# What are the four roots of the following complex polynomial?: P(z) = z^4 - 2z^2 + 4

Jun 14, 2018

$z = \pm \frac{\sqrt{6}}{2} \pm i \frac{\sqrt{2}}{2}$

#### Explanation:

${z}^{4} - 2 \cdot {z}^{2} + 4 = 0$

$\Delta = 4 - 4 \cdot 1 \cdot 4 = - 12$

${z}^{2} = \frac{2 \pm 2 i \sqrt{3}}{2}$

${z}^{2} = 2 \left(\frac{1}{2} \pm i \frac{\sqrt{3}}{2}\right)$

${z}^{2} = 2 \left(\cos \frac{\pi}{3} \pm i \sin \frac{\pi}{3}\right)$

$z = \pm \sqrt{2} \left(\cos \frac{\pi}{6} \pm i \sin \frac{\pi}{6}\right)$

$z = \pm \sqrt{2} \left(\frac{\sqrt{3}}{2} \pm \frac{i}{2}\right)$

$z = \pm \frac{\sqrt{6}}{2} \pm i \frac{\sqrt{2}}{2}$