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sqrt( ): (N x C') -> C' sqrtn(z) := <sqrtn(z0), z1/n>.
Proposition: For every n in N>0 and z in C', the n-th roots of z are sqrt^n(z) and the n-1 values that have the same distance from the origin and their angle shifted by multiples of 2 pi /n:
forall n in N>0, z in C': let r = sqrtn(z): (forall s in C': z = sn <=> exists i in N: s = <r0, shift(r1+2 pi i/n)>).