推荐回答（2个）
这是代数基本定理，第一个严格证明通常认为是高斯给出的(1799年在哥廷根大学的博士论文)。至于为什么，这应该主要是复数域的完备性原因，相应的，在实数域，有理数域就不一定有N个根。
参考百度百科：
http://baike.baidu.com/link?url=eQF6OTjaBtxX_BvrWgJl_FiCvEDxLJocxCkEjNCsaNAcGZiFIhAwybH5QidkhrBxvehcgdiLPgFtjpCKQ3xTcIX4BQBMGL5P3exNQu-NF3S
这个是代数基本定理,高斯最早给的证明
我只记得一个在抽象代数书上的证明
证明比较长
思路大概是
1
实系数奇数次方程有实根
(这只要用数学分析中连续函数的介值定理)
2
复系数2次方程有2复根
(配方法就行)
3
实系数方程有复根
证
(粗略的)
次数设为
2^MQ
Q为奇数
对M归纳
M=0时
由1
得证
若M>=K时成立
对M=K+1时
G(X)=X^N+A(N-1)X^(N-1)......+A0
(N=2^MQ)
为实域R上多项式
则
在某一拓域F上有N个根(用到域的拓张的知识
如果不懂
可以想象
取X1为
一个字
定义他满足上述方程
讲其加到
R上
得R上拓域记为R(X1)
当然这一点是要证明的
不过涉及知识比较多
理解一下就好
然后
原多项式可分解为
(X-X1)G1(X)
接着继续取G1(X)=0的根X2
得R(X1,X2)
一直做下去
可得
在某1拓域上
G(X)=0有N个根
X1,X2......XN)
设为
X1,X2,......XN
则G(X)=(X-X1)......(X-XN)
对实数C
有
作X-(XI+XJ+CXIXJ)
对每个N>=I>J>=0
将他们全部相乘
得H(X)
则H(X)
为
N(N+1)/2=2^(M-1)Q(N+1)次注意到
Q(N+1)为奇数
再看H(X)
易知
H(X)中每项系数都为
X1,X2......XN在R上的对称多项式
由
对称多项式基本定理
知
每项系数
都能写成
U1,U2......UN的多项式
其中
U1=X1+X2+...XN
U2=X1X2+X1X3+...X1XN+X2X3...X2XN...+XN-1XN
U3=X1X2X3+X1X2X4...XN-2XN-1XN
......
UN=X1X2...XN
由韦达定理(或者说由(X-X1)(X-X2)...(X-XN)=G(X)展开对比系数)知
U1=-A(N-1)
U2=A(N-2)
......
UN=(-1)^N
*A
所以
U1...UN为实数
所以H(X)为实系数多项式
所以由归纳假设知
H(X)=0有复根
所以存在某个
I,J有
XI+XJ+CXIXJ为复数
(注意到
I
J
是与C有关的
所以记为I(C)
J(C))
因为
(I,J)的数对只有有限多个
但C属于R有无穷多
所以
存在
C1不=C2有
(I(C1),J(C1))=(I(C2),J(C2))记为I
J
则
XI+XJ+C1XIXJ=A属于C
XI+XJ+C2XIXJ=B属于C
则
容易解得
XI+XJ=(C2A-C1B)/(C2-C1)属于C
XIXJ=(A-B)/(C1-C2)属于C
则
XI
XJ
为
复系数2次方程
X^2-
(C2A-C1B)/(C2-C1)X+(A-B)/(C1-C2)=0
的2根
由2知
XI
XJ为复数
所以F(X)=0有复根
4
复系数方程有复根
证
设F(X)为复系数多项式
F1(X)为他的共轭
则
G(X)=F(X)F1(X)为实系数多项式
所以
G(X)=0有复根X
则为F(X)=0或F1(X)=0的根
所以
X或X的共轭为F(X)=0的复根
5复系数N次方程有N个复根(计入重根)
(这是明显的
因为由5
知
N次复系数方程F1(X)=0有复根
设为X1则F可分解
有
F1(X)=(X-X1)F2(X)
其中F2为复系数N-1次多项式
所以有复根
X2
则
F1(X)=(X-X1)(X-X2)F3(X)
一直下去得
F(X)=(X-X1)(X-X2)......(X-XN)
所以有N个复根
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