complex plane (i.e., for an ƒ with no roots on the imaginary line, p + q = n). Thus, we have the condition that ƒ is a (Hurwitz) stable polynomial if and only if p − q = n (the proof is given below). Using the Routh–Hurwitz theorem, we can replace the condition on p and q by a condition on the generalized Sturm chain, which will give in turn a condition on the coefficients of ƒ. Using matrices Let f(z) be a complex polynomial. The process is as follows: Compute the polynomials and such that where y is a real number. Compute the Sylvester matrix associated to and . Rearrange each row in such a way that an odd row and the following one have the same number of leading zeros. Compute each principal minor of that matrix. If at least one of the minors is negative (or zero), then the polynomial f is not stable. Example Let (for the sake of simplicity we take real coefficients) where (to avoid a root in zero so that we can use the Routh–Hurwitz theorem). First, we have to calculate the real polynomials and : Next, we divide those polynomials to obtain the generalized Sturm chain: yields yields and the Euclidean division stops. Notice that we had to suppose b different from zero in the first division. The generalized Sturm chain is in this case . Putting , the sign of is the opposite sign of a and the sign of by is the sign of b. When we put , the sign of the first element of the chain is again the opposite sign of a and the sign of by is the opposite sign of b. Finally, -c has always the opposite sign of c. Suppose now that f is Hurwitz-stable. This means that (the degree of f). By the properties of the function w, this is the same as and . Thus, a, b and c must have the same sign. We have thus found the necessary condition of stability for polynomials of degree 2. Routh–Hurwitz criterion for second, third and fourth-order polynomials The second-degree polynomial has both roots with negative real part (and the system with characteristic equation is stable) if and only if both coefficients satisfy . The third-order polynomial has all roots in the open left half-plane if and only if , , and are positive