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I am asked to find all the solutions to $z^{42}=-1$. I go a head and square root both sides to produce $z^{21}= i$. Then I can write $z^{21}= r^{21} (\cos(21\ θ) + i\sin(21\ θ)) = 0+i. $ Hence $\cos(21\ θ)= 0$ and $\sin(21\ θ)=1$, as $r^{21} $ can't be equal to $0$. Therefore $21θ $ = ${π \over 2} $ and $ θ = {π\over42}$. Then using $\sin(21\ θ)=1$, I can write $r^{21}=1 $ and produce that $r\pm1$ $\ $so my answer would be $z=\pm \cos({π\over42}) + i\sin({π\over42}). $

My answer doesn't produce the cleanest numbers so I'm a little worried but I do believe the process is correct.

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2 Answers2

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You could just come to my consulting hour tomorrow :-) Hint: no need to square root. We did an example like this in class.

Michael Murray
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Hint: where $n$ is an integer, $$z^{42}=-1$$$$(re^{i\theta})^{42}=1e^{i \pi }$$ $$r^{42}e^{i 42\theta}=1e^{i(\pi+ 2 \pi n)}$$

Note $r\ge 0$ (by definition), and that $e^{i2\pi n}=1$.

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