代数的基本定理据/H1>据/HE.一种der>
用Facebook注册据/一种>据span class="or">或者据/span>
手动注册据/一种>据/div>
已经有一个帐户?据一种HR.E.F="//www.parkandroid.com/account/login/?next=/wiki/fundamental-theorem-of-algebra/" class="ax-click" data-ax-id="clicked_signup_modal_login" data-ax-type="link">这里登录。据/一种>据/P.>据/div>
相关......据/H4.>据ul class="unstyled">
代数据/span>>据/span>
贡献据/div>
这据strong>代数的基本定理据/strong>说任何不合适的据一种HR.E.F="//www.parkandroid.com/wiki/polynomials/" class="wiki_link" title="多项式GydF4y2Ba" target="_blank">多项式据/一种>和据一种HR.E.F="//www.parkandroid.com/wiki/complex-numbers/" class="wiki_link" title="复杂的GydF4y2Ba" target="_blank">复杂的据/一种>系数至少有一个复杂的据一种HR.E.F="//www.parkandroid.com/wiki/polynomial-roots/" class="wiki_link" title="根GydF4y2Ba" target="_blank">根据/一种>。定理意味着任何具有复杂度系数的多项式据span class="katex">
N据/span>已据span class="katex">
N据/span>复杂根,数量计数。一种据一种HR.E.F="//www.parkandroid.com/wiki/ring-theory/" class="wiki_link" title="场地GydF4y2Ba" target="_blank">场地据/一种>据span class="katex">
F据/span>随着每个不透视多项式与系数的财产据span class="katex">
F据/span>有一个根据span class="katex">
F据/span>叫做据strong>代数封闭据/strong>,因此代数的基本定理说明了这一点据/P.>据blockquote>
场据span class="katex">
C据/span>复杂数字是代数封闭的。据/P.>据/blockquote>
多项式据span class="katex">
X据/span>2据/span>+据/span>1据/span>没有真正的根,但它有两个复杂的根据span class="katex">
一世据/span>和据span class="katex">
-据/span>一世据/span>。据/span>
多项式据span class="katex">
X据/span>2据/span>+据/span>一世据/span>有两个复杂的根,即据span class="katex">
±据/span>2据/span>
1据/span>-据/span>一世据/span>。据/span>
人们可能期望具有复杂系数的多项式具有与实际多项式相似的根源不存在的问题;也就是说,猜测一些多项式的像这样的多项式是不合理的据span class="katex-display">
X据/span>3.据/span>+据/span>一世据/span>X据/span>2据/span>-据/span>(据/span>1据/span>+据/span>π据/span>一世据/span>)据/span>X据/span>-据/span>E.据/span>不会有一个复杂的根,并找到这样一个根源需要看一些包含复杂数字的更大字段。代数的基本定理表明,这不是这种情况:可以在复杂数内找到具有复杂系数的多项式的所有根。据/P.>据/div>
要约据/H2>据/HE.一种der>
本节给出了代数基本定理的不同等效形式的更精确的陈述。这需要定义多项式的多项的多个。据/P.>据blockquote class="definition">
这据strong>多重据/strong>根本据span class="katex">
R.据/span>多项式据span class="katex">
F据/span>(据/span>X据/span>)据/span>是最大的正整数据span class="katex">
K.据/span>这样据span class="katex">
(据/span>X据/span>-据/span>R.据/span>)据/span>K.据/span>划分据span class="katex">
F据/span>(据/span>X据/span>)据/span>。据/span>等效,是据E.m>最小的据/E.m>正整数据span class="katex">
K.据/span>这样据span class="katex">
F据/span>(据/span>K.据/span>)据/span>(据/span>R.据/span>)据/span>据/span>=据/span>0.据/span>那据/span>在哪里据span class="katex">
F据/span>(据/span>K.据/span>)据/span>表示这一点据span class="katex">
K.据/span>TH.据/span>衍生物据span class="katex">
F据/span>。据/span>
让据span class="katex">
F据/span>是一个领域。以下是等同的:据/P.>据P.>(1)每个不符合系数的多项式据span class="katex">
F据/span>有一个根据span class="katex">
F据/span>。据/span>
(2)每一个不合适的程度多项式据span class="katex">
