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The Fundamental Theorem of Algebra |
The Fundamental Theorem of Algebra says that an n-th degree polynomial having complex coefficients has, counting multiplicities, n complex roots. It was first successfully proven by the great German mathematician Karl Friederich Gauss.
The task of proving this theorem can be greatly simplified. If it can be proven that such a polynomial p(z) has at least one complex root, say r, then the polynomial can be divided by (z−r), where r is the root, to obtain a polynomial of degree one less. Repeated operations of this sort results in n roots counting multiplicities.
Let p(z) be polynomial in z of degree n
Let q(z) be the polynomial created by replacing all of the coefficients of p(z) with their complex conjugate.
Then w(z)=p(z)q(z) is a polynomial of degree 2n with all real coefficients.
Proof:
Let the conjugate of c_{j} be denoted as c_{j}*. Let
where the indices j and k run from 0 to n.
Then
But
Now consider the conjugate of a coefficient d_{m} of w(z),
But changing the summation index from h to g=m-h gives
which is identical except for the labeling of the summation index to the expression for d_{m}. Thus d_{m}=d_{m}* and hence the coefficients of w(z) are real.
If w(z_{0})=0 then either p(z_{0})=0 or q(z_{0})=0.
Let p(z) be Σ_{j=0}^{n}r_{j}z^{j}. Let z=x+iy where i is the square root of negative one and both x and y are real. Suppose all of the powers of z^{j} are expanded into terms of powers of x and iy. All of the terms involving even powers of iy will be real and all terms involving odd powers of iy will be imaginary. Thus
All terms in R(x,y) will involve even powers of y, including possibly a constant which involves y^{0}. All terms in I(x,y) involve y to an odd power, and thus y may be factored out leaving only terms with an even power of y. Therefore
A root of p(z) is an x+iy such that
The second condition requires either that y=0 and hence the root is a real number, or that G(x, y^{2}) = 0.
If y is not zero and some value x_{0} and y_{0}^{2} such that F=0 and G=0 then x_{0}+iy_{0} is a root and also x_{0}−iy_{0}.
Let p(z)=z²+bz+c. Then
A root would involve
If y=0 then x²+bx+c=0. This latter equation can be expressed as
This is the case of two real roots.
If 2x+b=0 then x=−b/2 and y²=x²+bx+c. The solution is straightforward but it is insightful to view the solution as the intersection of the two functions, x=−b/2 and y²=x²+bx+c, as shown below for b=1 and c=1.
Taking the square root of y² gives a display in terms of x and y.
Let p(z)=z³+bz²+cz+d. Then
Thus
Ignoring the case of y=0, the two functions for y² for the case of b=1, c=1 and d=1 are plotted below.
There is a singularity at x=−b/3 and the graph line at the singularity is an artifact of the graphing program and not truly a part of the function.
Let p(z)=z^{4}+bz³+cz²+dz+e. Then
Then
and there is a singularity at x=−1/4. The other condition for a root is
This is a quadratic equation in y² and is easily solved. The results are:
(To be continued.)
Suppose p(z) is polynomial of degree n, where n is an odd number, with real coefficients and the coefficient of z^{n} is unity. Then p(z) can be represented as b(1/z)z^{n} where b(1/z) for z real has a limit of 1 as real z increases without bound in either the positive direction or the negative direction.
For some sufficiently large value of real z,say z_{+}, p(z) is positive. For some sufficiently large negative real value of z, say z_{-}, p(z) is negative. Therefore, by the Intermediate Value Theorem, there is some value of z between z_{-} and z_{+} there is a value of z, say z_{0}, such that p(z_{0})=0.
(To be continued.)
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