Solve the complex number equation $$z^4 +4z^2 +16 = 0$$
Using the quadratic formula
$$ z^2 = \frac{-4 \pm \sqrt{4^2 -4(16)}}{2} = -2 \pm 2\sqrt{3} i $$
My teacher proceeded to write:
$$z = 2e^{(\pm \frac{2\pi}{3} +2k \pi)i/2}, \qquad k = 0,1$$
What does this last expression mean in solving for this question?
In general, the solution to
$$z^2 = r \times e^{i\theta} ~: ~r \in \Bbb{R^+}, ~\theta \in \Bbb{R}$$ is given by
$$z = \sqrt{r} \times e^{i\alpha} ~: \alpha \in \left\{\frac{\theta}{2}, ~\frac{\theta + \pi}{2} \right\}.$$
So,
$$-2 \pm \left[i \times 2\sqrt{3}\right]$$
may be re-written as
$$4 \times \left[-\frac{1}{2} \pm i\frac{\sqrt{3}}{2}\right]$$
$$= 4 \times [\cos(\theta) + i\sin(\theta)] = 4e^{i\theta} ~: ~\theta \in \left\{ \frac{2\pi}{3}, ~-\frac{2\pi}{3} \right\}.$$
Simply apply the result directly above to the analysis at the start of this answer.