In each case, find all numbers x. Find all numbers x
a) $x^4 = 16$
So i was given this question and i had no clue how to understand it when following my textbook. So by following a different question and trying to follow the steps i came up with this partial solution
If $x = re^{−i\theta}$ then $x^4 = 16$ becomes $r^4e^{4i\theta} = 16e^{i·0}$. Hence $r^4 = 16$ (whence $r = 2\sqrt 2$) and $4\theta = 0 + 2k\pi$; $0 = k\pi/2$ ,$k = 0, 1, 2, 3$.
I really do not understand the process of this and how it works.
 A: You are following the correct approach, but have a slight error when you go from $r^4=16$ to $r=2\sqrt 2$.  It should be $r=2$  Intuitively, when you raise a complex number to a power, the radius is raised to the power and the angle is multiplied by the power.  In this case, where the power is $4$ the angle $\theta$ of the original number goes to $4\theta$ in the power.  But remember, an angle of $2\pi$ is equivalent to $0$, so you can add $\frac 14$ (the inverse of the power) or any multiple times $2\pi$ and when you multiply by $4$ you get the same angle.  Try it with $30^\circ, 120^\circ, 210^\circ, 300^\circ$ and see what happens.
A: You wish to find all distinct complex numbers of the form $r\mathsf e^{i\theta}$, such that $$r^4 \mathsf e^{4i\theta}\; [0\leq \theta< 2\pi] = 16\mathsf e^{2k\,i\pi}\;[k\in\Bbb Z]$$


*

*We need a non-negative real number, $r$, such that $r^4 = 16$.  So $r=2$.  

*We need an angle, $\theta$ on the interval $[0;2\pi)$, such that $\theta = \frac{k\pi}{2}$ and $k$ is an integer.
Hence $r= 2$ and $\theta\in\{0, \tfrac {\pi} 2, \pi, \tfrac {3\pi}2\}$
Thus you want... $r\mathsf e^{i\theta}\in \{ 2\, \mathsf e^{0},  2\, \mathsf e^{i\pi /2},  2\, \mathsf e^{i\pi},  2\, \mathsf e^{3i\pi/2}\}$
Now simplify these results.
