高次多项函数write the equation of lowest degree with real coefficient if two of its roots are -1 and 1+i

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高次多项函数writetheequationoflowestdegreewithrealcoefficientiftwoofitsrootsare-1and1+i高次多项函数writetheequat

高次多项函数write the equation of lowest degree with real coefficient if two of its roots are -1 and 1+i
高次多项函数
write the equation of lowest degree with real coefficient if two of its roots are -1 and 1+i

高次多项函数write the equation of lowest degree with real coefficient if two of its roots are -1 and 1+i
因为为这个方称的根1+i为根,因为conjugate原理,1-i也为这个方程的根
(x+1)(x-1-i)(x-1+i)=0
(x+1)(x^2-x+ix-x+1-i-ix+i+1)=0
(x+1)(x^2-2x+2)=0
(x^3-2x^2+2x+x^2-2x+2)=0
x^3-x^2+2=0

实系数的话,因为有1+i为根,因此1-i也必为根
因此最低次的方程为:(x+1)[(x-1)^2+1]=0