The fundamental theorem of algebra states that every non constant polynomial having
coefficients in ℂ has a zero in ℂ. If ℂ is replaced by ℝ, this is not true because of the example,
x^{2} + 1 = 0. This theorem is a very remarkable result and notwithstanding its title, all the
best proofs of it depend on either analysis or topology. It was proved by Gauss in
1797. The proof given here follows Rudin [32]. See also Hardy [19] for a similar
proof, more discussion and references. You can also see the interesting article on
Wikipedia. You google fundamental theorem of algebra and go to this site. There
are many ways to prove it. This article claims the first completely correct proof
was done by Argand in 1806. The shortest proof is found in the theory of complex
analysis and is a simple application of Liouville’s theorem or the formula for counting
zeros.

Recall De Moivre’s theorem, Problem 15 on Page 459 from trigonometry which is listed
here for convenience.

Theorem 8.10.1Let r > 0 be given. Then if n is a positive integer,

[r(cost+ isin t)]n = rn(cosnt+ isin nt).

Recall that this theorem is the basis for proving the following corollary from trigonometry,
also listed here for convenience, see Problem 16 on Page 460.

Corollary 8.10.2Let z be a non zero complex numberand let k be a positiveinteger. Then there are always exactly k k^{th}roots of z in ℂ.

< ε. The function of
the lemma is just the sum of functions of this sort and so it follows that it is also
continuous.

Theorem 8.10.4(Fundamental theorem of Algebra) Let p

(z)

be anonconstantpolynomial. Then there exists z ∈ ℂ such that p

(z)

= 0.

Proof: Suppose the nonconstant polynomial p

(z)

= a_{0} + a_{1}z +

⋅⋅⋅

+ a_{n}z^{n},a_{n}≠0, has no
zero in ℂ. Since lim_{|z|
→∞}

|p(z)|

= ∞, there is a z_{0} with

|p(z0)| = min |p(z)| > 0
z∈ℂ

Then let q

(z)

=

p(pz(+z0z0))

. This is also a polynomial which has no zeros and the minimum of

|q(z)|

is 1 and occurs at z = 0. Since q

(0)

= 1, it follows q

(z)

= 1 + a_{k}z^{k} + r

(z)

where r

(z)

consists of higher order terms, exponent larger than k. Here a_{k} is the first coefficient
which is nonzero. Choose a sequence, z_{n}→ 0, such that a_{k}z_{n}^{k}< 0. For example, let
−a_{k}z_{n}^{k} =

(1∕n)

. Then

|q(zn)|

≤ 1 − 1∕n +

|r(zn)|

< 1 for all n large enough because the
higher order terms in r

(zn)

converge to 0 faster than z_{n}^{k}. This is a contradiction.
■