answer 50

I got it!!! Could you help me with the next one...? It's subtraction..

Answer 51

Notice that for the denominator, you don't have to actually do the multiplication: you can just write down the denominator! if your old denominator is \(3+4i\), that's \(a=3,b=4\) and the rationalized denominator is just \(a^2+b^2 = 9+16 = 25\)

Answer 52

okay, sure why not?

Answer 53

(11-9i)-(15-12i)

Answer 54

Okay, this is a bit tricky for those who aren't paying attention!

\[(11-9i)-(15-12i) = (11-9i)-1(15-12i)\]
Can you use the distributive property to remove the parentheses?

Answer 55

Yes I believe so.

Answer 56

Okay, will you please do so? :-)

Answer 57

-11+9i-15+12i?

Answer 58

oh, I was about to profess my undying love for you, but then I noticed you made a mistake in a spot where I wouldn't have expected it!

Answer 59

:-)

Answer 60

What? xD

Answer 61

it is difficult to believe, I admit, but could you check your work on the first ( ) expression?

Answer 62

Would it be 11-9i-15+12i

Answer 63

Yes, it would.
Now would you please simplify that?

Answer 64

-4+3i?

Answer 65

the odds-on bet is that if you give any random OpenStudy student a problem where they have to simplify something like \[-(2-3a)\]they'll get that part wrong because they don't apply the negative sign to everything. You sailed right through that part, good for you!

Yes, \[(11-9i)-(15-12i) = 11-9i-15-(-12i) = -4-9i+12i = -4+3i\]

Answer 66

Okay so.. Addition..

Answer 67

yes...what about it?

Answer 68

(4-10i)+(7-3i)

Answer 69

Okay, use the distributive property just like you did in the previous problem.

Answer 70

Wouldn't it be 4-10i+7-3i?

Answer 71

which simplifies to...

Answer 72

11-13i?

Answer 73

for all the marbles, is that your final answer?

Answer 74

Yes!

Answer 75

Ding ding ding! We have a winner!

Answer 76

Yay! now multiplying...
(5+3i)(4+2i)

Answer 77

Okay, use your FOIL skilz, and remember to substitute \(i^2=-1\) after the dust settles

Answer 78

here, we can a race, I'll do it in my head :-)

Go!

Answer 79

done yet? :-)

Answer 80

20+10i+12i+6i^2
20+10i+12i-6
14+22i!

Answer 81

wow, a tie! and we both got the right answer :-)

Answer 82

Wanna race again xD
(8-3i)(6+5i)

Answer 83

48+40i-18i-15i^2
48+40i-18i+15
63+22i!

Answer 84

very good!

Answer 85

I have another division one... Will you walk me through it one more time?

Answer 86

sure. what's the question?

Answer 87

\[\frac{ (6-i) }{ (5+4i)}\]

Answer 88

Okay, what is the conjugate of the denominator?

Answer 89

5+4i

Answer 90

???

Answer 91

No, that is the denominator. The conjugate of the denominator is the denominator with the sign of the imaginary portion (the part that has \(i\) in it\) changed...

Answer 92

oh so 5-4?

Answer 93

close, but not close enough...what happened to the \(i\)?

Answer 94

5-4i?

Answer 95

that's more like it. Now, you multiply your fraction by the fraction that is the conjugate of the denominator as both numerator and denominator. remember, anything (except 0) divided by itself = 1, so we can multiply by that fraction without changing the value of our original fraction...

Answer 96

30-24i-5i+4i^2
30-24i-5i-4?