To solve the equation Z ^ 2 + 2Z + 1-I = 0 in the complex range

Let z = x + Yi
Then x ^ 2-y ^ 2 + 2xyi + 2x + 2yi + 1-I = 0
(x^2-y^2+2x+1)+(2xy+2y-1)i=0
So (x + 1) ^ 2-y ^ 2 = 0 (1)
2xy+2y-1=0 (2)
From (1): y = ± (x + 1)
From (2) we know: (x + 1) y = 1 / 2 > 0
So y = x + 1, so (x, y) = (√ 2 / 2-1, √ 2 / 2) or (- √ 2 / 2-1, √ 2 / 2)
So z = √ 2 / 2-1 + √ 2 / 2I or - √ 2 / 2-1 - √ 2 / 2I

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