Lotsa roots

Write down all solutions of z^{42}+i=0.

We can rewrite this equation to day that z^{42} = -i. Furthermore, we can also write -i in polar form, -i = e^{i\frac{3\pi}{2}} = e^{i\frac{7\pi}{2}} = e^{i\frac{11\pi}{2}}, and so on, since adding 2\pi to the exponent only brings the coordinate back around to its original placement.

To find one root of z, we cold do some simple algebra for the equation z^{42} = e^{i\frac{3\pi}{2}}. We find that z = (e^{i\frac{3\pi}{2}})^{\frac{1}{42}} = e^{i\frac{3\pi}{84}}. We could do the same for any other polar form of -i. For example, z = (e^{i\frac{7\pi}{2}})^{\frac{1}{42}} = e^{i\frac{7\pi}{84}}.

The only thing that changed is that 2\pi was added to the exponent, only this time the coordinate rotated a small fraction of the way around the unit circle. Thus, any root of z^{42} + i = 0 can be expressed as e^{i\frac{(3 + 4n)\pi}{84}}.

Note that if \frac{4n\pi}{84} \geq 2\pi, we would have repeating roots, so the roots of z^{42} + i = 0 must be e^{i\frac{(3 + 4n)\pi}{84}} for n = 0, 1, 2, 3, ..., 41.

z^{42} + i = 0\\ \Leftrightarrow z^{42} = -i\\ \Leftrightarrow z^{42} = e^{(\frac{3\pi}{2}+2k\pi)i}, \text{ where }k\in\{0,1,2,...,41\}\\ \Leftrightarrow z= e^{(\frac{\pi}{28}+\frac{k\pi}{21})i}, \text{ where }k\in\{0,1,2,...,41\}.

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