The Student Room Group

Fp3 Help

https://2d31ca6ec4f6115572728c3a9168ad5dfc897f33.googledrive.com/host/0B1ZiqBksUHNYSjAyWFYyQmU5Tjg/OCR/FP3%20Complex%20Numbers.pdf


q1ii?

How am I supposed to draw it?
Like I don't know what 1+w is, if someone can explain how to work that out.
w=cos(2π5)+isin(2π5)w=\cos\left(\frac{2\pi}{5}\right)+i\sin\left(\frac{2\pi}{5}\right)
So 1+w=1+cos(2π5)+isin(2π5)1+w=1+\cos(\frac{2\pi}{5})+i\sin(\frac{2\pi}{5})
Use De Moivres Theorem such that
wn=cos(2nπ5)+isin(2nπ5)w^{n}=\cos(\frac{2n\pi}{5})+i \sin ( \frac{2n\pi}{5})

Then you just draw a nice spiral.
Original post by Super199
https://2d31ca6ec4f6115572728c3a9168ad5dfc897f33.googledrive.com/host/0B1ZiqBksUHNYSjAyWFYyQmU5Tjg/OCR/FP3%20Complex%20Numbers.pdf


q1ii?

How am I supposed to draw it?
Like I don't know what 1+w is, if someone can explain how to work that out.
(edited 7 years ago)
Reply 2
Original post by Cryptokyo
w=cos(2π5)+isin(2π5)w=\cos\left(\frac{2\pi}{5}\right)+i\sin\left(\frac{2\pi}{5}\right)
So 1+w=1+cos(2π5)+isin(2π5)1+w=1+\cos(\frac{2\pi}{5})+i\sin(\frac{2\pi}{5})
Use De Moivres Theorem such that
wn=cos(2nπ5)+I(2nπ5)w^{n}=\cos(\frac{2n\pi}{5})+I(\frac{2n\pi}{5})
Forget the n preceding the isin
Then you just draw a nice spiral.


What happens to the one after the De Moivre's bit?
w=cos(2π5)+isin(2π5)w=\cos\left(\frac{2\pi}{5}\right)+i\sin\left(\frac{2\pi}{5}\right)
So 1+w=1+cos(2π5)+isin(2π5)1+w=1+\cos(\frac{2\pi}{5})+i\sin(\frac{2\pi}{5})
Use De Moivres Theorem such that
wn=cos(2nπ5)+isin(2nπ5)w^{n}=\cos(\frac{2n\pi}{5})+i \sin ( \frac{2n\pi}{5})

Original post by Super199
What happens to the one after the De Moivre's bit?
Unparseable latex formula:

$LaTeX$

is so good to use....
Reply 5
Original post by Cryptokyo
w=cos(2π5)+isin(2π5)w=\cos\left(\frac{2\pi}{5}\right)+i\sin\left(\frac{2\pi}{5}\right)
So 1+w=1+cos(2π5)+isin(2π5)1+w=1+\cos(\frac{2\pi}{5})+i\sin(\frac{2\pi}{5})
Use De Moivres Theorem such that
wn=cos(2nπ5)+isin(2nπ5)w^{n}=\cos(\frac{2n\pi}{5})+i \sin ( \frac{2n\pi}{5})


No I get that bit. Thats what part 1 was asking. The bit I don't understand is how you draw 1+w.
Reply 6
Original post by Super199
What happens to the one after the De Moivre's bit?


Yeah, dunno what the other user is doing... just think of them as vectors. So you move 1 unit along the real plane to the right and then tack on the vector w and move in that direction, for w^2 you just move from your previous position along the direction of w^2. You should dint that you end up at the origin again once you've drawn the one with w^4. (think back to roots of unity and how they form vertices of special geometric shapes)
Reply 7
Original post by Zacken
Yeah, dunno what the other user is doing... just think of them as vectors. So you move 1 unit along the real plane to the right and then tack on the vector w and move in that direction, for w^2 you just move from your previous position along the direction of w^2. You should dint that you end up at the origin again once you've drawn the one with w^4. (think back to roots of unity and how they form vertices of special geometric shapes)


How do you know that 1+w causes the point to move to the right of 1 and that 1+w+w^2 causes it to move to the left of 1+w? I'm looking at the ms atm. I don't quite get how you work out the direction
Reply 8
Original post by Super199
How do you know that 1+w causes the point to move to the right of 1 and that 1+w+w^2 causes it to move to the left of 1+w? I'm looking at the ms atm. I don't quite get how you work out the direction


Angle of 2pi/5, that's obviously to the right. Angle of 4pi/5 is obviously to the left, etc...
The points you need to point are:
1=1+0i1=1+0i
1+w=1.31+0.95i1+w=1.31+0.95i
1+w+w2=0.5+1.54i1+w+w^{2}=0.5+1.54i
1+w+w2+w3=0.31+0.95i1+w+w^{2}+w^{3}=-0.31+0.95i
1+w+w2+w3+w4=0+0i1+w+w^{2}+w^{3}+w^{4}=0+0i
Essentially you are plotting the behaviour of the sine and cosine functions as the angle varies. That is the idea they want and not exact plotting but here are the points anyway.
I think I am going crazy :biggrin:
Reply 11
Original post by Zacken
Angle of 2pi/5, that's obviously to the right. Angle of 4pi/5 is obviously to the left, etc...


Damn, yh got it cheers man.
Although how do you know the last one has to go to the origin?
Reply 12
Original post by Cryptokyo
Then you just draw a nice spiral.


It's not a spiral, it's a polygon.

Quick Reply

Latest