Find all 4 complex roots to 3z^4-z^3+2z^2-z+3=0 using a trig substitution. A previous part

Zugdichte2r

Zugdichte2r

Answered question

2022-01-26

Find all 4 complex roots to 3z4z3+2z2z+3=0 using a trig substitution.
A previous part of the question made us prove that 2cosnθ=zn+zn and it wants us to prove that the polynomial can be expressed as 6cos2θ2cosθ+2=0. I'm assuming that using 6cos2θ2cosθ+2=0 we will be able to find all the roots of the equation but I'm unable to see how to get the polynomial into the trig equation and then how to solve the trig equation.

Answer & Explanation

oferenteoo

oferenteoo

Beginner2022-01-27Added 12 answers

Hint: The given equation is equivalent to
3(z2+1z2)(z+1z)+2=0
By letting z+1z=w we get
3(w22)w+2=0
Solve this quadratic equation for w and subsequently solve 1+1z=w.Since w is real and |w|<2, we can substitute z=eiθ, w=2cosθ and solve for θ
dodato0n

dodato0n

Beginner2022-01-28Added 9 answers

Use that
cos(2x)=2cos2(x)1

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