Need help badly... Complex Numbers (1 Viewer)

[currysauce]

please someone boost my motivation... this is hard for me

anyway.. i'll put the book and question if ur lazy to read..... Cambridge 2.4, Q7.8

anyway... all help is 100% appreciated

7. 1, w, w^2 are the three cube roots of unity. State the values of w^3 and 1+w+w^2. Hence simplify each of the expressions (1+ 3w+w^2)^2 and (1+w+3w^2)^2 and show that their sum is -4 and their product is 16.

OK WTF!!!

what is cube roots of unity!!!

8. Use De Moivre's Theorem to find in mod/arg for

a) the square roots of root3 + i
b) the cube roots of -2-2i

again WTF

 

[lfc_reds2003]

ok

Complex cube roots of unity just means

roots of z^3 = 1

where w is the first complex root

the sum of the roots = 0 ie. 1 + w + w^2 =0

w^3 = 1 of course so can worlk from there...

 

[lfc_reds2003]

n last two are easy:

fconvert to mod/arg form n raise to 1/2 AND 1/3 respectively using De Moivre

 

[currysauce]

sorry if u find this easy.. but this is difficult for me

can u explain them?

 

[mojako]

since I'm here I might as well answer ur Q.. (note: when I started typing aural hadnt replied)

7.
unity = 1 (also if u see the term "unit circle" later, it means circle of radius one, centred at origin)
cube root is.. the third root...
w is the cube root of 1
so w^3 = 1

let z^3 = 1 and the roots to this equation are 1, w and w^2
1+w+w^2 is the sum of roots
the equation is z^3 + 0 z^2 + 0 z - 1 = 0
sum of roots = 0
(there is another way of doing this)

(1+ 3w+w^2)^2 = (1+ w+w^2 + 2w)^2 = (0 + 2w)^2
...
(1+ 3w+w^2)^2 = 4w^2
(1+w+3w^2)^2 = 4w^4
sum: 4 w^2 (1 + w^2) = 4 w^2 (-w), since 1 + w^2 + w = 0
= -4w^3 = -4(1)
product: 16 w^6 = 16 (w^3)^2 = 16 (1)^2


8. a.
As aural said, convert root3 + i to the mod-arg form (thats the r cis@ form)
then raise to 1/2 using De Moivre's.
Call the result A. the roots will be +- A (just like if x^2 = y, x = +- root y)
but mod-arg form can't have negative r so for the -A you need to change the @... draw A in the Argand diagram and then draw -A (just change the signs both the x and y coordinates of A)... now its @ can be easily found
note that technically @ shoud be in the range -pi < @ <= pi

8. b.
Well for this one, you need to use the fact that complex roots (of unity or of anything else) are equally spaced in the Argand diagram... raising to 1/3 would give 1 of the 3 cube roots. The space or angle between each root is 2pi/3 (2pi is full rotation)....
(note: in the complex number field, there are 3 cube roots of a number.... 4 fourth roots of a number, etc)

Alternative approach to Q8:
8. a.
let z be the square roots and z=x+iy
z^2 = root3 + i
(x+iy)^2 = root3 + i
x^2-y^2 + 2xyi = root3 + i
equating real and imaginary parts,
x^2-y^2 = root3
2xy = 1
now solve simultaneously... you'll get two values of z, A and -A
then convert to mod-arg form now.

8. b.
use similar approach like in part a.
EDIT: Actually I don't think this would work out easily.. just use the De Moivre's approach...


BTW most solutions to cambridge are available on the net somewhere.. forgot the link...
or you can email Nathan here: http://www.boredofstudies.org/community/showthread.php?t=48242

 

[currysauce]

(1+w+3w^2)^2 = 4w^4

this part of your working i don't understand for q7

i got it.. lol

go me! +2 points

 

[mojako]

(1+w+3w^2)^2
= (1+w+w^2 + 2w^2)^2
= (0 + 2w^2)^2
= 4w^4

 

[currysauce]

i know i sound like a complete bitch now...

but could u work me through 8a)

then i'll try to do 8b?

