Roots in CDate: 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/ |
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