How do you find the zeros of f(z)=z4+2z3−5z2−7z+20 ?
1 Answer
Use a numerical method to find approximations to the zeros:
x1,2≈1.49358±0.773932i
x3,4≈−2.49358±0.921833i
Explanation:
f(z)=z4+2z3−5z2−7z+20
By the rational roots theorem, any rational zeros of
That means that the only possible rational zeros are:
±1 ,±2 ,±4 ,±5 ,±10 ,±20
None of these works (in fact they all give positive values), so
This is a fairly typical quartic with two complex conjugate pairs of roots. It is possible to solve algebraically, but gets very messy.
The way I would approach it algebraically would be:
-
Use a Tschirnhaus transformation to derive a simplified quartic of the form:
t4+pt2+qt+r=0 -
Consider a factorisation of this simplified quartic of the form
(t2−at+b)(t2+at+c) -
Equate coefficients and use
(b+c)2=(b−c)2+4bc to get a cubic ina2 -
If the resulting cubic has three Real irrational roots then use a
t=kcosθ substitution to find the solutions in trigonometric form. Otherwise use Cardano's method to find the one Real zero. -
Use the positive zero of the cubic as
a2 and the positive square root of that asa . -
Derive
b andc to get two quadratics to solve. -
Solve the quadratics using the quadratic formula.
As I say, this (usually) gets rather messy.
Alternatively we can use a numerical method such as Durand Kerner to find approximations to the zeros:
x1,2≈1.49358±0.773932i
x3,4≈−2.49358±0.921833i
Here's the C++ program I used (implementing the Durand Kerner algorithm)...
