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The fundamental theorem of algebra shows that any non-zero polynomial has a number of roots at most equal to its degree, and that the number of roots and the degree are equal when one considers the complex roots (or more generally, the roots in an algebraically closed extension) counted with their multiplicities. [3]
Graph of x 3 + 2x 2 − 7x + 4 with a simple root (multiplicity 1) at x=−4 and a root of multiplicity 2 at x=1. The graph crosses the x axis at the simple root. It is tangent to the x axis at the multiple root and does not cross it, since the multiplicity is even.
This proves Bézout's theorem, if the multiplicity of a common zero is defined as the multiplicity of the corresponding linear factor of the U-resultant. As for the preceding proof, the equality of this multiplicity with the definition by deformation results from the continuity of the U -resultant as a function of the coefficients of the f i ...
Rather, the degree of the zero polynomial is either left explicitly undefined, or defined as negative (either −1 or −∞). [10] The zero polynomial is also unique in that it is the only polynomial in one indeterminate that has an infinite number of roots. The graph of the zero polynomial, f(x) = 0, is the x-axis.
The solutions of the system are in one-to-one correspondence with the roots of h and the multiplicity of each root of h equals the multiplicity of the corresponding solution. The solutions of the system are obtained by substituting the roots of h in the other equations. If h does not have any multiple root then g 0 is the derivative of h.
If p(z 0) is nonzero, it follows that if a is a k th root of −p(z 0)/c k and if t is positive and sufficiently small, then |p(z 0 + ta)| < |p(z 0)|, which is impossible, since |p(z 0)| is the minimum of |p| on D. For another topological proof by contradiction, suppose that the polynomial p(z) has no roots, and consequently is never equal to 0 ...
Technically, a point z 0 is a pole of a function f if it is a zero of the function 1/f and 1/f is holomorphic (i.e. complex differentiable) in some neighbourhood of z 0. A function f is meromorphic in an open set U if for every point z of U there is a neighborhood of z in which at least one of f and 1/ f is holomorphic.
If f has a zero of order m at z 0 then for every small enough ρ > 0 and for sufficiently large k ∈ N (depending on ρ), f k has precisely m zeroes in the disk defined by |z − z 0 | < ρ, including multiplicity. Furthermore, these zeroes converge to z 0 as k → ∞. [1]