Question

Find the polynomial equation over R of lowest degree which is satisfied by (1-i) and (3+2i).

Answer 1

Answer on Question #86079 – Math – Algebra

Question

Find the polynomial equation over $R$ of lowest degree which is satisfied by (1-i) and $(3+2i)$.

Solution

When the polynomial equation over $R$ has an integrated root, then the conjugate number to the root is also the root. So $(1 + i)$ and $(3 - 2i)$ are the roots of the equation.

Since we have 4 roots, the polynomial equation over $R$ of lowest degree is of the fourth degree. Then

$$\begin{array}{l} (x - (1 - i))(x - (1 + i))(x - (3 + 2i))(x - (3 - 2i)) = 0, \\ (x^2 - (1 + i)x - (1 - i)x + (1 - i)(1 + i))(x^2 - (3 - 2i)x - (3 + 2i)x + (3 + 2i)(3 - 2i)) = \\ = 0, \\ (x^2 - 2x + 1 - i^2)(x^2 - 6x + 9 - 4i^2) = 0, \\ (x^2 - 2x + 2)(x^2 - 6x + 13) = 0, \\ x^4 - 6x^3 + 13x^2 - 2x^3 + 12x^2 - 26x + 2x^2 - 12x + 26 = 0, \\ x^4 - 8x^3 + 27x^2 - 38x + 26 = 0. \end{array}$$

Answer: $x^4 - 8x^3 + 27x^2 - 38x + 26 = 0$ is a polynomial equation over $R$ of the lowest degree which is satisfied by (1-i) and $(3+2i)$.

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