Sketch of Second Proof (after Cauchy)

Consider the quantity s(z) = |P(z)|, where P(z) = a_nz ⁿ + a_n -1z ^n -1 + ... + a₁z¹ + a₀ is a polynomial of degree n. Clearly, s(z)0. If s(z) assumes a global minimum value, say, , on C, then it suffices to prove that  = 0, or equivalently, it suffices to derive a contradiction from the alternative assumption that > 0.

Suppose, then, that s(z₀) = > 0. It is convenient to "re-center" the arithmetic by setting Q(z) = P(z + z₀ ), so |Q(0)| = |P(z₀ )| = , where is likewise a global minimum for |Q(z)|.

Expanding out the definition of |Q(z)|, we obtain a set of new coefficients: Q(z) = b_nz ⁿ + b_n -1z ^n -1 + ... + b₁z¹ + b₀, where b_n = a_n0, and b₀ = P(z₀ )0. Let m be the exponent of the lowest power of z in Q(z) whose coefficient b_m is not zero. Now consider the behavior of Q(z) for points z whose absolute value is very small, say the points z =  lying on a small circle centered at the origin of radius . As observed above, as z sweeps once around this small circle, Q(z) closely approximates the behavior of b_mz^m + b₀, which sweeps out a small circle (of radius |b_m|^m) around the point b₀ = P(z₀ ). (In fact, Q(z) sweeps around this circle m times.) By choosing sufficiently small, we may ensure that the radius of this circle (that is, the magnitude of Q(z) - b₀) is smaller than . Such a circle will necessarily intersect the line segment connecting the origin to the point b₀ = P(z₀ ), at a point, say Q(z₁ ), nearer the origin than b₀ = P(z₀ ). But then |Q(z₁ )| = |P(z₁ + z₀ )| <, contradicting our choice of as a global minimum of s(z) = |P(z)|. (Actually, Q(z) may only lie near this circle, not on it. Fortunately, the discrepancy consists of the remaining terms of Q(z) (if any), all of which contain powers of higher than m. By taking even smaller, if necessary, we can guarantee that this discrepancy does not wreck the geometry – see Figure 5.)

Figure 5.Behavior of Q(z) for "small" values of z. If z lies on the small circle of radius centered at the origin of the z-plane, then Q(z)bmzm + b0, which is the larger circle in the figure, centered at Q(0) = b0, with radius |bm|m. For sufficiently small, the remaining terms of (if any) of Q(z) sum to a vector whose magnitude (proportional to n, n>m, is less than the radius of the smaller circle in the figure.) As z sweeps out the small circle of radius on the z-plane, the locus of Q(z) loops around Q(0) = b0 (in fact, m times), and necessarily intersects the vector drawn from the origin of the w-plane to Q(0).

Thus the assumption that  > 0 is untenable, and we conclude that  = 0, that is, that P(z) has a complex root.

• Perfect numbers are complex, complex numbers might be perfect
• Fundamental Theorem of Algebra: Statement and Significance
• What's in a proof?
• More about proofs
• Axiomatics
• Intuition and Rigor
• How to Prove Bolzano's Theorem
• Early attempts
• Proofs of the Fundamental Theorem of Algebra
  • Remarks on Proving The Fundamental Theorem of Algebra
    • Sketch of a Proof by Birkhoff and MacLane
    • Sketch of Second Proof (after Cauchy)
    • Details of the Cauchy's Proof
    • Note on the Extreme Value Theorem
    • Sketch of Proof by the methods of the theory of Complex Variables (after Liouville)
    • Maximum Modulus Theorem
  • A Proof of the Fundamental Theorem of Algebra: Standing on the shoulders of giants
  • Yet Another Proof of the Fundamental Theorem of Algebra
  • Fundamental Theorem of Algebra - Yet Another Proof
  • A topological proof, going in circles and counting