N据/span>有系数据span class="katex">
F据/span>已据span class="katex">
N据/span>根in.据span class="katex">
F据/span>那据/span>用多个程度计数。据br>(3)每个不符合系数的多项式据span class="katex">
F据/span>完全分裂为带有系数的线性因子的产物据span class="katex">
F据/span>。据/span>
显然(3)据span class="katex">
⇒据/span>(2)据span class="katex">
⇒据/span>(1),所以唯一的非活动部分是(1)据span class="katex">
⇒据/span>(3)。要查看此项,请参阅学位据span class="katex">
N据/span>的据span class="katex">
F据/span>(据/span>X据/span>)据/span>。据/span>基本情况据span class="katex">
N据/span>=据/span>1据/span>清楚了。现在假设结果具有多项式的程度据span class="katex">
N据/span>-据/span>1据/span>。据/span>然后让我们据span class="katex">
F据/span>(据/span>X据/span>)据/span>是一种多项式据span class="katex">
N据/span>。据/span>(1),据span class="katex">
F据/span>(据/span>X据/span>)据/span>有一个根据span class="katex">
一种据/span>。据/span>标准据一种HR.E.F="//www.parkandroid.com/wiki/division-algorithm/" class="wiki_link" title="划分算法GydF4y2Ba" target="_blank">划分算法据/一种>参数显示据span class="katex">
X据/span>-据/span>一种据/span>是一个因素据span class="katex">
F据/span>(据/span>X据/span>)据/span>:据/P.>据P.>划分据span class="katex">
F据/span>(据/span>X据/span>)据/span>经过据span class="katex">
X据/span>-据/span>一种据/span>要得到据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>(据/span>X据/span>-据/span>一种据/span>)据/span>问:据/span>(据/span>X据/span>)据/span>+据/span>R.据/span>那据/span>在哪里据span class="katex">
R.据/span>是一个恒定的多项式。堵塞据span class="katex">
一种据/span>双方给予据span class="katex">
0.据/span>=据/span>(据/span>一种据/span>-据/span>一种据/span>)据/span>问:据/span>(据/span>一种据/span>)据/span>+据/span>R.据/span>那据/span>所以据span class="katex">
R.据/span>=据/span>0.据/span>。据/span>所以据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>(据/span>X据/span>-据/span>一种据/span>)据/span>问:据/span>(据/span>X据/span>)据/span>。据/span>但据span class="katex">
问:据/span>(据/span>X据/span>)据/span>是一种程度的多项式据span class="katex">
N据/span>-据/span>1据/span>那据/span>因此,它通过归纳假设分裂成线性因素的产物。所以据span class="katex">
F据/span>(据/span>X据/span>)据/span>也是如此。所以结果是通过诱导证明的。据span class="katex">
□据/span>
代数的基本定理说这个领域据span class="katex">
C据/span>复数有属性(1),因此在上面的定理必须具有属性(1),(2)和(3)。据/P.>据blockquote class="example">
如果据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>X据/span>4.据/span>-据/span>X据/span>3.据/span>-据/span>X据/span>+据/span>1据/span>那据/span>然后可以将复杂的根部一对一,直到多项式完全是:据span class="katex">
F据/span>(据/span>1据/span>)据/span>=据/span>0.据/span>那据/span>所以据span class="katex">
X据/span>4.据/span>-据/span>X据/span>3.据/span>-据/span>X据/span>+据/span>1据/span>=据/span>(据/span>X据/span>-据/span>1据/span>)据/span>(据/span>X据/span>3.据/span>-据/span>1据/span>)据/span>。据/span>然后据span class="katex">
1据/span>是一个根源据span class="katex">
X据/span>3.据/span>-据/span>1据/span>那据/span>所以据span class="katex-display">
