A rather annoying polynomial question. (1 Viewer)

[lucifel]

I do have a solution to this, it's just I want to see if there is some other way of doing this. Anyways here's the question:

solve: (x+1)⁶ + (x-1)⁶ = 0 over C (this *is* the Ext-2 forum)

 

[maths > english]

solution attached to message

 

[maths > english]

another method attached

 

[KFunk]

The second method is quite ingenious.

 

[maths > english]

thanks .

 

[Antwan23q]

Is that a HSC question?
Id never think to use e^iθ.

 

[FinalFantasy]

the e^ix is not in the syllabus

 

[haboozin]

it isnt, i havnt seen a question like this either...

e^ix gets mentioned in most of the text books and your teacher might have talked about it.

 

[lucifel]

that e^ix method is beyond the course, but it's nice. the way I got it was slighty different, and the answers are also different to what you all have, but I have a feeling it'll boil down to the same thing. The textbook says: +/- isqrt of (7+ sqrt of 48) ,+/- i, +/- isqrt (7-sqrt 48) and I got similar answers, but in terms of cot:

x = -icot ((2k+1)pi)/12) where k = 0,1,2,3,4,5

i got that answer by a similar method to maths>english. The first one, not the 2nd.

 

[who_loves_maths]

Originally Posted by haboozin
it isnt, i havnt seen a question like this either...

if you have or use Mathematics Extension 2 by S.K. Patel (the 2nd edition one), then go to Exercise 4J (page 145), Questions 5 and 6.

Originally Posted by lucifel
The textbook says: +/- isqrt of (7+ sqrt of 48) ,+/- i, +/- isqrt (7-sqrt 48) and I got similar answers, but in terms of cot:

x = -icot ((2k+1)pi)/12) where k = 0,1,2,3,4,5
...

hey lucifel, all the methods the others have so far used are all very good. but most questions in tests or exams of this style will ask you to give the answers in the form itan(x) or icot(x) {there's a reason for this too, but i won't discuss it here}, so the following is something you should try to remember for your convenience or for the next time you meet a question like this in an exam, it also provides a much faster and elegant fashion in solving equations of this type:

use this identity: (1+ itan(x))^n + (1- itan(x))^n = (2Cos(nx))/((Cos(x))^n)

for all positive integral 'n' and all real 'x'. This identity is not hard to prove at all, and i'll leave that to you (start with the LHS, and don't expand).

so for your question: (z+1)^6 + (z-1)^6 = 0 ---> 0 = (1+z)^6 + (1-z)^6
now let z=itan(x), since the question requires it to be of the form -icot(x) which is the same as itan(x).
therefore, (1 +itan(x))^6 + (1 -itan(x))^6 = 0 = (2Cos(6x))/(Cos(x)^6)
-----> Cos(6x) = 0 -----> 6x = pi/2 + (k)pi ----> x = pi/12 + kpi/6 = (2k+1)pi/12

hence, z = itan[(2k+1)pi/12] = -icot[(2k+1)pi/12] as required.


hope that simplifies matters for you

 

[lucifel]

wow that is really really neat, beats what I did, by a mile. Hmm that identity, i should get proving it. thanks.