Answer
$\color{blue}{\pm\sqrt{1\pm i\sqrt{3}}}$
or
$\color{blue}{\dfrac{\sqrt{6}}{2} +i\dfrac{\sqrt{2}}{2},\ -\dfrac{\sqrt{6}}{2} -i\dfrac{\sqrt{2}}{2},\ \dfrac{\sqrt{6}}{2} -i\dfrac{\sqrt{2}}{2},\ -\dfrac{\sqrt{6}}{2} +i\dfrac{\sqrt{2}}{2}}$
Work Step by Step
Since we can rewrite the given equation as $(z^2)^2 - 2(z^2) +4=0$, if we let $u=z^2$ then the given equation becomes $u^2-2u+4=0$ and may be considered as a quadratic equation in $u$. We can then use the quadratic equation to find solutions for $u$.
$\begin{align*}
u &= \frac{-(-2) \pm \sqrt{(-2)^2-4(1)(4)}}{2(1)} \\
&= \frac{2\pm \sqrt{4-16}}{2} \\
&= \frac{2\pm \sqrt{-12}}{2} \\
&= \frac{2\pm 2i\sqrt{3}}{2} \\
u &= 1 \pm i\sqrt{3}
\end{align*}$
Since $u=z^2$,
$\begin{align*}
z^2 &= 1 \pm i\sqrt{3} \\
z &= \color{blue}{\pm \sqrt{1 \pm i\sqrt{3}}}
\end{align*}$
The four roots may also be obtained by using DeMoivre's Theorem to extract the square roots of each value for $u$.
$u = 1\pm i\sqrt{3} = x+ iy \implies x=1, y=\pm\sqrt{3}$
$r = \sqrt{x^2+y^2} = \sqrt{1^2 +(\sqrt{3})^2}= \sqrt{1+3} = \sqrt{4} = 2$
$\tan\theta = \pm y/x = \pm \sqrt{3}/1 = \pm \sqrt{3} \implies \theta = \pm\pi/3$
$\begin{align*}
u &= r\,\text{cis}\,\theta \\
u &= 2\,\text{cis}\,(\pi/3), \color{blue}{2\,\text{cis}\,(-\pi/3)} \\
z^2 &= 2\,\text{cis}\, (\pi/3), \color{blue}{2\,\text{cis}\,(-\pi/3)} \\
z &= \sqrt{2}\,\text{cis}\,(\pi/6 + k\pi), \color{blue}{\sqrt{2}\,\text{cis}\,(-\pi/6+ k\pi)},\ k=0,1 \\
&= \sqrt{2}\,\text{cis}\,(\pi/6), \sqrt{2}\,\text{cis}\,(7\pi/6), \color{blue}{\sqrt{2}\,\text{cis}\,(-\pi/6), \sqrt{2}\,\text{cis}\,(5\pi/6)} \\
&= \sqrt{2}(\cos(\pi/6) + i\sin(\pi/6)), \sqrt{2}(\cos(7\pi/6) +i\sin(7\pi/6)), \\
& \qquad\qquad \color{blue}{\sqrt{2}(\cos(-\pi/6) +i\sin(-\pi/6)), \sqrt{2}(\cos(5\pi/6) + i\sin(5\pi/6))} \\
&= \sqrt{2}(\sqrt{3}/2 + i(1/2)), \sqrt{2}(-\sqrt{3}/2 - i(1/2)), \\
& \qquad\qquad \color{blue}{\sqrt{2}(\sqrt{3}/2 - i(1/2)), \sqrt{2}(-\sqrt{3}/2 + i(1/2))} \\
z &= \sqrt{6}/2 +i(\sqrt{2}/2)), -\sqrt{6}/2 -i(\sqrt{2}/2)), \\
& \qquad\qquad \color{blue}{\sqrt{6}/2 -i(\sqrt{2}/2)), -\sqrt{6}/2 +i(\sqrt{2}/2))}
\end{align*}$
