stupidness (1 Viewer)

[fashionista]

k this question will prob seem totally obvious immediately after my typing up this thread but wut tha hey
im kinda confused in this example thingo wit complex numbers where it goes
find (sq.rt3 + i)^8 + (sq.rt3 - i)^8 in the form a+ib
rite so theres all the working out ra ra ra...convert into modarg form etc etc (letting z=(sq.rt3 + i)
so it ends up being
z^8=2^8(cis((4pi)/3))
then z^8 + (conjugate z)^8 = z^8 + conjugate(z^8)
then in the answers it goes to this other step where it sez the above = 2 Re (z^8)
can anyone tell me where the 2 came from????

 

[fashionista]

wait ..y is it 2 Re(z^8) in general? i mean when u add z and conjugate z both to the power of 8 u get
2^8(cis0)
so shudnt that just be 2^8?

 

[BlackJack]

Hmmm.. no matter what, z^8 is an actual complex number.
When you add a cplx and its conjugate, (draw a little cplx plane for yourself and try adding them like they teach you,) you'll get twice the real component.

(In other words, let's say:
z^8 = y.
Then, it becomes
y + conjugate(y).
THe two Im(y)'s cancel out, by definition, and we're left with the two identical Re(y)'s. Hence.

So, 2^8 is the answer.

 

[Supra]

yeh that sounds rite, the best way to verify is the argand diagram...the resultant will b along the real axis, therefore it has to b completely real

 

[KeypadSDM]

Or just do it through algebra.

z^8 + (conjugate z)^8
= z^8+ conjugate(z^8)
= Re(z^8) + i * Im(z^8) + Re(conjugate(z^8)) + i * Im(conjugate(z^8))
= Re(z^8) + i * Im(z^8) + Re(z^8) - i * Im(z^8)
= 2Re(z^8)

 

[Grey Council]

nice, do as Keypad did. It doesn't take long, and will more likely than not be quicker than an argand diagram.

 

[fashionista]

thank u!! omg i feel so stupid now. see we havnt been taught much in the way of complex numbers and graphs apart from....up until i dunno..sumthing ...mainly due to tha fact that our teacher deserted us for 4 weeks..n we only have him 3 periods a week
all good now
thanking u!! (everyone!!)

 

[Grey Council]

Don't feel stupid. Whats there to feel stupid about? You don't know how to do a question, thats fine! All of us are learning, even Turtle and Keypad. Its life. Ask a question now and feel 'stupid' ( )or don't ask the question and remain "stupid" forever. Opportunity cost.

Hehh, its all a part of learning.

 

[fashionista]

me likes ur way of tinking
thanks for ur helpings
i have another questionnn to ask
can u please please please xplain roots of unity??? we didnt do it in class so im trying to splain it to myself but i dunno combination of lack of botheredness and lack of comprehendingness

 

[Rorix]

Uh, what about them? It's best to put up an example question you are having trouble with

 

[fashionista]

umm jus the concept...cud u please post a simple question n answer? jus so i can get the jist kindof.......like how wud u noe wen to use them? do the questions say 'roots of unity...' in the question? or do u have to assume that u'll have to use em?
oooh im so lost

 

[fashionista]

ooh i found summit
it goes if w is a non real cube root of unity show that
1 + w + ww = 0
yeh so the solution goes
the cube roots of unity satisfy xxx-1=0.
But xxx-1 = (x -1)(xx + x + 1)
Hence w cannot equal 1 -----> ww+w+1=0
that bit is that all u have to do? say that w cannot equal zero and therefore ww+w+1 must equal zero? becuase to me the statement reads
w cannot equal 1 otherwise ww+w+1 would equal zero...it doesnt seem to show that it actually does = 0

 

[ND]

All you really need to do is:

w^3-1=0
(w-1)(w^2+w+1)=0
w^2+w+1=0 since w cannot equal 1.

 

[fashionista]

ritio..i getchya
thanxxaloott!

 

[Rorix]

z^3 - 1 = 0
z^3 - 1 = (z - 1)(1 + z + z^2)
(z - 1)(1 + z + z^2) = 0
Which gives us either z - 1 = 0 or 1 + z + z^2 = 0
and since w is complex, w =/= 1
Therefore 1 + w + w^2 = 0.

You can also do this topic using GP
First term is 1, common ratio is w and there's 3 terms
So 1 + w + w^2 = (w^3 - 1)/(w - 1)
and since w =/= 1, and w^3 = 1 (as w is a cubed root), the sum is 0.

 

[fashionista]

YAY!!! thanku!!!!! thank u thank u!!!!
omg i think i actually like em now!!

 

[ND]

Just to get you more excited, i'll post up another method :

w^3-1=0
the roots of this eqn are 1,w,w^2 (you're allowed to write this without proof)
and the sum of roots of this eqn is 0 (as the coefficient of w^2 is 0).
.'. w^2+w+1=0

 

[freaking_out]

Originally posted by ND
Just to get you more excited, i'll post up another method :

w^3-1=0
the roots of this eqn are 1,w,w^2 (you're allowed to write this without proof)
and the sum of roots of this eqn is 0 (as the coefficient of w^2 is 0).
.'. w^2+w+1=0

LoL, u got this from the terry lee book didn't ya?

 

[ND]

I returned all my books to the school after the 3u exam... Maybe i saw it in T. Lee when i was doing complex numbers... i don't remember.

edit: but it's a pretty standard thing.

 

[freaking_out]

Originally posted by ND
...but it's a pretty standard thing.

yep, once u've done either 3u or 4u polynomials that is.