proof of the fundamental theorem of algebra (Liouville’s theorem)
Let be a polynomial, and suppose has no root in . We will show is constant.
Let . Since is never zero, is defined and holomorphic on (ie. it is entire). Moreover, since is a polynomial, as , and so as . Then there is some such that whenever , and is continuous and so bounded on the compact set .
So is bounded and entire, and therefore by Liouville’s theorem is constant. So is constant as required.
|Title||proof of the fundamental theorem of algebra (Liouville’s theorem)|
|Date of creation||2013-03-22 12:18:59|
|Last modified on||2013-03-22 12:18:59|
|Last modified by||Evandar (27)|