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Why does an equation of degree n have n complex roots? As for how to solve (or can solve) needless to say Of course, the more superficial the better
Solve the equation Z2 = Z (= followed by the conjugate complex number of Z), Z is a complex number
The equation you want to solve is the conjugate of Z ^ 2 = Z, right?
Let z = a + bi (B ≠ 0), then the conjugate of Z = a-bi, substituting into
a^2-b^2+2abi=a-bi
Then a ^ 2-B ^ 2 = a
2ab=-b
The solution is a = - 1 / 2, B = √ 3 / 2 or B = - √ 3 / 2
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