在复数范围内解方程z^2+2z+1-i=0
人气:343 ℃ 时间:2019-12-07 05:26:30
解答
设z=x+yi
则x^2-y^2+2xyi+2x+2yi+1-i=0
(x^2-y^2+2x+1)+(2xy+2y-1)i=0
所以(x+1)^2-y^2=0 (1)
2xy+2y-1=0 (2)
由(1)知:y=±(x+1)
由(2)知:(x+1)y=1/2>0
所以y=x+1,所以(x,y)=(√2/2-1,√2/2)或(-√2/2-1,-√2/2)
所以z=√2/2-1+√2/2i或-√2/2-1-√2/2i
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