The n-th root of non-zero complex R (COS θ + isin θ) is n complex numbers Is this something for Shanghai college entrance examination?

The nth root of R (COS θ + isin θ)
=R ^ (1 / N) * [(COS θ + isin θ) n-th root]
=R ^ (1 / N) * [the nth root of e ^ (I * (2k * PI + θ)]
=r^(1/n) * e^(i*(2k*pi+θ)/n)
=r^(1/n) * (cos((2k*pi+θ)/n)+isin((2k*pi+θ)/n))
Where k = 0,1,2,... (n-1)
So: there are n complex numbers

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