X据/span>4.据/span>-据/span>X据/span>3.据/span>-据/span>X据/span>+据/span>1据/span>=据/span>(据/span>X据/span>-据/span>1据/span>)据/span>(据/span>X据/span>-据/span>1据/span>)据/span>(据/span>X据/span>2据/span>+据/span>X据/span>+据/span>1据/span>)据/span>那据/span>现在据span class="katex">
X据/span>2据/span>+据/span>X据/span>+据/span>1据/span>有两个复杂的根,即据一种HR.E.F="//www.parkandroid.com/wiki/primitive-roots-of-unity/" class="wiki_link" title="原始的第三根统一GydF4y2Ba" target="_blank">原始的第三根统一据/一种>据span class="katex">
ω.据/span>和据span class="katex">
ω.据/span>2据/span>那据/span>在哪里据span class="katex">
ω.据/span>=据/span>E.据/span>2据/span>π据/span>一世据/span>/据/span>3.据/span>。据/span>所以据span class="katex-display">
X据/span>4.据/span>-据/span>X据/span>3.据/span>-据/span>X据/span>+据/span>1据/span>=据/span>(据/span>X据/span>-据/span>1据/span>)据/span>2据/span>(据/span>X据/span>-据/span>ω.据/span>)据/span>(据/span>X据/span>-据/span>ω.据/span>2据/span>)据/span>。据/span>有三个不同的根,但四根具有多重的根,自根本据span class="katex">
1据/span>有多个据span class="katex">
2据/span>。据/span>
定理的应用据/H2>据/HE.一种der>
在复数号上考虑任何多项式的能力会降低到其他字段(例如,实数)上的许多困难的非线性问题到在复数上的线性。例如,复杂数字上的每个方形矩阵都有一个复杂的据一种HR.E.F="//www.parkandroid.com/wiki/eigenvalues-and-eigenvectors/" class="wiki_link" title="特征值GydF4y2Ba" target="_blank">特征值据/一种>,因为特征多项式总是有根。这不是真实数字的真实,例如,矩阵据span class="katex-display">
(据/span>0.据/span>-据/span>1据/span>1据/span>0.据/span>)据/span>那据/span>旋转真实坐标平面据span class="katex">
9.据/span>0.据/span>∘据/span>那据/span>没有真正的特征值。据/P.>据P.>另一个普遍应用是代数几何形状的领域,或者对多项式方程的解决方案的研究。假设多项式方程的系数位于代数封闭的场中,这对于简化和加强理论是必不可少的,因为它保证该领域“足够大”以包含多项式的根。例如,具有实际系数的多项式方程的复杂解决方案的组复溶液通常具有比真实解决方案集更自然和有用的特性。据/P.>据P.>另一个值得一提的申请是据一种HR.E.F="//www.parkandroid.com/wiki/integration-with-partial-fractions/" class="wiki_link" title="与部分分数集成GydF4y2Ba" target="_blank">与部分分数集成据/一种>。通过实数,存在涉及分母的不可缩短的二次因素的尴尬案例。通过在复杂数字上使用部分分数来简化代数(有一些警告据一种HR.E.F="//www.parkandroid.com/wiki/complex-analysis/" class="wiki_link" title="复杂分析GydF4y2Ba" target="_blank">复杂分析据/一种>需要解释结果的积分)。据/P.>据/div>
多项式超越真实数字据/H2>据/HE.一种der>
让据span class="katex">
P.据/span>(据/span>X据/span>)据/span>是具有真实系数的多项式。这是真的据span class="katex">
P.据/span>(据/span>X据/span>)据/span>可以在复杂数字中进行线性因素,但如果需要具有真实系数,则分解稍微复杂。据/P.>据P.>例如,多项式据span class="katex">
X据/span>2据/span>+据/span>1据/span>可以是因素据span class="katex">
(据/span>X据/span>-据/span>一世据/span>)据/span>(据/span>X据/span>+据/span>一世据/span>)据/span>在复杂的数字上,但是它的实际数字是据strong>不可减少据/strong>:它不能作为具有真实系数的两个不合作多项式的产物写成。据/P.>据blockquote class="theorem">
每一个多项式据span class="katex">
P.据/span>(据/span>X据/span>)据/span>具有真实系数可以被赋予具有真实系数的线性和不可缩短的二次因素的产物。据/P.>据!-- end-theorem -->
indinuct据span class="katex">
N据/span>。据/span>基本情况是据span class="katex">
N据/span>=据/span>1据/span>和据span class="katex">
N据/span>=据/span>2据/span>那据/span>这是微不足道的。现在假设定理对于多项式是真实的据span class="katex">
