Solving roots of higher order

By KS Teng

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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