complex polynomials (1 Viewer)

kr73114 · Feb 19, 2011

a) use de moivres theorem to show that cos5A= 16cos^5A-20cos^3A+5cosA
b)hence solve 16x^4-20x^3+5=0
c) hence determine the exact value of cos^2(pi/10)cos^2(3pi/10)

ohexploitable · Feb 19, 2011

$\\(\cos\theta+i\sin\theta)^5=\cos5\theta+i\sin5\theta$

$\\(\cos\theta+i\sin\theta)^5=\cos^5\theta+5i\cos^4\theta\sin\theta-10\cos^3\theta\sin^2\theta-10i\cos^2\theta\sin^3\theta+5\cos\theta\sin^4\theta+i\sin^5\theta$

$\\\text{Equating real parts,}\\\cos5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta\\\cos5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta$

kr73114 said:
b)hence solve 16x^4-20x^3+5=0

$\text{Are you sure it's not }16x^4-20x^2+5?\text{ If so...}$

$\\16x^4-20x^2+5=0\\16x^5-20x^3+5x=0\\\text{Let }x=\cos\theta\\16\cos^5\theta-20\cos^3\theta+5\cos\theta=0\\\cos5\theta=0\\\theta=\frac{\pi}{10},\,-\frac{\pi}{10},\,\frac{3\pi}{10},\,-\frac{3\pi}{10}\\x=\cos\frac{\pi}{10},\,-\cos\frac{\pi}{10},\,\cos\frac{3\pi}{10},\,-\cos\frac{3\pi}{10},\,\cos\frac{\pi}{2}$

$\\\text{Product of roots four at a time, the ones with }\cos\frac{\pi}{2}\text{ get cancelled out},\\\cos\frac{\pi}{10}\times-\cos\frac{\pi}{10}\times\cos\frac{3\pi}{10}\times-\cos\frac{3\pi}{10}=\frac{5}{16}\\\cos^2\frac{\pi}{10}\cos^2\frac{3\pi}{10}=\frac{5}{16}$

shaon0 · Feb 19, 2011

1) cis5A=(cos(A)+isin(A))^5
Expand the RHS and then equate the reals. It's better if you do it as it'll give you some practice.
2) Let x=cos(A) and cos(5A)=0.
<=> cos(5A)=0=16x^5-20x^3+5x (divide by x here)
<=> cos(5A)=0=16x^4-20x^2+5 (you have a typo here)

=> cos(5A)=0=cos(+-(2n+1)*pi/2) where n E Z+
=> 5A=+-(2n+1)*pi/2
=> A=+-(2n+1)*pi/10
Now, just sub in n values to get x=cos(+-pi/10,...,+-9pi/10)
x=+-cos(+-pi/10,+-3pi/10,+-pi/2,+-pi/10,+-3pi/10) [by symmetry of cos(x) ie. cos(x)=-cos(pi-x)]
x=+-cos(+-pi/10,+-3pi/10,+pi/2) [but x=/=0 and hence x=/=+-cos(+-pi/2)]
x=+-cos(pi/10,3pi/10) [cos(x)=cos(-x)]

3) 16x^4-20x^2+5=0
=> x^2(16x^2-20+5x^-2)=0
=> 16 (x^2-5/8)^2-5/4=0
=> (x^2-5/8)^2=5/64
=> x^2=5/8+-sqrt(5)/8
=> x=+-(1/2)sqrt((1/2)(5+-sqrt(5))

By product of roots;
=> (+-1)^2[cos(pi/10)cos(3pi/10)]^2=[(1/2)sqrt((1/2)(5+sqrt(5))]^2[(1/2)sqrt((1/2)(5-sqrt(5))]^2
=> [cos(pi/10)cos(3pi/10)]^2=(1/4)^2(0.25(25-5))=(1/16)(0.25(20))=5/16

ohexploitable said:
$\\(\cos\theta+i\sin\theta)^5=\cos5\theta+i\sin5\theta$

$\\(\cos\theta+i\sin\theta)^5=\cos^5\theta+5i\cos^4\theta\sin\theta-10\cos^3\theta\sin^2\theta-10i\cos^2\theta\sin^3\theta+5\cos\theta\sin^4\theta+isin^5\theta$

$\\\text{Equating real parts,}\\\cos5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta\\\cos5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta$

$\text{Are you sure it's not }16x^5?\text{ If so...}$

$\\16x^5-20x^3+5=0\\\text{Let }x=\cos\theta\\16\cos^5\theta-20\cos^3\theta+5\cos\theta=0\\\cos5\theta=0\\\theta=\frac{\pi}{10},\,-\frac{\pi}{10},\,\frac{3\pi}{10},\,-\frac{3\pi}{10}\\x=\cos\frac{\pi}{10},\,-\cos\frac{\pi}{10},\,\cos\frac{3\pi}{10},\,-\cos\frac{3\pi}{10}$

$\\\text{Product of roots,}\\\cos\frac{\pi}{10}\times-\cos\frac{\pi}{10}\times\cos\frac{3\pi}{10}\times-\cos\frac{3\pi}{10}=\frac{5}{16}\\\cos^2\frac{\pi}{10}\cos^2\frac{3\pi}{10}=\frac{5}{16}$

In part 2): it's 16x^4-20x^2+5 as you need a symmetric quartic to obtain the final part

cutemouse · Feb 19, 2011

should be 20x^2 i think...

kr73114 · Feb 19, 2011

ohexploitable said:
$\\(\cos\theta+i\sin\theta)^5=\cos5\theta+i\sin5\theta$

$\\(\cos\theta+i\sin\theta)^5=\cos^5\theta+5i\cos^4\theta\sin\theta-10\cos^3\theta\sin^2\theta-10i\cos^2\theta\sin^3\theta+5\cos\theta\sin^4\theta+isin^5\theta$

$\\\text{Equating real parts,}\\\cos5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta\\\cos5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta$

$\text{Are you sure it's not }16x^5?\text{ If so...}$

$\\16x^5-20x^3+5=0\\\text{Let }x=\cos\theta\\16\cos^5\theta-20\cos^3\theta+5\cos\theta=0\\\cos5\theta=0\\\theta=\frac{\pi}{10},\,-\frac{\pi}{10},\,\frac{3\pi}{10},\,-\frac{3\pi}{10}\\x=\cos\frac{\pi}{10},\,-\cos\frac{\pi}{10},\,\cos\frac{3\pi}{10},\,-\cos\frac{3\pi}{10}$

$\\\text{Product of roots,}\\\cos\frac{\pi}{10}\times-\cos\frac{\pi}{10}\times\cos\frac{3\pi}{10}\times-\cos\frac{3\pi}{10}=\frac{5}{16}\\\cos^2\frac{\pi}{10}\cos^2\frac{3\pi}{10}=\frac{5}{16}$

thanks

ohexploitable · Feb 19, 2011

lols, I made some mistakes, check my fixed solution.

complex polynomials (1 Viewer)

kr73114

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ohexploitable

hey

shaon0

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cutemouse

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kr73114

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ohexploitable

hey

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