proof of fundamental theorem of algebra (argument principle)
The fundamental theorem of algebra can be proven using the argument principle. Not only is this proof interesting because it demonstrates an important result, it also serves as an example of how to use the argument principle. Since it is so simple, it can be thought of as a “toy model” (see toy theorem) for theorems on the zeros of analytic functions. For a variant of this proof using Rouché’s theorem (which is a consequence of the argument principle) please see the proof of the fundamental theorem of algebra (Rouché’s theorem).
Proof. Consider the rational function
Denote the degree of the polynomial by . Then we can write
This makes it clear that . Hence there exists a real constant such that whenever .
Consider the integral
This can be rewritten as
Split the integral into two parts, writing where
The integral is easy: . As for , we shall bound it using our bound for .
Since polynomials are analytic functions in the whole complex plane, is an analytic function of when , so the argument principle applies and we conclude that must equal the number of zeros of , counted with multiplicity. Among other things, this means that must be an integer. By explicit computation, we already know that is also an integer. Hence is an integer. But
Now, the only integer smaller that in absolute value is , so we must have . This implies that has zeros (counting with multiplicity) when . (By the way we chose , whenever , so has exactly n zeros in the whole complex plane.)
|Title||proof of fundamental theorem of algebra (argument principle)|
|Date of creation||2013-03-22 14:36:14|
|Last modified on||2013-03-22 14:36:14|
|Last modified by||rspuzio (6075)|