Purely empirically, for 0<r<1/2, EllipticK[4 E^(I ArcTan[1/2 Sqrt[-4 + 1/(1 - r) + 1/r]]) Sqrt[(1- r) r]] == E^(I ArcTan[Sqrt[r/(1 - r)]]) EllipticK[r] E.g., for r=1/4 EllipticK[Sqrt[3] E^(I*π/6)] == E^(I*π/6) EllipticK[1/4] For r=1/2, it gets the conjugate of the right answer, which is EllipticK[2] == -(-1)^(3/4) EllipticK[1/2] which can be found by listing all eight sign variations of the three square roots. Then we can try complex r: EllipticK[1/2 + I/2] == (-3 + 2 Sqrt[2])^(1/4) EllipticK[4 - 2 Sqrt[2]] which puts the lhs in polar form. And for r = I, EllipticK[(8 + 8 I) - 4 Sqrt[-1 + 7 I]] == (2^(1/4) E^(-(I*π/8)) - I E^(I*π/4)) EllipticK[I] None of the eight sign guesses works for Re(r)>1/2. If no one finds a proof of this stuff, imagine some future Mathematica just trying everything and choosing the numerically plausible one. --rwg