Proving the fundamental theorem of algebra by using Rouche's theorem.

Related BrainMass Solutions

• Rouche's Theorem

  315727 Assessing Rouche's Theorem Use Rouche's Theorem to determine the number of zeros, counting multiplicities, of the polynomials inside the given regions.

• Complex variable ( Rouche's Theorem)

  This solution uses Rouche's Theorem to find the location of roots of a complex polynomial with step-by-step calculations.

• Complex Variables : Analytic Functions, Contours, Polynomials and Rouche's Theorem

  45970 Analytic Functions, Contours, Polynomials and Rouche's Theorem Please see the attached file for the fully formatted problems. --- - Suppose the function f is analytic inside... - Determine the number of zeros... - Write f(z) - z^n and...

• Fundamental theorem of algebra

  Again by using the Fundamental Theorem of Algebra, Q(x) has at least one root, say z_2 in C, then Q(x) = (x - z_2)R(x), where R(x) is a polynomial with degree n-2. So P(x) = (x - z_1)(x - z_2) R(x).

• Linear Algebra - Four Fundamental Subspaces

  (c) According to Theorem 4.7, we have (by using result of (a)) Thus we get . Done. This solution is comprised of a detailed explanation to answer Linear Algebra - Four Fundamental Subspaces.

• Fundamental Theorem of Algebra : How Many Solutions?

  216167 Fundamental Theorem of Algebra: How Many Solutions? Based on the fundamental theorem of algebra, how many solutions would the following equation have: x^3 + 2x^2 - 3x = 0?

• The composition of linear maps and their underlying spaces.

  The step-by-step solution to this problem illustrates the use of fundamental concepts in linear algebra.

• Polynomials and real zeros

  Please see the attached file for the complete solution. Thanks for using BrainMass. According to the Fundamental Theorem of Algebra, if is a polynomial with degree , then the equation has roots. These roots may be real or complex.

• Complex Analysis / Singularities / Argument Principle : Let f be meromorphic on the region G and not constant; show that neither the poles nor the zeros of f have a limit point in G.

  l fn(z)-f(z) / f(z) l < 1 , n >n0 or l f (z) l > l fn(z)- f(z) l , n>n0 Therefore, writing fn(z) as the sum of the two function as fn(z) = f(z) + { fn(z) - f(z)} and using Rouche's theorem, we get that functions fn(z), n >n0 have