Finding the solutions to $z^4 =-16$ in $\mathbb{C}$

Question

I am currently reading for my first course in Abstract Algebra and came across the following example: $z^4 =-16$ in $\mathbb{C}$. However, I am stuck on how the author found that 4$\theta$ is $\pi+n(2\pi)$. I have included the following statement here:

Answer 1

All you need is the following:

If $\cos \varphi=-1$ and $\sin\varphi =0$, then one has $$ \varphi =(2n+1)\pi,\quad n\in\mathbb{Z}. $$

One can check this fact by looking at the graphs of the functions $f(x)=\cos x$ and $=\sin x$.

Answer 2

Here is another way to approach it. Notice that

\begin{align*} z^{4} = -16 \Longleftrightarrow z^{4} + 16 = 0 \Longleftrightarrow (z^{2} + 4i)(z^{2} - 4i) = 0 \end{align*}

Thus your problem is reduced to solve $z^{2} = \pm 4i$. You may do it by setting

$$z^{2} = x^{2} - y^{2} + 2xyi = \pm 4i$$

Or you can approach it as suggested, that is to say, write $z^{4}$ in polar coordinates. More precisely, we have $z = \rho(\cos(\theta) + i\sin(\theta))$. Therefore, according to the De Moivre's theorem, it follows \begin{align*} & \rho^{4}(\cos(4\theta) + i\sin(4\theta)) = 16(\cos(\pi + 2k\pi) + i\sin(\pi + 2k\pi)) \Longleftrightarrow\\\\ & \rho = 2\,\,\wedge\,\,\theta = \frac{\pi}{4} + \frac{k\pi}{2},\,\,k\in\textbf{Z}. \end{align*}