N据/span>-据/span>2据/span>和据span class="katex">
N据/span>-据/span>1据/span>。据/span>让据span class="katex">
F据/span>(据/span>X据/span>)据/span>是一种多项式据span class="katex">
N据/span>那据/span>然后让据span class="katex">
一种据/span>是一个复杂的根据span class="katex">
F据/span>(据/span>X据/span>)据/span>(由代数的基本定理存在)。有两种情况。据/P.>据P.>如果据span class="katex">
一种据/span>那是真的,然后据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>(据/span>X据/span>-据/span>一种据/span>)据/span>问:据/span>(据/span>X据/span>)据/span>对于多项式据span class="katex">
问:据/span>(据/span>X据/span>)据/span>具有真实系数的程度据span class="katex">
N据/span>-据/span>1据/span>。据/span>通过归纳假设,据span class="katex">
问:据/span>(据/span>X据/span>)据/span>可以考虑到线性和不可减少的二次因素的产品中,所以据span class="katex">
F据/span>(据/span>X据/span>)据/span>也可以。据/P.>据P.>如果据span class="katex">
一种据/span>不是真的,然后让据span class="katex">
一种据/span>是据一种HR.E.F="//www.parkandroid.com/wiki/complex-conjugates/" class="wiki_link" title="复杂共轭GydF4y2Ba" target="_blank">复杂共轭据/一种>的据span class="katex">
一种据/span>。据/span>注意据span class="katex">
一种据/span>据/span>=据/span>一种据/span>。据/span>写据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>C据/span>N据/span>X据/span>N据/span>+据/span>⋯据/span>+据/span>C据/span>1据/span>X据/span>+据/span>C据/span>0.据/span>那据/span>然后据/P.>据P.>据span class="katex-display">
F据/span>(据/span>X据/span>)据/span>=据/span>C据/span>N据/span>X据/span>N据/span>+据/span>⋯据/span>+据/span>C据/span>1据/span>X据/span>+据/span>C据/span>0.据/span>=据/span>C据/span>N据/span>X据/span>N据/span>+据/span>⋯据/span>+据/span>C据/span>1据/span>X据/span>+据/span>C据/span>0.据/span>=据/span>C据/span>N据/span>X据/span>N据/span>+据/span>⋯据/span>+据/span>C据/span>1据/span>X据/span>+据/span>C据/span>0.据/span>=据/span>C据/span>N据/span>X据/span>N据/span>+据/span>⋯据/span>+据/span>C据/span>1据/span>X据/span>+据/span>C据/span>0.据/span>=据/span>F据/span>(据/span>X据/span>)据/span>通过复杂缀合物的性质(并且是因为据span class="katex">
C据/span>一世据/span>是实数)。因此,如果据span class="katex">
F据/span>(据/span>一种据/span>)据/span>=据/span>0.据/span>那据/span>然后据span class="katex">
F据/span>(据/span>一种据/span>)据/span>=据/span>F据/span>(据/span>一种据/span>)据/span>=据/span>0.据/span>=据/span>0.据/span>。据/span>结论是据strong>具有真实系数的多项式的非真实根源来自复杂的共轭对据/strong>。据/P.>据P.>写据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>(据/span>X据/span>-据/span>一种据/span>)据/span>问:据/span>(据/span>X据/span>)据/span>那据/span>在哪里据span class="katex">
问:据/span>(据/span>X据/span>)据/span>具有复杂的系数,并插入据span class="katex">
一种据/span>双方。然后据span class="katex">
问:据/span>(据/span>一种据/span>)据/span>=据/span>0.据/span>。据/span>(这是参数使用的地方据span class="katex">
一种据/span>据/span>=据/span>一种据/span>。据/span>)据/span>所以据span class="katex">
问:据/span>(据/span>X据/span>)据/span>=据/span>(据/span>X据/span>-据/span>一种据/span>)据/span>H据/span>(据/span>X据/span>)据/span>那据/span>所以据span class="katex">
