Find all solutions to the equation, $$(\omega^2+1)^4=\omega$$ In complex numbers!
I tried the substitution $\omega=z^4$ but wasn't helpful...and the equation becomes more complicated by this way!
2026-04-29 10:09:11.1777457351
Finding all solutions to an equation in complex numbers!
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If you take the fourth root of each side: $$\begin{align} &\omega^2+1=\omega^{\frac{1}{4}} \\ \iff &z^8+1=z \\ \iff &z^8-z+1=0 \\ \iff &(z^2-z+1)(z^6+z^5-z^3-z^2+1) =0 \end{align}$$ Hope this helps.