Solve $\;$ $x^4 + 4 = 0$
Given: $\;$ $x^4 + 4 = 0$
i.e. $\;$ $\left(x^2 + 2i\right) \left(x^2 - 2i\right)= 0$
$\implies$ $\left(x^2 + 2i\right) = 0$ $\;$ OR $\;$ $\left(x^2 - 2i\right) = 0$
Now, $x^2 - 2i = 0$ $\implies$ $x = \left(2i\right)^{\frac{1}{2}}$
Let $2i = r \left(\cos \theta_1 + i \sin \theta_1\right)$
$\implies$ $r \cos \theta_1 = 0$, $\;\;$ $r \sin \theta_1 =2 $
$\therefore$ $\;$ $r = \sqrt{\left(0\right)^2 + \left(2\right)^2} = 2$
$\therefore$ $\;$ $\cos \theta_1 = 0$, $\;$ $\sin \theta_1 = 1$ $\implies$ $\theta_1 = \dfrac{\pi}{2}$
$\therefore$ $\;$ $2i = 2 \left[\cos \left(\dfrac{\pi}{2}\right) + i \sin \left(\dfrac{\pi}{2}\right)\right]$
$\begin{aligned}
\therefore \; x = \left(2 i\right)^{\frac{1}{2}} & = 2^{\frac{1}{2}} \left[\cos \left(\dfrac{\pi}{2}\right) + i \sin \left(\dfrac{\pi}{2}\right)\right]^{\frac{1}{2}} \\\\
& = \sqrt{2} \left[\cos \left(2 k \pi + \dfrac{\pi}{2}\right) + i \sin \left(2 k \pi + \dfrac{\pi}{2}\right)\right]^{\frac{1}{2}} \\\\
& = \sqrt{2} \left[\cos \left(4 k + 1\right) \dfrac{\pi}{4} + i \sin \left(4 k + 1\right) \dfrac{\pi}{4}\right] \;\;\;\; k = 0,1 \\\\
& = \sqrt{2} \left(\cos \dfrac{\pi}{4} + i \sin \dfrac{\pi}{4}\right), \; \sqrt{2} \left(\cos \dfrac{5 \pi}{4} + i \sin \dfrac{5 \pi}{4}\right)
\end{aligned}$
Now, $x^2 + 2i = 0$ $\implies$ $x = \left(-2i\right)^{\frac{1}{2}}$
Let $-2i = R \left(\cos \theta_2 + i \sin \theta_2\right)$
$\implies$ $R \cos \theta_2 = 0$, $\;\;$ $R \sin \theta_2 = - 2 $
$\therefore$ $\;$ $R = \sqrt{\left(0\right)^2 + \left(-2\right)^2} = 2$
$\therefore$ $\;$ $\cos \theta_2 = 0$, $\;$ $\sin \theta_2 = - 1$ $\implies$ $\theta_2 = 2 \pi - \dfrac{\pi}{2} = \dfrac{3 \pi}{2}$
$\begin{aligned}
\therefore \; x = \left(-2 i\right)^{\frac{1}{2}} & = 2^{\frac{1}{2}} \left[\cos \left(\dfrac{3\pi}{2}\right) + i \sin \left(\dfrac{3\pi}{2}\right)\right]^{\frac{1}{2}} \\\\
& = \sqrt{2} \left[\cos \left(2 k \pi + \dfrac{3\pi}{2}\right) + i \sin \left(2 k \pi + \dfrac{3\pi}{2}\right)\right]^{\frac{1}{2}} \\\\
& = \sqrt{2} \left[\cos \left(4 k + 3\right) \dfrac{\pi}{4} + i \sin \left(4 k + 3\right) \dfrac{\pi}{4}\right] \;\;\;\; k = 0,1 \\\\
& = \sqrt{2} \left(\cos \dfrac{3\pi}{4} + i \sin \dfrac{3\pi}{4}\right), \; \sqrt{2} \left(\cos \dfrac{7 \pi}{4} + i \sin \dfrac{7 \pi}{4}\right)
\end{aligned}$
$\therefore$ $\;$ $x = \sqrt{2} \; cis \left(\dfrac{\pi}{4}\right), \; \sqrt{2} \; cis \left(\dfrac{3\pi}{4}\right), \; \sqrt{2} \; cis \left(\dfrac{5\pi}{4}\right), \; \sqrt{2} \; cis \left(\dfrac{7\pi}{4}\right)$