已经有一个帐户?据一种HR.E.F="//www.parkandroid.com/account/login/?next=/wiki/fundamental-theorem-of-algebra/" class="ax-click" data-ax-id="clicked_signup_modal_login" data-ax-type="link">这里登录。据/一种>据/P.>据/div>
相关......据/H4.>据ul class="unstyled">
这据strong>代数的基本定理据/strong>说任何不合适的据一种HR.E.F="//www.parkandroid.com/wiki/polynomials/" class="wiki_link" title="多项式GydF4y2Ba" target="_blank">多项式据/一种>和据一种HR.E.F="//www.parkandroid.com/wiki/complex-numbers/" class="wiki_link" title="复杂的GydF4y2Ba" target="_blank">复杂的据/一种>系数至少有一个复杂的据一种HR.E.F="//www.parkandroid.com/wiki/polynomial-roots/" class="wiki_link" title="根GydF4y2Ba" target="_blank">根据/一种>。定理意味着任何具有复杂度系数的多项式据span class="katex"> N据/span>已据span class="katex"> N据/span>复杂根,数量计数。一种据一种HR.E.F="//www.parkandroid.com/wiki/ring-theory/" class="wiki_link" title="场地GydF4y2Ba" target="_blank">场地据/一种>据span class="katex"> F据/span>随着每个不透视多项式与系数的财产据span class="katex"> F据/span>有一个根据span class="katex"> F据/span>叫做据strong>代数封闭据/strong>,因此代数的基本定理说明了这一点据/P.>据blockquote>
场据span class="katex"> C据/span>复杂数字是代数封闭的。据/P.>据/blockquote>
多项式据span class="katex"> X据/span>2据/span>+据/span>1据/span>没有真正的根,但它有两个复杂的根据span class="katex"> 一世据/span>和据span class="katex"> -据/span>一世据/span>。据/span>
多项式据span class="katex"> X据/span>2据/span>+据/span>一世据/span>有两个复杂的根,即据span class="katex"> ±据/span>2据/span> 1据/span>-据/span>一世据/span>。据/span>
人们可能期望具有复杂系数的多项式具有与实际多项式相似的根源不存在的问题;也就是说,猜测一些多项式的像这样的多项式是不合理的据span class="katex-display"> X据/span>3.据/span>+据/span>一世据/span>X据/span>2据/span>-据/span>(据/span>1据/span>+据/span>π据/span>一世据/span>)据/span>X据/span>-据/span>E.据/span>不会有一个复杂的根,并找到这样一个根源需要看一些包含复杂数字的更大字段。代数的基本定理表明,这不是这种情况:可以在复杂数内找到具有复杂系数的多项式的所有根。据/P.>据/div>
要约据/H2>据/HE.一种der>
本节给出了代数基本定理的不同等效形式的更精确的陈述。这需要定义多项式的多项的多个。据/P.>据blockquote class="definition">
这据strong>多重据/strong>根本据span class="katex"> R.据/span>多项式据span class="katex"> F据/span>(据/span>X据/span>)据/span>是最大的正整数据span class="katex"> K.据/span>这样据span class="katex"> (据/span>X据/span>-据/span>R.据/span>)据/span>K.据/span>划分据span class="katex"> F据/span>(据/span>X据/span>)据/span>。据/span>等效,是据E.m>最小的据/E.m>正整数据span class="katex"> K.据/span>这样据span class="katex"> F据/span>(据/span>K.据/span>)据/span>(据/span>R.据/span>)据/span>据/span>=据/span>0.据/span>那据/span>在哪里据span class="katex"> F据/span>(据/span>K.据/span>)据/span>表示这一点据span class="katex"> K.据/span>TH.据/span>衍生物据span class="katex"> F据/span>。据/span>
让据span class="katex"> F据/span>是一个领域。以下是等同的:据/P.>据P.>(1)每个不符合系数的多项式据span class="katex"> F据/span>有一个根据span class="katex"> F据/span>。据/span>
