Find the four roots of the equation $z^{4} + 16 = 0$ and them to factorize $z^{4} + 16$ into quadratic factors with real coefficients.

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Solution

$z^{4} + 16 = 0$
$z^{4} = - 16$
$z^{4} = 16 i^{2}$
$z^{4} = \pm 4 i$
$4 i = 2 (2 i)$
$= 2 (1 + 2 i - 1)$
$= 2 (1 + 2 i + i^{2})$
$= 2 {(1 + i)}^{2}$
So, $- 4 i = 2 {(1 - i)}^{2}$
$z^{2} = 2 {(1 \pm i)}^{2}$
$z = \pm \sqrt{2} (1 \pm i)$
Four roots $= \sqrt{2} (1 + i), \sqrt{2} (1 - i), - \sqrt{2} (1 + i)$ & $- \sqrt{2} (1 - i)$

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