Polynomials From Fitzpatrick (1 Viewer)

[AFGHAN22]

HEY guys, can i please have some help with polynomials from page 196 from the fitzpatrick book:

1. Find real values of a for which ai is a solution of the polynomial equation z^4 -2 z^3 +7 z^2 -4 z + 10 = 0. Hence find all the solutions of this equation.

Answer:
a = plus or minus sqrt2, z = plus or minus sqrt 2i, 1 plus or minus 2i

2. if w= x + x^-1, prove that x^4 +x^3+x^2+x+1= x^2(w^2+w-1) = (x^2 +1/2 x(1+sqrt 5) + 1) ( x^2 + 1/2 x (1-sqrt5) + 1)
show that the roots of x^4 +x^3 +x^2 + x+1=0 are the four complex roots of x^5=1. deduce that cos 72 degrees = 1/4(sqrt5-1), cos 36 degrees=1/4(sqrt 5 +1)

3. Find the real number k such that z=ki is a root of the equation z^3 + (2+i) z^2 + (2+2i)z + 4=0. hence, or otherwise, find the three roots of the equation.

answer:
k=1 or -2, i, -2i, -2

thanks in advance and waiting in anticipation
this is a bit urgent and i need it before 6 o clock january the 8th of january

 

[Riviet]

HEY guys, can i please have some help with polynomials from page 196 from the fitzpatrick book:

1. Find real values of a for which ai is a solution of the polynomial equation z^4 -2 z^3 +7 z^2 -4 z + 10 = 0. Hence find all the solutions of this equation.

Answer:
a = plus or minus sqrt2, z = plus or minus sqrt 2i, 1 plus or minus 2i

Firstly you start by substituting ai into the equation, rearrange a little and you obtain:
a⁴+3+(2a³-4a)i=0
Now we quickly observe that if we equate the real parts of boths sides, there won't be real solutions for a⁴=-3, so we go straight to equating imaginary parts,

a(a²-2)=0

Therefore, a=+sqrt2

So 2 of the roots are +(sqrt2)i

EDIT: Here's the rest.

If we had to factorise LHS into linear factors, let

(z-sqrt2i)(z+sqrt2i) (z-a)(z-b)=0, where a and b are the other two roots

We can find the quadratic formed by (z-a)(z-b) by dividing p(z) by the quadratic of (z-sqrt2i)(z+sqrt2i)

So expanding (z-sqrt2i)(z+sqrt2i) we obtain z²+2

Then applying long division of dividing p(z) by z²+2, we obtain the quadratic z²-2z+5, which when solved by the quadrtaic formula, gives the roots 1+2i

That's one down.

 

[Riviet]

Number 3 is very similar, basically the same method as before.

Substitute z=ki into p(z), expand, then factorise the i terms to obtain:

(-2k²-2k+4) + (-k³-k²+2k)i=0

If we equate the real parts of both sides, we end up with the quadratic k²+k-2=0, which has no real roots, so we equate the imaginary part of both sides to obtain:

k³+k²-2k=0
k(k+2)(k-1)=0
k=-2,1

.: Two roots of p(z) are -2i, i

Now we create our factorised cubic like this:

(z+2i)(z-i)(z-a)=0, where a is the third root.

Now we observe that the last term in p(z) is 4, and that 2i.-i.-a=4, since this is the last term when you expand (z+2i)(z-i)(z-a). So by inspection a=-2.

That's 2 down.

 

[Stan..]

Q3

Just sub in ki for z. To get K= 1.
Then just factorise(Over C) to get the roots.

 

[AFGHAN22]

thanks for your replies everyone, but can someone please try and give question two an attempt please

 

[Riviet]

I did actually work out the first bit for 2, just couldn't finish the rest of it, but I'll show you what I already have:

LHS=x⁴+x³+x²+x+1

=x²(x²+x+1+x^-1+x^-2)

=x²(w²-2+w+1), using the results found in the previous bit

=x²(w²+w-1)

=x²(w²+w+1/4-5/4)

=x²[(w+1/2)²-(sqrt5/2)²]

=x(w+[1+sqrt5]/2).x(w+[1-sqrt5]/2)

=x(x+x^-1+[1+sqrt5]/2).x(x+x^-1+[1-sqrt5]/2)

=(x²+[1+sqrt5]x/2+1)(x²+[1-sqrt5]x/2+1)

 

[Roobs]

Just a bump to this thread--i just got insanely stuck on "q2" --question 59ii of (36a) p196 of fitzpatrick:

