Complex Polynomial Question (1 Viewer)

[nrlwinner]

Can someone help me out with these questions.

1)Use z^n + z^-n = 2cosn@ to solve 2z^4 + 3z^3 + 5z^2 + 3z +2 =0

2) Solve (z-1)^4 = 16(z+1)4

 

[shaon0]

Can someone help me out with these questions.

1)Use z^n + z^-n = 2cosn@ to solve 2z^4 + 3z^3 + 5z^2 + 3z +2 =0

2) Solve (z-1)^4 = 16(z+1)4

1) Use De Moivre's for 1st part of q1, then you know the poly is symmetric so factor z^2 out:
2z^4+3z^3+5z^2+3z+2=0
z^2(2z^2+3z+5+3z^-1+2z^-2)=0
Then, group conj. pairs:
z^2(2(z^2+z^-2)+3(z+z^-1)+5)=0
z^2(4cos2@+6cos@+5)=0
2z^2(4(cos@)^2+3(cos@)+1/2)=0
z=0 OR cos@=...(using quadratic formula)
but z=/=0...so, only two symmetric solutions.

2) (z-1)^4=16(z+1)^4
(z-1)^4-(2(z+1))^4=0
((z-1)^2-(2(z+1))^2)((z-1)^2+(2(z+1))^2)=0
-(3z+1)(z+3)(z^2-2z+1+4z^2+8z+4)=0
-(3z+1)(z+3)(5z^2+6z+5)=0
(3z+1)(z+3)(5z^2+6z+5)=0
(3z+1)(z+3)z(10cos@+6)=0
(3z+1)(z+3)z(5cos@+2)=0
z=-1/3,-3,0,* for which * is z sol for @=acos(-2/5)
but z=/=0.
So, z=-1/3,-3,*.
EDIT: * are conj, roots: 1/5(-3+-4i)