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### Roots in C

```
Date: 01/01/99 at 13:13:08
From: ILAN FRAIMAN
Subject: Complex numbers

How do you prove the theorem that says that every polynomial has a root
in C?
```

```
Date: 01/01/99 at 16:27:33
From: Doctor Anthony
Subject: Re: Complex numbers

This is known as the 'Fundamental Theorem of Algebra' and you will find
proofs in books on Advanced Algebra.

The various proofs are rather long-winded and the most common ones
develop the idea of considering the behaviour of w in the W plane as z
traces out a closed rectangular path in the Z plane where:

w = a z^n + b z^(n-1) + c z^(n-2) + ..... + g z + h

If the path of w encloses the origin, when z completes its closed path
the value of arg(w) will have changed by a multiple of 2pi * w, or will
be unchanged if the path of w does not enclose the origin.

We choose the original z-path so that w does enclose the origin. We
then subdivide the original z path into 4 equal squares and again let z
follow the boundaries of each smaller square until we get one for which
w does enclose the origin (with subsequent change of arg(w) by a factor
of 2pi).

Carry out further subdivisions of the z path, each time choosing a
smaller z-square for which w encloses the origin. This process
continues until the size of the z-square is delta where delta -> 0. But
if w does circle the origin arg(w) must change by a multiple of 2pi.
However in the limit the change in z -> zero and from the equation
connecting w and z the change in arg(w) must also -> zero. But this
contradicts the fact that arg(w) should change by a multiple of 2pi.
It follows that the path of w must pass through the origin and
this means that there is a zero to the z polynomial.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Imaginary/Complex Numbers
High School Imaginary/Complex Numbers

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