(2)每一个不合适的程度多项式据span class="katex"> N据/span>有系数据span class="katex"> F据/span>已据span class="katex"> N据/span>根in.据span class="katex"> F据/span>那据/span>用多个程度计数。据br>(3)每个不符合系数的多项式据span class="katex"> F据/span>完全分裂为带有系数的线性因子的产物据span class="katex"> F据/span>。据/span>
显然(3)据span class="katex"> ⇒据/span>(2)据span class="katex"> ⇒据/span>(1),所以唯一的非活动部分是(1)据span class="katex"> ⇒据/span>(3)。要查看此项,请参阅学位据span class="katex"> N据/span>的据span class="katex"> F据/span>(据/span>X据/span>)据/span>。据/span>基本情况据span class="katex"> N据/span>=据/span>1据/span>清楚了。现在假设结果具有多项式的程度据span class="katex"> N据/span>-据/span>1据/span>。据/span>然后让我们据span class="katex"> F据/span>(据/span>X据/span>)据/span>是一种多项式据span class="katex"> N据/span>。据/span>(1),据span class="katex"> F据/span>(据/span>X据/span>)据/span>有一个根据span class="katex"> 一种据/span>。据/span>标准据一种HR.E.F="//www.parkandroid.com/wiki/division-algorithm/" class="wiki_link" title="划分算法GydF4y2Ba" target="_blank">划分算法据/一种>参数显示据span class="katex"> X据/span>-据/span>一种据/span>是一个因素据span class="katex"> F据/span>(据/span>X据/span>)据/span>:据/P.>据P.>划分据span class="katex"> F据/span>(据/span>X据/span>)据/span>经过据span class="katex"> X据/span>-据/span>一种据/span>要得到据span class="katex"> F据/span>(据/span>X据/span>)据/span>=据/span>(据/span>X据/span>-据/span>一种据/span>)据/span>问:据/span>(据/span>X据/span>)据/span>+据/span>R.据/span>那据/span>在哪里据span class="katex"> R.据/span>是一个恒定的多项式。堵塞据span class="katex"> 一种据/span>双方给予据span class="katex"> 0.据/span>=据/span>(据/span>一种据/span>-据/span>一种据/span>)据/span>问:据/span>(据/span>一种据/span>)据/span>+据/span>R.据/span>那据/span>所以据span class="katex"> R.据/span>=据/span>0.据/span>。据/span>所以据span class="katex"> F据/span>(据/span>X据/span>)据/span>=据/span>(据/span>X据/span>-据/span>一种据/span>)据/span>问:据/span>(据/span>X据/span>)据/span>。据/span>但据span class="katex"> 问:据/span>(据/span>X据/span>)据/span>是一种程度的多项式据span class="katex"> N据/span>-据/span>1据/span>那据/span>因此,它通过归纳假设分裂成线性因素的产物。所以据span class="katex"> F据/span>(据/span>X据/span>)据/span>也是如此。所以结果是通过诱导证明的。据span class="katex"> □据/span>
代数的基本定理说这个领域据span class="katex"> C据/span>复数有属性(1),因此在上面的定理必须具有属性(1),(2)和(3)。据/P.>据blockquote class="example">
如果据span class="katex"> F据/span>(据/span>X据/span>)据/span>=据/span>X据/span>4.据/span>-据/span>X据/span>3.据/span>-据/span>X据/span>+据/span>1据/span>那据/span>然后可以将复杂的根部一对一,直到多项式完全是:据span class="katex"> F据/span>(据/span>1据/span>)据/span>=据/span>0.据/span>那据/span>所以据span class="katex"> X据/span>4.据/span>-据/span>X据/span>3.据/span>-据/span>X据/span>+据/span>1据/span>=据/span>(据/span>X据/span>-据/span>1据/span>)据/span>(据/span>X据/span>3.据/span>-据/span>1据/span>)据/span>。据/span>然后据span class="katex"> 1据/span>是一个根源据span class="katex"> X据/span>3.据/span>-据/span>1据/span>那据/span>所以据span class="katex-display"> X据/span>4.据/span>-据/span>X据/span>3.