Question

Simple Complex Number::
http://i289.photobucket.com/albums/ll211/zephyrxX_Y/ScreenShot2013-04-01at70546PM_zpsfebc8e68.png

I don't understand how the fraction turned into the final answer.

I thought that you're supposed to do 2pi-(-4pi/3)

Please help explain!

Answer 1

simple complex number lol

Answer 2

lol I just realised hahaha.

Answer 3

Well, this seems straightforward enough :)

Answer 4

How did they get the answer O_o?

Answer 5

Hang on, let me write it out for you :)
\[\huge \frac{8e^{i2\pi}}{4e^{\frac{4\pi}{3}}}\]

Answer 6

you forgot the minus and the i for the denominator =)

Answer 7

Me? Forget? hah ^.^
But yeah, glad you caught that :)

First, let's deal with the "normal" numbers 8 and 4...
\[\huge \frac{8e^{i2\pi}}{4e^{-i\frac{4\pi}{3}}}\]
\[\huge 2\times \frac{e^{i2\pi}}{e^{-i\frac{4\pi}{3}}}\]

Answer 8

OK ;3

Answer 9

Now, normally, I'd tell you to use the laws of exponents, but before anything else...
\[\huge e^{i\theta}=\cos\theta \ + \ i\sin\theta\]

Answer 10

Aha. Got that ;3.

Answer 11

Well, what does that mean for
\[\huge e^{i2\pi}\]?

Answer 12

urhm. cos2pi+isin2pi

Answer 13

Yes. And?

Answer 14

andddd~? O__________O

Answer 15

And?
What's cos 2π ?
What's sin 2π ?

Answer 16

oh. cos2pi is 1 and sin2pi is 0

Answer 17

That's right. So in the end
\[\huge e^{i2\pi}=\cos 2\pi \ + \ i\sin2\pi \ = \ 1 +i 0 = 1\]

Answer 18

okaayy =3.

Answer 19

You still don't know where t go from here? :/

Answer 20

so I do it for the denominator as well

Answer 21

No need. But you do know now that the numerator is just 1.

Answer 22

OH RIGHT O____O

Answer 23

And work from there, silly :)

Answer 24

LOLLLL

Answer 25

HAHAHHAHAHAHAHAHHAH. OK. GET IT NOW. NOT EVEN FUNNY TO MISS THAT I DON"T HAVE TO DO ANYTHING ELSE AFTER KNOWING THE NUMERATOR IS 1.

Answer 26

Thankk youu! =DDD

Answer 27

Alright. Thanks, Pan xD. I'm Mai. And, for this form of complex number, can't I just do the normal power of numerator minus power of denominator thing? @_@?

Answer 28

You can do that, but it might take longer. Call me Peter ^.^

Answer 29

\[\Large 2e^{i(2\pi+\frac{4\pi}{3})}=2\left[\cos\left(2\pi+\frac{4\pi}{3}\right)+i\sin\left(2\pi+\frac{4\pi}{3}\right)\right]\]

Answer 30

And I already don't want to proceed :)

Answer 31

O___O wow. ok.

Answer 32

Thanks, Peter! Gotta go now. Mocks are tmr T^T.

Answer 33

bye :)