88 THE ROYAL SOCIETY OF CANADA 



and let this dissected plane, including the point at oo , be called X. In the 

 figure the banks are drawn slightly apart and d is replaced by d^, d", d"^ 



A branch of the integrand is selected 

 and the path from x^ to x is confined to X; 

 this branch is finite at a, b, c, d, oo in virtue 

 of the restrictions on a, j8, 7, 8. 



Now consider the map of X on the 

 u-plane. The points a, b, c map into A, B, 

 C; the points d\ d'\ d"^ into D^, D'\ D"^ 

 Points on opposite banks of a line L map 

 ^ into distinct points on the u-plane; the two 



banks of the cut which meet at a at an angle 2ir map into two curves 

 of like form which meet at A at an angle 2aT. This is easily seen by 

 taking h as initial point and connecting it with k by the loop hpqk 

 round a, and remembering that near a 



u - u. 



(x - a) Ko+ Kj (x - a) +K2 (x - a)^+ , Kq + o 



and that (x - a) is multiplied by e after the description of the 



small circle round a. 



Similar considerations apply to the maps of the banks of the other 

 cross-cuts. The boundary, then, of X maps into a curvilinear hexagon 

 AD"^ BD^ CD" A. The two sides which meet at a corner A, are the 

 same curve, but in different positions owing to the rotation at A. 

 Similarly for B, C. The angles at A, B, C are 2aT, 2/37r, 27T, and 

 the sum of the angles at D*, D", D"^ is 28t. The region that maps X 

 does not contain u= œ, and it is unbranched, since du/dx4=o when x is 

 finite and within X, and since x = 00 does not furnish a branch point. 



Let us now pass to the case of the ehiptic integral by putting 

 a =l^=ir = d = ^. We wish to show that the cuts da, db , dc can be 

 so chosen that their maps shall be straight lines. The curvilinear 

 hexagon is now replaced by a triangle D^ D" D"^ whose sides are bisected 

 by A, B, C; e.g. the half sides D" A, D"' A arise from one another by 

 a rotation t about A. 



By hypothesis a circle through three of a, b, c, d will not pass 

 through the fourth. We can, therefore, by a bilinear transformation 

 project any three into 0, 1, 00 and the 4th into some point z off the 

 real axis and discuss the inversion problem for the case 



dx 



u„ = 



1/x (x - 1) (x - z) 



Let z lie in the negative half of the x-plane; also let the lines L (o, z), 

 L (1, z), L (00, z), into which the earlier cuts have been transformed, 



