Question

Divide: 8-2i/5+i

Answer 1

\[\frac{ 8 - 2i }{ 5 + i }\]

Answer 2

no, that's just the beginning :)

Answer 3

Oh, sorry

Answer 4

do you know about the complex conjugate when you are dividing complex numbers like this?

Answer 5

How do you type that in like that on the computer?

Answer 6

You can use the equation button below your response area. It takes a little practice but it can be useful so you can be more clear in your equations.

Answer 7

Are you a teacher?

Answer 8

Not formally :) Just enjoy math.

For dividing complex (real + imaginary) numbers, you need to multiply the top and bottom of the fraction by the complex conjugate.

Answer 9

Ohhhh that makes sense! What grade are you in?

Answer 10

Me and Miranda are in the same class

Answer 11

Nice to meet you. I'm not in a grade... graduated already. But helping with math helps me to not forget what I know, or to re-learn what I may have forgotten :)

Answer 12

Thats good! What college do you plan to go to?

Answer 13

Can you please show us how to do this problem?

Answer 14

Ummm... let's get back to the problem :)

The complex conjugate is the real part and the imaginary part, but with the imaginary sign reversed... here, from a + to a -
\[\frac{ 8 - 2i }{ 5 + i } \times \frac{ 5 - i }{ 5 - i }\]

Answer 15

Do you see how that works? You had (5 + i) in the bottom of the fraction to start, so the complex conjugate is just 5 - i.... you just reverse the sign on the "i" term.

Answer 16

Okay

Answer 17

So, once you've set it up like that, you need to multiply out the top and the bottom.\[\frac{ 8-2i }{ 5+i }\times \frac{ 5-i }{ 5-i } = \frac{ (8-2i)(5-i) }{ (5-i)(5-i) }\]

Answer 18

Then what next?

Answer 19

you surely have learned that when you multiply i * i, you get a "-1", right?

Answer 20

yeah i^2

Answer 21

Yes

Answer 22

so on the top, you have (8 - 2i)(5 - i) = 40 -10i -8i +2i^2

that simplifies to 40 - 18i + 2i^2, but the i^2 is -1, so...
40 -18i - 2
or
38 - 18i

That's the numerator part.

Answer 23

Okay thank you

Answer 24

Ohhh that makes sense.. basically to get the numerator part you just box the equation

Answer 25

I guess... I don't know the term "box the equation", but it seems like you get the idea.

Now you need the denominator part...

(5 + i) (5 - i) = 25 +5i - 5i - i^2
= 25 - i^2
= 25 + 1
= 26

This shows the whole reason for doing that complex conjugate thing... it allows you to get rid of the imaginary number stuff in the denominator.

Answer 26

Thanks

Answer 27

Thank you!! I love you!! <3

Answer 28

Glad to help you both :) Thanks for the medal...

Answer 29

\[\sqrt{5} \over 9+\sqrt{7}\]

Answer 30

lol can you explain this problem?

Answer 31

On that last problem, you probably need to reduce the fraction...
(38 - 18i) / 26

you can divide it all through by 2...

(19 - 9i) / 13

Also, the best way to show your answer is to break it up into real and imaginary parts...

so 19/13 - (9/13)i

Answer 32

So, on your next problem, it's a similar idea... just as you don't want to have an imaginary number in the denominator, you don't want any roots or radicals in the denominator on this one.

So, you start with a denominator of 9 + sqrt(7)

You want to reverse the sign on the square root to get the conjugate version...
9 - sqrt(7)

\[\frac{ \sqrt{5} }{ 9 +\sqrt{7} } \times \frac{ 9 - \sqrt{7} }{ 9 - \sqrt{7} }\]

Answer 33

Then what next?

Answer 34

multiply :) Similar to last problem... multiply and simplify.

Answer 35

Okay! thank you :)

Answer 36

So, for a little help/hint, the bottom part should end with no radicals in it...

(9 + sqrt(7) ) (9 - sqrt(7) ) = 81 - 7 (because the square root terms drop out)
= 74

Answer 37

gotta go... sorry to run, but good luck! :)

Answer 38

Alright ! bye