 

[Euler]

1. convert \sqrt 3 + i into mod arg form first
the answer is 2 cis (pi / 6)

2. Recall De Moivre's theorem
which is (r cis(theta))^n =r^n cis (n.theta)

3. Apply to answer in part 1 with n = 1/2
you get
\sqrt 2 cis {(pi/6 + 2kpi)/2}
putting k=0 and k=2 gives you the answers you want:
\sqrt 2 cis(pi/12) and \sqrt 2 cis(13pi/12)

typing math in text is hopeles...

 

[mojako]

root3 + i
--
converting to mod-arg form:
mod = root ([root3]^2 + 1^2) = 2
arg = inverse tan of 1/root3 = pi/6 (btw if it was -root3 + i, the arg is 5pi/6.., just want to make sure you know how to find arg)
--

so we have 2 cis pi/6
De Moivre's theorem says:
(rcis@)^k = r^k cis(k@)

(2 cis pi/6)^(1/2)
= root2 cis pi/12

now I'll just use the logic which you'll need to use for part b.
A is one of the roots of 2cispi/6. There are two roots.
These 2 roots are equally spaced in the Argand diagram so the angle between them is (2pi)/2=pi or 180 degrees...
[if there were 3 roots then the spacing is (2pi)/3.. draw them if you don't really understand... you have 3 roots so you draw 3 lines on the same diagram... (2pi)/3 is the angle between any two lines...]

One root is "root2 cis pi/12", the other is "root2 cis (pi/12 + pi)" = "root2 cis 13pi/12" or "root2 cis -11pi/12".

oh... before I said to reflect in the x-axis.. I was wrong sorry. I'll edit it.

Oh yeah u can use Euler's notation of +2kpi/2... its been a while since I did this stuff so I forgot.
complex n-th roots of a number can be represented by
r cis (@ + k 2pi/n) , k is any integer
as you try different values of k, you'll find that there are only n distinct roots because the values of "r cis (@ + k 2pi/n)" will be repeated as you change the k.

 

[Raginsheep]

mojako, I don't think its (@+k 2pi/n), rather just (@+k 2pi) where k is any integer

 

[mojako]

mojako, I don't think its (@+k 2pi/n), rather just (@+k 2pi) where k is any integer

if it is k 2pi then they all correspond to the same angle and the exact same solution, hehe...

maybe we're thinking of different things

 

[Raginsheep]

with you're "n" is it all over the power of n or somethin?

cause what im thinking of is somethin like "find the cube roots of -1."

therefore: z=cis ((pi+k 2pi)/3) where k is an integer.

 

[mojako]

yeap, n is the power
so for cube roots n is 3

but the cube roots of -1 is
z=cis (pi + k 2pi/3) where k is an integer
(the 3 is inside the brackets)

 

[Raginsheep]

hmm.........but isn't it like:

z^3=-1
z^3=cis pi
z^3=cis (pi+k 2pi)
therefore: z=cis (pi+k 2pi)^(1/3)
using De Moivre's, the arg is multiplied by 1/3 therefore the whole arg in the brackets is (pi+k 2pi)/3 isn't it?

Hmm.....I'll go look at my text book i guess...

 

[mojako]

actually that one works too, sorry.
its just that I've never done it that way before so I thought it was wrong.

z = r cis (@ + k 2pi/n) where @ is any argument for which z = r cis @ is one of the roots.
in z = cis (pi + k 2pi/3), @=pi and z = cis pi = -1 is one of the roots
in z = cis (pi + k 2pi)/3 = cis (pi/3 + k 2pi/3), @=pi/3 and z = cis pi/3 is one of the roots

 

[Raginsheep]

kool.......i see what you mean 2...well i guess we both learn't somethin then