Solve z^4-2z^2+3=0

$${ z }^{ 4 } -2 { z }^{ 2 } +3=0$$

Solve for z

$$z=\sqrt[4]{3}e^{\frac{-\arctan(\sqrt{2})i+2\pi i}{2}}\approx -1.168770894+0.605000334i$$

$$z=\sqrt[4]{3}e^{-\frac{\arctan(\sqrt{2})i}{2}}\approx 1.168770894-0.605000334i$$

$$z=\sqrt[4]{3}e^{\frac{\arctan(\sqrt{2})i+2\pi i}{2}}\approx -1.168770894-0.605000334i$$

$$z=\sqrt[4]{3}e^{\frac{\arctan(\sqrt{2})i}{2}}\approx 1.168770894+0.605000334i$$

Steps Using the Quadratic Formula

Substitute t for z^{2}.

$$t^{2}-2t+3=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -2 for b, and 3 for c in the quadratic formula.

$$t=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\times 3}}{2}$$

Do the calculations.

$$t=\frac{2±\sqrt{-8}}{2}$$

Solve the equation t=\frac{2±\sqrt{-8}}{2} when ± is plus and when ± is minus.

$$t=1+\sqrt{2}i t=-\sqrt{2}i+1$$

Since z=t^{2}, the solutions are obtained by evaluating z=±\sqrt{t} for each t.

$$z=\sqrt[4]{3}e^{\frac{\arctan(\sqrt{2})i+2\pi i}{2}} z=\sqrt[4]{3}e^{\frac{\arctan(\sqrt{2})i}{2}} z=\sqrt[4]{3}e^{-\frac{\arctan(\sqrt{2})i}{2}} z=\sqrt[4]{3}e^{\frac{-\arctan(\sqrt{2})i+2\pi i}{2}}$$