Question

if a and b are the roots of x^2-2x+4=0 ; then a^9 +b^9 = ?

Answer 1

sum of roots, $a+b=2$
product of roots, $ab=4$
$a^2+b^2=(a+b)^2-2ab=2^2-2(4)=-4$

$x^2-2x+4=0\implies x^2=2x-4$
Since $a,b$ are roots, for $k>2$ we have :
$a^2=2a-4\implies a^k=2a^{k-1}-4a^{k-2}$
$b^2=2b-4\implies b^k=2b^{k-1}-4b^{k-2}$

Add them and get a recurrence relation :
$a^k+b^k=2(a^{k-1}+b^{k-1})-4(a^{k-2}+b^{k-2})$

Plugin $k=3,4,5,\ldots$ successively in above recurrence relation :$a^3+b^3=2(a^2+b^2)-4(a^1+b^1)=2(-4)-4(2)=-16$$a^4+b^4=2(a^3+b^3)-4(a^2+b^2)=2(-16)-4(-4)=-16$$a^5+b^5=2(a^4+b^4)-4(a^3+b^3)=2(-16)-4(-16)=32$$a^6+b^6=2(a^5+b^5)-4(a^4+b^4)=2(32)-4(-16)=128$$a^7+b^7=2(a^6+b^6)-4(a^5+b^5)=2(128)-4(32)=128$$a^8+b^8=2(a^7+b^7)-4(a^6+b^6)=2(128)-4(128)=-256$$a^9+b^9=2(a^8+b^8)-4(a^7+b^7)=2(-256)-4(128)=-1024$

Answer 2

              x² - 2 x + 4 = 0
       Given that    a and b are the roots.
   =>  a² - 2 a + 4 = 0        and      b² - 2 b + 4 = 0

Using complex numbers method

   a = [ 2 + √(4-16) ] / 2 = 1 + i √3
      = 2 (Cos π/3 + i Sin π/3 ) = 2 $e^{i\frac{\pi}{3}}$

   b = [ 2 - √(4-16) ] / 2 = 1 - √3  i
      = 2 (Sin π/3 -  i Sin π/3 ) = 2 $e^{-i\frac{\pi}{3}}$

$a^9 + b^9 = 2^9 * [ e^{i\frac{9\pi}{3} } + e^{i\frac{-9\pi}{3} } ]$

So  a⁹+b⁹
       = 2⁹ * [ Cos 3π + i Sin 3π + Cos 3π - i Sin 3π ]
       =  2⁹ * 2 * (-1)
         = - 1024
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Another simpler method:
              x² - 2 x + 4 = 0

So sum of roots:   a + b = 2      and      product of roots:   a b = 4

  a³ + b³ = (a+b)³ - 3 a b (a+b)        ---- (1)
            =  8 - 3 * 4 * 2 = - 16

  a⁹ + b⁹ = (a³ + b³)³ - 3 a³ b³ (a³ + b³)            applying the same principle as in (1)
             = (-16)³ - 3 (4)³ (-16)
             = - 16³  + 3 * 4 * 16²
             = 16² * [ -16 + 12 ]
             = - 1024