Question

Complex numbers help. Will give medal and fan. *Question attached below*

Answer 1

Simplify, without the use of tables:

Can I have a step-by-step approach as to how to solve this, please? I think the use of de Moivre's will be required?

Answer 2

Lemme write it out a sec :)

Answer 3

\[\Large\rm \frac{\left(\cos\frac{\pi}{7}-\mathcal i \sin\frac{\pi}{7}\right)^3}{\left(\cos\frac{\pi}{7}+\mathcal i \sin\frac{\pi}{7}\right)^4}\]

Answer 4

Have you learned about the complex exponential form yet?
This would be much easier to handle as exponentials.
If not, that's ok.

Answer 5

Taxi! Where you at >.<

Answer 6

Oh I'm sorry! I got caught up in some other homework :C
We haven't really done that much on complex exponential :O @zepdrix

Answer 7

Then I guess..... yes.. De'Moivre's Theorem is probably a good place to start :)

Answer 8

\[\Large\rm \frac{\cos\frac{3\pi}{7}-\mathcal i \sin\frac{3\pi}{7}}{\cos\frac{4\pi}{7}+\mathcal i \sin\frac{4\pi}{7}}\]So we get something like that after the first step, yes?

Answer 9

Yes :)

Answer 10

Mmmmmm ok.
This... Mmm yer not gonna like this.. but..
oh oh oh no no no, we can avoid that actually.. we have another nice trick we can pull here.

Answer 11

err no maybe we can't.. mmm thinking.
the problem is the subtraction in the numerator.
ok ok ok let's try to deal with it.. I hope your head doesn't explode here.

Answer 12

cosine is an `even` function, \(\Large\rm \cos(x)=\cos(-x)\)
while sine is an `odd` function, \(\Large\rm -\sin(x)=\sin(-x)\)

Answer 13

We can rewrite our numerator:\[\large\rm \cos\frac{3\pi}{7}-\mathcal i \sin\frac{3\pi}{7}\quad =\quad\cos\left(-\frac{3\pi}{7}\right)+\mathcal i\sin\left(-\frac{3\pi}{7}\right)\]

Answer 14

Chew on that for a sec, I'll be back.

Answer 15

Alrighty :)

Answer 16

So our problem is currently:\[\Large\rm \frac{\cos\left(-\frac{3\pi}{7}\right)+\mathcal i\sin\left(-\frac{3\pi}{7}\right)}{\cos\frac{4\pi}{7}+\mathcal i \sin\frac{4\pi}{7}}\]What the last step ok? :d

Answer 17

Was*

Answer 18

Yes the last step was fine

Answer 19

There is some rule for dividing complex trig expressionssssss
mmmm I can't remember what it's called, but it looks like this:\[\Large\rm \frac{a(\cos(x)+\mathcal i \sin(x))}{b(\cos(y)+\mathcal i \sin(y))}=\frac{a}{b}\cos(x-y)+\mathcal i \sin(x-y)\]

Answer 20

See how we can use that rule? :O

Answer 21

Hmm, yes I see that :)

Answer 22

So apply it!
What do you get? c:

Answer 23

Give me a sec :)

Answer 24

Woops, the a/b was supposed to be multiplying both terms :O
Hope that was clear,\[\Large\rm \frac{a(\cos(x)+\mathcal i \sin(x))}{b(\cos(y)+\mathcal i \sin(y))}=\frac{a}{b}\left(\cos(x-y)+\mathcal i \sin(x-y)\right)\]

Answer 25

Oh yeah...I see it! Thanks for the correction :)