Fundamental theorem of algebrapolynomial of degree n has exactly n zeroes, counted with multiplicity. More formally, if
The theorem had been conjectured in the 17th century but could not be proved since the complex numbers had not yet been firmly grounded. The first rigorous proof was given by Carl Friedrich Gauss in 1799. (An almost complete proof had been given earlier by d'Alembert.) Gauss produced several different proofs throughout his lifetime.
All proofs of the fundamental theorem necessarily involve some analysis, or more precisely, the concept of continuity of real or complex polynomials. The main difficulty in the proof is to show that every non-constant polynomial has at least one zero. We mention approaches via complex analysis, topology, and algebra:
- Find a closed disk D of radius r centered at the origin such that |p(z)| > |p(0)| whenever |z|≥r. The minimum of |p(z)| on D is therefore achieved at some point z0 in the interior of D. If |p(z0)| = m > 0, then 1/p(z) is a holomorphic function in the entire complex plane. Applying Liouville's theorem which states that a bounded entire function must be constant, we conclude that a polynomial without complex zeroes must be constant. As an alternative to Liouville's theorem, we can take a Taylor series expansion of p(z) at z0: for some k>0 and some non-zero constant ck, we have p(z)=p(z0)+ck(z-z0)k+... It follows that for positive ε sufficiently small,
- Choose a positive number R such that for |z|=R, the leading term zn of p(z) dominates all other terms combined. As z traverses the circle |z|=R once counter-clockwise, p(z), like zn, winds n times counter-clockwise around 0. If p(z) has no zeroes, winding number remains unchanged as the loop followed by z is continuously deformed through circles of smaller and smaller radius from the original circle to the constant circle |z|=0. This is absurd.
- Replacing p(z) by its product with its complex conjugate, it suffices to check that the fundamental theorem is true for all polynomials with real coefficients. This can be proved by induction on the highest power of 2 dividing the degree of n. For n odd, a real polynomial of degree n has a real root by the intermediate value theorem. For n even, the number of two element subsets of an n element set is divisible by one less factor of 2 than n. We can therefore apply the induction hypothesis to the polynomials whose roots are given by symmetric functions in pairs of roots of p(z). If we know zi+zj and zizj are both complex numbers, then we can use the quadratic formula to show that zi and zj are in C.