Complex Numbers Question (1 Viewer)

[martin]

z^4 - 3z^2 + 1 = 0
Show that the roots are 2cos(Pi/5), 2cos(2Pi/5), 2cos(3Pi/5) and 2cos(4Pi/5).

I've got a solution but its pretty long and indirect. I wonder if anybody can come up with a nice way of showing it.

 

[dawso]

from quickly lookin at this....reduce the equation to a quadratic (z squared = m etc...) and then u would hav 2 use quadratic formula to find z^2, and then find the 4 values for z...how did u do it?

 

[martin]

Yeah thats right,
z=± sqrt((3 ± √ 5)/2)

Then using a calculator we can see sqrt((3 + √ 5)/2) = 2cos(Pi/5) and similar results.

But why is this? Surely there is a proof.

 

[Slidey]

z^4 - 3z^2 + 1 = 0
Show that the roots are 2cos(Pi/5), 2cos(2Pi/5), 2cos(3Pi/5) and 2cos(4Pi/5).

I've got a solution but its pretty long and indirect. I wonder if anybody can come up with a nice way of showing it.

Let:
2cos(Pi/5) = z1
2cos(2Pi/5) = z2
2cos(3Pi/5) =z3=-z2
2cos(4Pi/5) = z4=-z3

From this craft quadratic equations then multiply them out to get the required equation.

You could note that if
w=cos(pi/5)+i.sin(pi/5), then
2cos(Pi/5) = w+1/w
2cos(2Pi/5) = w^2+1/w^2
2cos(3Pi/5) = w^3+1/w^3 = -w^2-1/w^2
2cos(4Pi/5) = w^4+1/w^4 = -w-1/w

But, I don't know where that would get you.

 

[Sirius Black]

z^4 - 3z^2 + 1 = 0
Show that the roots are 2cos(Pi/5), 2cos(2Pi/5), 2cos(3Pi/5) and 2cos(4Pi/5).

I've got a solution but its pretty long and indirect. I wonder if anybody can come up with a nice way of showing it.

Martin-are you sure it's the right question coz i substituted roots back to the eqn, it didn't give me zero in the end.

 

[Sirius Black]

Let:
2cos(Pi/5) = z1
2cos(2Pi/5) = z2
2cos(3Pi/5) =z3=-z2
2cos(4Pi/5) = z4=-z3

From this craft quadratic equations then multiply them out to get the required equation.

You could note that if
w=cos(pi/5)+i.sin(pi/5), then
2cos(Pi/5) = w+1/w
2cos(2Pi/5) = w^2+1/w^2
2cos(3Pi/5) = w^3+1/w^3 = -w^2-1/w^2
2cos(4Pi/5) = w^4+1/w^4 = -w-1/w

But, I don't know where that would get you.

Maybe you can use the relationship b/w coefficients of the terms by writting eqn as:
z⁴+0*z³-3z²+0*z+1=0
since 2[cos(pi/5)+cos(2pi/5)+cos(3pi/5)+cos(4pi/5)]=2[cos(pi/5)+cos(2pi/5)-cos(2pi/5)-cos(pi/5)=0

 

[martin]

Martin-are you sure it's the right question coz i substituted roots back to the eqn, it didn't give me zero in the end.

Yeah it looks right to me,
2cos(pi/5) = 1.618...
then 1.618...^4 - 3*1.618...^2 + 1 = 0 (it might come up as something * 10^-5 but that's just rounding errors).

The way I did it in the end was that used in another thread on this board.
Using De Moivres theorem
cos5t + isin5t = (cost + isint)^5
Expand RHS using binomial theorem, then equate real parts, use (sint)^2 = 1 - (cost)^2 to get cos5t as a polynomial in cost.
then can find solutions of cos5t = 1 and cos5t = -1 which gives you surd expressions for cos(Pi/5), cos(2Pi/5), ...
Then you can compare the solutions to z^4-3z^2+1=0 to these expressions and see that they are the same.

So as I said its pretty long and not all that illuminating. I like what Slide Rule said about w+1/w = 2cos(argw) but I don't quite see how to do it.

 

[Archman]

if w=cos(pi/5)+i.sin(pi/5), then w^5+1=0 and w^4-w^3+w^2-w+1=0
(z-(w+1/w))(z+(w+1/w))(z-(w^2+1/w^2)(z+(w^2+1/w^2))
=(z^2-(w+1/w)^2)(z^2-(w^2+1/w^2)^2)
=(z^2-(w^2+1/w^2+2))(z^2-(w^4+1/w^4+2))
=z^2-(w^2-w^3+w^4-w+4)z^2+(w^2+1/w^2+2)(w^4+1/w^4+2)
=z^2-(4-1)z^2+(2(w^2-w^3+w^4-w+4)+4+(-w-w^3+w^2+w^4)) yes... quite a bit of multiplying but not too bad
=z^2-3z^2+2*-1+4-1=z^2-3z^2+1

 

[Slidey]

Nice, Ivan. Aside the length, I like that solution (certainly beats others proposed).

 

[Will Hunting]

2cos(4Pi/5) = z4=-z3

z4=-z1, not -z3

 

[JamiL]

cant u jus substitue in... n if they equal 0 then it stay true.. the question said show, not prove...

 

[JamiL]

cos(pi/5)=-cos(4pi/5) and
cos(2pi/5)=-cos(3pi/5)
therefore
2cos(pi/5)+2cos(2pi/5)+2cos(3pi/5)+2cos(4pi/5)=0
sum of the roots=a+b+c+d
=-b/a
=0
i dunno if this is sufficent proof thou cos i could say -2, -1, 1, 2 sum is zero then they must bet the roots...

 

[ngai]

i dunno if this is sufficent proof thou cos i could say -2, -1, 1, 2 sum is zero then they must bet the roots...

i think you just answered your own question

 

[thunderdax]

Can't you just say the equation has 4 roots since its of degree 4 and then sub in each of the roots to show that they satisfy the equation?

 

[pull up]

yeh but its not as easy as it seems