*you can show that roots of x^4+x^3+x^2+x+1 are complex roots of x^5-1=0 by taking out the factor of x-1, and observing that x^5=1 has only one real solution

* as for "deducing" values of cos72 and cos36 the only way i could find was 3 pages long and involved completely factorising x^4+x^3+x^2+x+1, then finding roots in polar form, equating whiohc roots were which (ie which quadrant everything lies on when plotted) then equating the real parts of each method to get the reqired result--

Doesnt "deduce" usually mean a quick and simple 1 mark inference?

thanks for any help/better solutions

 

[haque]

roots of x^4+x^3+x^2+x+1=0 are cis(+/-72), cis(+/- 144). sum of roots is -1
so cos(72) +cos(144)=-1/2 but cos144=2cos^2(72) -1, this results in us obtaining a quadratic in cos 72
4Cos^272+2Cos72 -1=0 solving and taking only positive square root(since cos 72 is positive) gives cos72=1/4(sq root5 -1) we can again apply double angle to get cos36 and get the answer but we know that the roots of this quadratic are cos144 and cos 72 from which we deduce cos144 is the other negative root from the quadratic. cos144=-cos36 and from this we find cos36=1/4(sqroot5+1). Alternatively we could factor the the expression into two quadratics by grouping in conjugate pairs giving us (x^2-2xCos72 +1)(x^2-2xCox144 +1). equating the coefficients of the quadratics with the expressions in the quadratic (x^2+1/2x(1+sqroot5+1) +1)(x^2+1/2x(1-sqroot5 )+1) gives the required result. i think the second method is quicker but the its ur choice.

 

[Roobs]

hmm..the second method sure looks quicker, but im shaky on

"factor the the expression into two quadratics by grouping in conjugate pairs giving us (x^2-2xCos72 +1)(x^2-2xCox144 +1)"
how did you arrive at this expression?, as it doesnt seem like the most obvious factorisation.....was it something to do with first writing the quartic as the sums and products of its roots? im kinda confused....

and also, for the sake of completeness, with the first method when using the double angle formula to find cos 36, i ran into a nasty "square root of square root" expression..did you get this?

thanks

 

[haque]

Find the roots of x^5-1=0, then using the factor theorme write the polynomial x^4+x^3+x^2+x+1 as (x-a)(x-b)(x-c)(x-d) where a=cis(72) b=cis(-72) c=cis(144)
d=cis(-144) as for the square root under a square root, i actually turned the (3+sqroot5)/8 into (6+2sqroot5)/16 in which i saw that the root 5 had to have a two at the front for it to turn into a perfect square. the 6+2sqroot5 i wrote as 5+2sqroot5+1 which is 1+sqroot5 squared.

 

[haque]

Or u can just use the fact that cos(144) is the other root of the quadratic, u know the -1-sqroot5/4 is cos(144), but cos(144)=-cos36 and from here u can find the solution to cos 36-no square root problems(remember the quadratics coeffiecients satisfy cos(72) +cos(144)=-1/2 cos(72)cos(144)=-1/4)(by equating coefficients after expansion-am i a bit convoluted-tell me if u have any problems with what i'm saying.

 

[Roobs]

thanks heaps..im set now

 

[haque]

no worries roobs

 

[OmegaSTealth]

Not to flog a dead horse here, but for Q2 you could show that P(x)=x^5-1 can be factrorised to P(x)=(x-1).Q(x)=(x-1)(x^4+x^3+x^2+x+1) through long division, and it can be observed that P(x)=0 has only opne real root, x=1, which is clearly shown as the root given by the factor (x-1). It is also known from the theory of 'Roots of Unity' that x^5=1 has four complex roots, found by solving Q(x)=x^4+x^3+x^2+x+1=0. We do not actually need to solve Q(x)=0 at this stage, but we know that the complex roots form a regular pentagon, with one vertex at x=1, and centred at x=0. Thus, knowing that a regular pentagon makes angles of 360/5=72 degrees at its centre, we know that the first root is x=cis(72 deg), so we can find the roots of the equation Q(x)=0, by splitting it up into two quadratics and manipulating the resulting fractions.
The final part, however, i suppose would be the first complex root of x^10=1, but this would lead to having to repeat the whole process again, and while I wouldn't put it past a 4U question to do that, reallistically speaking I think there's an easier way - maybe through double angle trig formulae? This would lead to
2A^2-1=1/4 (sqrt(5)-1)
where A=cos(36 deg)
so, we have A=sqrt(1/2 (sqrt(5) +3))
I presume if you manipulate that you will get the right result....

Any thoughts?