据/span>-据/span>X据/span>+据/span>1据/span>=据/span>(据/span>X据/span>-据/span>1据/span>)据/span>(据/span>X据/span>-据/span>1据/span>)据/span>(据/span>X据/span>2据/span>+据/span>X据/span>+据/span>1据/span>)据/span>那据/span>现在据span class="katex"> X据/span>2据/span>+据/span>X据/span>+据/span>1据/span>有两个复杂的根,即据一种HR.E.F="//www.parkandroid.com/wiki/primitive-roots-of-unity/" class="wiki_link" title="原始的第三根统一GydF4y2Ba" target="_blank">原始的第三根统一据/一种>据span class="katex"> ω.据/span>和据span class="katex"> ω.据/span>2据/span>那据/span>在哪里据span class="katex"> ω.据/span>=据/span>E.据/span>2据/span>π据/span>一世据/span>/据/span>3.据/span>。据/span>所以据span class="katex-display"> X据/span>4.据/span>-据/span>X据/span>3.据/span>-据/span>X据/span>+据/span>1据/span>=据/span>(据/span>X据/span>-据/span>1据/span>)据/span>2据/span>(据/span>X据/span>-据/span>ω.据/span>)据/span>(据/span>X据/span>-据/span>ω.据/span>2据/span>)据/span>。据/span>有三个不同的根,但四根具有多重的根,自根本据span class="katex"> 1据/span>有多个据span class="katex"> 2据/span>。据/span>
定理的应用据/H2>据/HE.一种der>
在复数号上考虑任何多项式的能力会降低到其他字段(例如,实数)上的许多困难的非线性问题到在复数上的线性。例如,复杂数字上的每个方形矩阵都有一个复杂的据一种HR.E.F="//www.parkandroid.com/wiki/eigenvalues-and-eigenvectors/" class="wiki_link" title="特征值GydF4y2Ba" target="_blank">特征值据/一种>,因为特征多项式总是有根。这不是真实数字的真实,例如,矩阵据span class="katex-display"> (据/span>0.据/span>-据/span>1据/span>1据/span>0.据/span>)据/span>那据/span>旋转真实坐标平面据span class="katex"> 9.据/span>0.据/span>∘据/span>那据/span>没有真正的特征值。据/P.>据P.>另一个普遍应用是代数几何形状的领域,或者对多项式方程的解决方案的研究。假设多项式方程的系数位于代数封闭的场中,这对于简化和加强理论是必不可少的,因为它保证该领域“足够大”以包含多项式的根。例如,具有实际系数的多项式方程的复杂解决方案的组复溶液通常具有比真实解决方案集更自然和有用的特性。据/P.>据P.>另一个值得一提的申请是据一种HR.E.F="//www.parkandroid.com/wiki/integration-with-partial-fractions/" class="wiki_link" title="与部分分数集成GydF4y2Ba" target="_blank">与部分分数集成据/一种>。通过实数,存在涉及分母的不可缩短的二次因素的尴尬案例。通过在复杂数字上使用部分分数来简化代数(有一些警告据一种HR.E.F="//www.parkandroid.com/wiki/complex-analysis/" class="wiki_link" title="复杂分析GydF4y2Ba" target="_blank">复杂分析据/一种>需要解释结果的积分)。据/P.>据/div>
多项式超越真实数字据/H2>据/HE.一种der>
让据span class="katex">
P.据/span>(据/span>X据/span>)据/span>是具有真实系数的多项式。这是真的据span class="katex">
P.据/span>(据/span>X据/span>)据/span>可以在复杂数字中进行线性因素,但如果需要具有真实系数,则分解稍微复杂。据/P.>据P.>例如,多项式据span class="katex">
X据/span>2据/span>+据/span>1据/span>可以是因素据span class="katex">
(据/span>X据/span>-据/span>一世据/span>)据/span>(据/span>X据/span>+据/span>一世据/span>)据/span>在复杂的数字上,但是它的实际数字是据strong>不可减少据/strong>:它不能作为具有真实系数的两个不合作多项式的产物写成。据/P.>据blockquote class="theorem">
每一个多项式据span class="katex">
P.据/span>(据/span>X据/span>)据/span>具有真实系数可以被赋予具有真实系数的线性和不可缩短的二次因素的产物。据/P.>据!-- end-theorem -->
indinuct据span class="katex">
N据/span>。据/span>基本情况是据span class="katex">
N据/span>=据/span>1据/span>和据span class="katex">
N据/span>=据/span>2据/span>那据/span>这是微不足道的。现在假设定理对于多项式是真实的据span class="katex">
N据/span>-据/span>2据/span>和据span class="katex">
