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 most straightforward proofs depend on either analysis or topology. It was first mostly proved by Gauss in 1797. The first complete proof was given by Argand in 1806. The proof given here follows Rudin [28]. See also Hardy [15] for a similar proof, more discussion and references. The shortest proof is found in the theory of complex analysis. First I will give an informal explanation of this theorem which shows why it is reasonable to believe in the fundamental theorem of algebra.
Theorem 1.10.1 Let p
To begin with, here is the informal explanation. Dividing by the leading coefficient a_{n}, there is no loss of generality in assuming that the polynomial is of the form

If a_{0} = 0, there is nothing to prove because p

It follows that z^{n} is some point on the circle of radius
Denote by C_{r} the circle of radius r in the complex plane which is centered at 0. Then if r is sufficiently large and
For example, consider the polynomial x^{3} + x + 1 + i. It has no real zeros. However, you could let z = r

Expanding this expression on the left to write it in terms of real and imaginary parts, you get on the left

Thus you need to have both the real and imaginary parts equal to 0. In other words, you need to have (0,0) =

for some value of r and t. First here is a graph of this parametric function of t for t ∈ [0,2π] on the left, when r = 4. Note how the graph misses the origin 0 + i0. In fact, the closed curve is in the exterior of a circle which has the point 0 + i0 on its inside.
Next is the graph when r = .5. Note how the closed curve is included in a circle which has 0 + i0 on its outside. As you shrink r you get closed curves. At first, these closed curves enclose 0 + i0 and later, they exclude 0 + i0. Thus one of them should pass through this point. In fact, consider the curve which results when r = 1.386 which is the graph on the right. Note how for this value of r the curve passes through the point 0 + i0. Thus for some t, 1.386
Now here is a short rigorous proof for those who have studied analysis. The needed analysis will be presented later in the book. You need the extreme value theorem for example.
Proof: Suppose the nonconstant polynomial p

Then let q

for all n large enough because