so Eq. (B1) gives with which obviously obeys Eq. (B4). ¿From Eqs. (B6) and (B5) we find Equation (B2b) can then be rewritten as For the given analytic function C ( z ) it determines the reparametrizing function t ∗ ( s ). For the circle and ellipse it is more convenient to use the unit-disk parametrization, when For the ellipse C ( ω ) is given by the conformal map (where ν is the same as s in Eq. (B26) of [6]) and from Eq. (B11) we obtain The final step is to extract d θ ∗ ( σ ) / d σ from Eq. (B10). Remembering that we find which for ω = e i σ reproduces Eq. (B32) of [6]. 15 d θ d ω = 1 √$_{ν}$ − ω 2 √$_{1}$ − νω 2 (B15) C ( θ ) = a cos θ + i b sin θ, (B14) Φ( ω ) = √$_{a}$ 2 − b 2 π 2 K ( ν ) ( ω + 1) 2 √$_{ν}$ − ω 2 √$_{1}$ − νω 2 cos [ π 2 K ( ν ) F ( ω √$_{ν}$ , ν )] = ( ω 2 + 1) d C ( ω ) d ω . (B13) C ( ω ) = √$_{a}$ 2 − b $^{2}$sin [ π 2 K ( ν ) F ( ω √$_{ν}$ , ν )] (B12) Φ( ω ) = ( ω + 1) 2 ∮ S 1 d ω 2 π i [ C ( ω ) − C ( ω ′ )] ( ω ω ′ ) 2 . (B11) − i 2 Φ( s ) = d C ( t ∗ ( s )) d s = ˙ C ( t ∗ ( s )) d t ∗ ( s ) d s . (B10) Φ( s ) ≡ Φ$_{1}$( s ) + iΦ$_{2}$( s ) = 2Φ$_{1}$( s ); (B9a) Im Φ$_{1}$( s ) + iIm Φ$_{2}$( s ) = − i 2 Φ( s ) . (B9b) and from Eq. (B2b) conclude that t ∗ ( s ) = s , as it should be for the circle. For the functions obeying Eq. (B7), we always have Φ$_{1}$( s ) = iΦ$_{2}$( s ) and Im Φ$_{2}$( s ) = 1 − s 2 (1 + s $^{2}$)$_{2}$, 1 2 ˙ x$_{2}$ ( t ) = 1 − t 2 (1 + t $^{2}$)2 (B8b) Im Φ$_{1}$( s ) = − 2 s (1 + s $^{2}$)$_{2}$, 1 2 ˙ x$_{1}$ ( t ) = − 2 t (1 + t $^{2}$)$_{2}$; (B8a) ∑ µ Φ 2 $_{µ}$( s ) = 0 , (B7) Φ$_{1}$( z ) = iΦ$_{2}$( z ) = 2 (i + z ) 2 (B6)