N据/span>-据/span>1据/span>。据/span>让据span class="katex">
F据/span>(据/span>X据/span>)据/span>是一种多项式据span class="katex">
N据/span>那据/span>然后让据span class="katex">
一种据/span>是一个复杂的根据span class="katex">
F据/span>(据/span>X据/span>)据/span>(由代数的基本定理存在)。有两种情况。据/P.>据P.>如果据span class="katex">
一种据/span>那是真的,然后据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>(据/span>X据/span>-据/span>一种据/span>)据/span>问:据/span>(据/span>X据/span>)据/span>对于多项式据span class="katex">
问:据/span>(据/span>X据/span>)据/span>具有真实系数的程度据span class="katex">
N据/span>-据/span>1据/span>。据/span>通过归纳假设,据span class="katex">
问:据/span>(据/span>X据/span>)据/span>可以考虑到线性和不可减少的二次因素的产品中,所以据span class="katex">
F据/span>(据/span>X据/span>)据/span>也可以。据/P.>据P.>如果据span class="katex">
一种据/span>不是真的,然后让据span class="katex">
一种据/span>是据一种HR.E.F="//www.parkandroid.com/wiki/complex-conjugates/" class="wiki_link" title="复杂共轭GydF4y2Ba" target="_blank">复杂共轭据/一种>的据span class="katex">
一种据/span>。据/span>注意据span class="katex">
一种据/span>据/span>=据/span>一种据/span>。据/span>写据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>C据/span>N据/span>X据/span>N据/span>+据/span>⋯据/span>+据/span>C据/span>1据/span>X据/span>+据/span>C据/span>0.据/span>那据/span>然后据/P.>据P.>据span class="katex-display">
F据/span>(据/span>X据/span>)据/span>=据/span>C据/span>N据/span>X据/span>N据/span>+据/span>⋯据/span>+据/span>C据/span>1据/span>X据/span>+据/span>C据/span>0.据/span>=据/span>C据/span>N据/span>X据/span>N据/span>+据/span>⋯据/span>+据/span>C据/span>1据/span>X据/span>+据/span>C据/span>0.据/span>=据/span>C据/span>N据/span>X据/span>N据/span>+据/span>⋯据/span>+据/span>C据/span>1据/span>X据/span>+据/span>C据/span>0.据/span>=据/span>C据/span>N据/span>X据/span>N据/span>+据/span>⋯据/span>+据/span>C据/span>1据/span>X据/span>+据/span>C据/span>0.据/span>=据/span>F据/span>(据/span>X据/span>)据/span>通过复杂缀合物的性质(并且是因为据span class="katex">
C据/span>一世据/span>是实数)。因此,如果据span class="katex">
F据/span>(据/span>一种据/span>)据/span>=据/span>0.据/span>那据/span>然后据span class="katex">
F据/span>(据/span>一种据/span>)据/span>=据/span>F据/span>(据/span>一种据/span>)据/span>=据/span>0.据/span>=据/span>0.据/span>。据/span>结论是据strong>具有真实系数的多项式的非真实根源来自复杂的共轭对据/strong>。据/P.>据P.>写据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>(据/span>X据/span>-据/span>一种据/span>)据/span>问:据/span>(据/span>X据/span>)据/span>那据/span>在哪里据span class="katex">
问:据/span>(据/span>X据/span>)据/span>具有复杂的系数,并插入据span class="katex">
一种据/span>双方。然后据span class="katex">
问:据/span>(据/span>一种据/span>)据/span>=据/span>0.据/span>。据/span>(这是参数使用的地方据span class="katex">
一种据/span>据/span>=据/span>一种据/span>。据/span>)据/span>所以据span class="katex">
问:据/span>(据/span>X据/span>)据/span>=据/span>(据/span>X据/span>-据/span>一种据/span>)据/span>H据/span>(据/span>X据/span>)据/span>那据/span>所以据span class="katex">