Fundamental theorem of algebra

liyi

Theorem Every function $f(x) = x^n + a_1 x^{n-1} + a_2 x^{n-2} + \cdots + a_n$ has a real or complex root.

Sketch of Proof
Suppose $x=r(\cos\theta+i\sin\theta)$ ( $x^k=r^k(\cos k\theta+i\sin k\theta)$ ) such that
$f(x) = P+Qi$
where
$P = r^n \cos n\theta + \cdots,\quad Q=r^n \sin\theta + \cdots$

If there is a pair of $(r,\theta)$ such that $P^2+Q^2=0$ , the theorem is proved.

Consider the function $U = \arctan \frac{P}{Q}$
then
$\frac{\partial U}{\partial r}=\frac{\frac{\partial P}{\partial r}Q-P\frac{\partial Q}{\partial r}}{P^2+Q^2}\quad \frac{\parti...$
and
$\frac{\partial^2 U}{\partial r\partial \theta} = \frac{H(r,\theta)}{(P^2+Q^2)^2}$
where $H(r,\theta)$ is a continuous function of $r,\theta$ .

Finally, let us consider
$I_1 = \int_0^R dr \int_0^{2\pi} \frac{\partial^2 U}{\partial r\partial \theta}d\theta$ and $I_2 = \int_0^{2\pi} d\theta \int_0^R \frac{\partial^2 U}{\partial r\partial \theta}dr$
where $R$ is positive.

If $P^2+Q^2\neq 0$ , then the integrand is continuous and it must holds that $I_1=I_2$ . Then we prove that the equality fails for $R$ large enough.

Anyone who likes to complete the proof?

**Posted:** Tue Apr 12, 2005 5:46 am

liyi

Hint:

It is clear that $I_1 = 0$
Now try to prove that $\lim_{r\to\infty} \frac{\partial U}{\partial \theta} = -n$ and it tends uniformly to $-n$ about $\theta$ .
Then show that $\lim_{R\to\infty}I_2 = -2\pi n$

**Posted:** Tue Apr 12, 2005 6:16 am