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关于X^7—1的因式分解 在复数范围内
人气:460 ℃ 时间:2019-11-11 18:24:07
解答
先求:X^7—1=0的7个根.
X1=1
X2=cos(2PI/7)+i*sin(2PI/7)
X3=cos(4PI/7)+i*sin(4PI/7)
X4=cos(6PI/7)+i*sin(6PI/7)
X5=cos(8PI/7)+i*sin(8PI/7)
X6=cos(10PI/7)+i*sin(10PI/7)
X7=cos(12PI/7)+i*sin(12PI/7)
X^7—1
=(x-X1)*(x-X2)*(X-X3)*(X-X4)*(X-X5)*(X-X6)*(X-X7)
=(X-1)*[X-cos(2PI/7)+i*sin(2PI/7)]*[X-cos(4PI/7)+i*sin(4PI/7)]*[X-cos(6PI/7)+i*sin(6PI/7)]*[X-cos(8PI/7)+i*sin(8PI/7)]*[X-cos(10PI/7)+i*sin(10PI/7)]*[X-cos(12PI/7)+i*sin(12PI/7)]
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