The reason the complex numbers are so significant in linear algebra is that they are algebraically complete. This means that every polynomial ∑ _{k=0}^{n}a_{k}z^{k}, n ≥ 1,a_{n}≠0, having coefficients a_{k} in ℂ has a root in in ℂ. I will give next a simple explanation of why it is reasonable to believe in this theorem followed by a legitimate proof. The first completely correct proof of this theorem was given in 1806 by Argand although Gauss is often credited with proving it earlier and many others worked on it in the 1700’s.
Theorem 1.6.1 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.
Proof: Suppose the nonconstant polynomial p

Then let q

for all n large enough because