z^4 + 4z^3 -z +1 =0 solve

Question

z^4  + 4z^3 -z +1 =0  

 

 

we know that z1+z2+z3+z4=(-4)

&z1*z2*z3*z4=1

Answer 1

A graph of the quartic reveals that there are two real and two complex roots.

One real root is close to -1, the other close to -4. Newton's iterative method:

z_n+1=x_n-f(z_n)/f'(z_n) gives us successive iterations for z which converge towards a solution. Two separate calculations in which z₀=-1 and z₀=-4 give the roots z=-0.833487 and -3.918241 approximately. f(z) is the given quartic, and f'(z) its first derivative.

If we call these real roots A and B then we can combine them into a quadratic:

(z-A)(z-B)=z²-(A+B)z+AB then divide it into f(z).

A+B=-4.751728, AB=3.265803.

                          z²+(4+A+B)z+(4+A+B)(A+B)-AB                           (quotient)

z²-(A+B)z+AB ) z⁴       +4z³                                                      -z+1

                          z⁴-(A+B)z³                    +ABz²

                              (4+A+B)z³                 -ABz²                         -z

                              (4+A+B)z³-(4+A+B)(A+B)z²     +AB(4+A+B)z

                                        ((4+A+B)(A+B)-AB)z²-(AB(4+A+B)+1)z

We need go no further because we have the required quadratic in the quotient: z²+(4+A+B)z+(4+A+B)(A+B)-AB.

This can be solved using the quadratic formula and substituting in the values of A and B, remembering that the roots are complex so we are expecting the square root of a negative quantity.

The quadratic is z²+px+q where p=4+A+B and q=(A+B)p-AB.

The roots are ½(-p±√(p²-4q)).

p=-0.751728, q=0.306203; √(p²-4q)=√-0.659719=0.812231i. So the two complex roots are 0.375864±0.406115.

So the four roots are: -0.833487, -3.918241, 0.375864+0.406115i, 0.375864-0.406115i (to 6 decimal places).