(a) Substitute \(z = -1 + \sqrt{2}i\) into the equation:
\((-1 + \sqrt{2}i)^4 + 3(-1 + \sqrt{2}i)^2 + 2(-1 + \sqrt{2}i) + 12 = 0\).
Calculate \((-1 + \sqrt{2}i)^2 = 1 - 2\sqrt{2}i - 2 = -1 - 2\sqrt{2}i\).
Calculate \((-1 + \sqrt{2}i)^4 = ((-1 + \sqrt{2}i)^2)^2 = (-1 - 2\sqrt{2}i)^2 = 1 + 4\cdot2 - 4\sqrt{2}i = 9 - 4\sqrt{2}i\).
Substitute back: \(9 - 4\sqrt{2}i + 3(-1 - 2\sqrt{2}i) + 2(-1 + \sqrt{2}i) + 12 = 0\).
Simplify: \(9 - 4\sqrt{2}i - 3 - 6\sqrt{2}i - 2 + 2\sqrt{2}i + 12 = 0\).
Combine terms: \(16 - 8\sqrt{2}i = 0\), which is true.
(b) The conjugate root is \(-1 - \sqrt{2}i\).
Find a quadratic factor: \((z + 1 - \sqrt{2}i)(z + 1 + \sqrt{2}i) = z^2 + 2z + 3\).
Divide \(z^4 + 3z^2 + 2z + 12\) by \(z^2 + 2z + 3\) to find the other quadratic factor:
\(z^4 + 3z^2 + 2z + 12 = (z^2 + 2z + 3)(z^2 - 2z + 4)\).
Solve \(z^2 - 2z + 4 = 0\) using the quadratic formula:
\(z = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{2 \pm \sqrt{-12}}{2} = 1 \pm \sqrt{3}i\).
The other roots are \(1 + \sqrt{3}i\) and \(1 - \sqrt{3}i\).
