I've stumbled upon an interesting problem. The task is to find all complex numbers $z$ such that $$|z| = 1$$ and $$\Im((z+1)^{2020}) = 0$$ So far I found, that it's possible to follow these steps: $$u = z+1$$ $$\Im(u^{2020}) = 0$$ $$\sin(2020x) = 0$$ $$x = \dfrac{\pi n}{2020}$$ Which basically gives me all complex numbers $u$ that follow $\Im(u^{2020}) = 0$. However, I don't know how to continue from this point.
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I would explore the fact that $z+1$ is a scalar times a 2020th root of unity – Alexandru Ionut Nov 12 '20 at 01:30
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Hint: if $|z| = 1$ and $z \ne -1$, then $\operatorname{arg}(z+1) = \frac{1}{2} \operatorname{arg}(z)$. (To see this, draw a parallelogram with sides $1$ and $z$.) – Daniel Schepler Nov 12 '20 at 01:37
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From $\Im((z+1)^{2020}) = 0$, it follows that $$\operatorname{arg}(z+1) = \frac{2k\pi}{2020} $$ $z=-1$ is one solution. If not, from $|z| = 1$, it holds that $$\operatorname{arg}(z+1) = \frac{1}{2} \operatorname{arg}(z)$$ Combining the two $$\operatorname{arg}(z) = \frac{k\pi}{505}$$
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@Alexander Ionut: OK, lemme dwell on this for a while. Very engaging approach! *Cheers!* – Robert Lewis Nov 12 '20 at 02:28