Solve \(z^4=i\).
Solution
First find a principal root: \(z^4=i = e^{i\pi\over 2} \implies z_0 = e^{i\pi \over 8}\). Now all of the roots are \(\ts{z_k = z_0 \zeta_4^k \st k=0,1,2,3}\) where \(\zeta_4=e^{2\pi i \over 4} = e^{i\pi \over 2}\), so
\(z_0 = e^{i\pi \over 8} e^{0i\pi \over 4} = e^{i\pi \over 8}\) corresponding to \({1\over 8} + {0\over 4}\),
\(z_1 = e^{i\pi \over 8} e^{1i\pi \over 4} = e^{3i\pi \over 8}\) corresponding to \({1\over 8} + {1\over 4} = {3\over 8}\),
\(z_2 = e^{i\pi \over 8} e^{2i\pi \over 4} = e^{5i\pi \over 8}\) corresponding to \({1\over 8} + {2\over 4} = {5\over 8}\),
\(z_3 = e^{i\pi \over 8} e^{3i\pi \over 4} = e^{7i\pi \over 8}\) corresponding to \({1\over 8} + {3\over 4} = {7\over 8}\).