# How determine 2nd degree equation if i know one root z= (1-i`sqrt3` )/2?

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### 1 Answer

You should remember that if you know the complex solution to an equation, you automatically know the next root, that is the complex conjugate of the given complex root.

Hence, by definition, if `z_1 = (1 - i*sqrt3)/2` , then `z_2 = bar z_1 = (1 + i*sqrt3)/2` .

You need to find the quadratic equation whose solutions are `z_1` and `z_2` , hence, you may use its factored form, such that:

`(z - z_1)(z - z_2) = (z - (1 - i*sqrt3)/2)(z - (1 + i*sqrt3)/2)`

`(z - z_1)(z - z_2) = (z - 1/2 + isqrt3/2)(z - 1/2 - isqrt3/2)`

Converting the product into a difference of squares, yields:

`(z - z_1)(z - z_2) = (z - 1/2)^2 - (isqrt3/2)^2`

Expanding the squares yields:

`(z - z_1)(z - z_2) = z^2 - 2*1/2*z + 1/4 - 3i^2/4`

Since `i^2 = -1` yields:

`(z - z_1)(z - z_2) = z^2 - z + 1/4 + 3/4`

`(z - z_1)(z - z_2) = z^2 - z + 4/4`

`(z - z_1)(z - z_2) = z^2 - z + 1`

**Hence, evaluating the quadratic equation, whose solutions are complex conjugates `z_(1,2) = (1 +- i*sqrt3)/2` , yields **`z^2 - z + 1 = 0.`