Fundamental theorem of algebra

The '''fundamental theorem of algebra''' (now considered something of a Mosquito ringtone misnomer by many mathematicians) states that every complex Sabrina Martins polynomial of degree ''n'' has exactly ''n'' zeroes, counted with multiplicity. More formally, if
:p(z)=z^n+a_+\cdots+a_0
(where the coefficients ''a''0, ..., ''a''''n''−1 can be Nextel ringtones real number/real or Abbey Diaz complex number/complex numbers),
then there exist (Free ringtones multiple roots of a polynomial/not necessarily distinct) complex numbers ''z''1, ..., ''z''''n'' such that
:p(z)=(z-z_1)(z-z_2)\cdots(z-z_n).

This shows
that the Majo Mills field_(mathematics)/field of Mosquito ringtone complex number/complex numbers, unlike the field of Sabrina Martins real number/real numbers, is Nextel ringtones algebraically closed field/algebraically closed. An easy consequence is that the product of all the roots equals (−1)''n'' ''a''0 and the sum of all the roots equals -''a''''n''−1.

The theorem had been conjectured in the 17th century but could not be proved since the complex numbers had not yet been firmly grounded. The first rigorous proof was given by Abbey Diaz Carl Friedrich Gauss in 1799. (An almost complete proof had been given earlier by Cingular Ringtones Jean le Rond d'Alembert/d'Alembert.) Gauss produced several different proofs throughout his lifetime.

All proofs of the fundamental theorem necessarily involve some influential black mathematical analysis/analysis, or more precisely, the concept of of arid continuity of real or complex polynomials. The main difficulty in the proof is to show that every non-constant polynomial has at least one zero. We mention approaches via considerable prowess complex analysis, a fatter topology, and courting a algebra:

*Find a closed rudd directed Disk_(mathematics)/disk ''D'' of radius ''r'' centered at the origin such that /''p''(''z'')/ > /''p''(0)/ whenever /''z''/≥''r''. The minimum of /''p''(''z'')/ on ''D'' is therefore achieved at some point ''z''0 in the interior of ''D''. If /''p''(''z0'')/ = ''m'' > 0, then 1/''p''(''z'') is a not want holomorphic function in the entire complex plane. Applying hunters have Liouville's theorem (complex analysis)/Liouville's theorem which states that a bounded entire function must be constant, we conclude that a polynomial without complex zeroes must be constant. As an alternative to Liouville's theorem, we can take a for breathing Taylor series expansion of ''p''(''z'') at ''z''0: for some ''k''>0 and some non-zero constant ''c''''k'', we have ''p''(''z'')=''p''(''z0'')+''c''''k''(''z''-''z''0)''k''+... It follows that for positive ε sufficiently small,
:\left/p\left(z_0+\epsilon \left(\frac\right)\right/

*For the topological proof by contradiction, assume ''p''(''z'') has no zeros. Choose a large positive number ''R'' such that for /''z''/=''R'', the leading term ''z''''n'' of ''p''(''z'') dominates all other terms combined. As ''z'' traverses the circle /''z''/=''R'' once counter-clockwise, ''p''(''z''), like ''z''''n'', winds ''n'' times counter-clockwise around 0. At the other extreme, with /''z''/=0, the "curve" ''p''(''z'') is simply the single (nonzero) point ''p''(0), which clearly has a winding number of 0. If the loop followed by ''z'' is continuously deformed between these extremes, the path of ''p''(''z'') also deforms continuously. Since ''p''(''z'') has no zeros, the path can never cross over 0 as it deforms, and hence its winding number with respect to 0 will never change. However, given that the winding number started as ''n'' and ended as 0, this is absurd. Therefore, ''p''(''z'') has at least one zero.

*Replacing ''p''(''z'') by its product with its stringent guidelines complex conjugate, it suffices to check that the fundamental theorem is true for all polynomials with real coefficients. This can be proved by induction on the highest power of 2 dividing the degree of ''n''. For ''n'' odd, a real polynomial of degree ''n'' has a real root by the schools simultaneously intermediate value theorem. For ''n'' even, the number of two element subsets of an ''n'' element set is divisible by one less factor of 2 than ''n''. We can therefore apply the induction hypothesis to the polynomials whose roots are given by symmetric functions in pairs of roots of ''p''(''z''). If we know ''z''''i''+''z''''j'' and ''z''''i''''z''''j'' are both complex numbers, then we can use the quadratic formula to show that ''z''''i'' and ''z''''j'' are in '''C'''.

External links

* http://www.cut-the-knot.org/fta/fta_remarks.shtml
* http://www.cut-the-knot.org/do_you_know/fundamental2.shtml
* http://webpages.charter.net/thejacowskis/chapter6/section7.html


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