Question

any help is appreciated!! simplify 2-i/2+5i

Answer 1

Hey, buddy.

Answer 2

Okay, so there's this thing for complex numbers called the "complex conjugate." And that's a real fancy sounding name, but all it means is that we take the complex number, leave the real part the same, and change the sign of the imaginary part.
Here's an example:
2+4i has the complex conjugate
2-4i

1-3i has the complex conjugate
1+3i

-5 +i has the complex conjugate
-5-i

understand? Pretty simple stuff.

Answer 3

ya looks good to me!...basically the flip flop of the original equation

Answer 4

Yeah, you can think of it that way. So why do we care about the complex conjugate? Well, it's actually pretty cool. It's cool because any time we take a complex number and multiply it by its complex conjugate, the result is just a real number! Sweet, right?
So I said that the complex conjugate of 2 +4i is 2-4i. Let's see what happens when we multiply those two things. I'm going to basically FOIL them.
(2+4i)(2-4i) = 4 - 8i +8i -16i^2
= 4-16i^2 because the -8i and the +8i cancel
=4-16(-1) because i^2 =-1
=4+16
=20
So we multiplied them and just got a regular ol' number. Sweet!

Answer 5

lol ya i suppose that is pretty sweet!

Answer 6

You're probably thinking, "Great, but what does that have to do with this problem?"
Well, we have this fraction
\[\frac{2-i}{2+5i}\]
and we'd like to simplify it. The WAY we'll simplify it is by rewriting it so that it no longer has a complex number on the bottom. The complex conjugate gives us a really cool way to do that.
We know that if we multiply by the same thing on top and bottom, it doesn't change the value of the fraction. ALSO, we know that if we multiply a complex number by its complex conjugate, we'll end up with just a regular number.

Answer 7

So, you tell me what we're going to do.

Answer 8

hmm...good question i suppose...multiply the complex number on the bottom by the top and bottom in order to cancel it out on the bottom and move it to the top and then solve for the top by using the complex conjugate?...im i kinda right or am i making up a whole new math lol

Answer 9

am i*

Answer 10

You might be mildly making a new math. Let's keep it simple.
We have a complex number on the bottom. How do we make it into NOT a complex number?

Answer 11

wave a magic wand lol.....im thinking cancel it out somehow..which i would normally do by multiplying the i by the top and bottom

Answer 12

Did you forget about the complex conjugate?

Answer 13

no...im just looking at some notes i got from my teacher...the complex conjugate still includes the i though...so how would that help if you were trying to get rid of it?...there's obviously a reason my user name is mathdummy haha....i tried to solve this early and i got (1/29)-(12/29)i...am i at least close??

Answer 14

oo wait scratch that i see y you need the conjugate

Answer 15

Well, maybe. Let's talk about the process though. That's the important part. How do we GET to the answer. Not WHAT IS the answer.

Answer 16

multiple the conjugate of 2+5i by the top and bottom

Answer 17

The key fact to understand here is that if we have a complex number and we want it to become just a real number we
multiply it by its complex conjugate.

Answer 18

Good, good.
So we look at the bottom, say to ourselves, "Let's turn that into a real number by multiplying it by its conjugate."

But THEN, we say to ourselves, "Oh wait, since I'm multiplying on bottom, I have to multiply on top to keep it the same."

Answer 19

agreed..so when working it out you factor the denominator in this case (2+5i)(2-5i) and you get 4-10i+10i-25i...correct?

Answer 20

and the 10i cancel out...so you end up with 4-(-25i)...or at least i did

Answer 21

Tiny, but important mistake
4-10i+10i-25i^2
Then the -10i+10i cancel and you get just
4-25i^2

Answer 22

oo right..oops..and in this case...in know the i^2=-1....but because its attached to the 25 can u still do that?....im assuming not...sadly...that would make my life so much easier haha

Answer 23

You totally can do that! And you should =)

Answer 24

oo yes! finally something about math that goes my way! haha...okay so i did that..and for my shiny new answer i got -1/29-12/29i...

Answer 25

Okay, so I guess you already did the work for the top. Here it is, just to confirm it for you:
(2-i)(2-5i) =
4 -10i-2i +5i^2 =
4-12i +5(-1) =
4 -12i -5 =
-1-12i

And then, since the bottom was 29, we have
\[\frac{-1-12i}{29} = \frac{-1}{29}-\frac{12}{29}i\]

Answer 26

yep that's what i got!..thank you so much haha if you aren't a teacher already then you should seriously consider becoming one

Answer 27

Any time, my little pumpkin. =) And I am already a teacher! I subbed for this past year, and this year I'll have my own high school math classroom! c:

Answer 28

well your sure ready for it! hahaha thanks again..im sure ill be in need of your assistance again soon:)