This is one question that I know ALL my students can excel in doing, that is solve $z^4 = -16$. So I’m not interested in showing you how to solve such problems, but I want to explain a particular step which you introduce in your working. So let’s take a quick look at the solution first.

$z^4 = -16 = 16 e^{i\pi}$

$z^4 = 16 e^{i(\pi + 2k\pi)}$ for $k = 0, \pm1, -2$

So here, we note that we introduced $e^{i(\pi)} = e^{i(\pi + 2k \pi)}$, but how is $e^{i\theta} = e^{i(\theta + 2k\pi)}$?

Intuitively, $2k\pi$ for $k \in \mathbb{Z}$ is simply full circle. So we are really just turning full circles about the same point here, which is why we are still referring to the same number.

I will do a simple mathematical proof here too.

$e^{i(2k\pi + \theta)}$

$= cos(2k\pi + \theta) + isin(2k\pi + \theta)$

$= cos(2k\pi)cos\theta - sin(2k\pi)sin\theta + i[sin(2k\pi)cos\theta + cos(2k\pi)sin\theta]$ (using formulas in th MF15)

For $k \in \mathbb{Z}, cos(2k\pi)=1 \mathrm{~and~} sin(2k\pi)=0$

$\therefore, e^{i(2k\pi + \theta)} = e^{i\theta}$

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