A.._/|* = ;e,, ■/ 





1-a 



^ (1-e ) 



1 



Since a < 1, the exponent of r, is positive. Hence, as the semicircle is shrunk down onto 

 t = a and r^ -» 0, As -> 0. On the 2-plane, therefore, the two lines must meet at a point. 



3y proceeding in this manner it can be shown that, as t traverses its real axis from 

 - oo to + oo, 3 moves along a broken line with corners corresponding to ^ = a , a . . . a^, at 

 which exterior angles a 77, a, 77 . . . a^n occur. To form a finite polygon, the ends of this 

 broken line must coincide. 



Now the distance between the ends is equal to f_^ (dz/dt) dt along the entire real 

 axis, calculated with avoidance of all the singular points in the manner just described. The 

 value of the integral can be found more easily by the following indirect method. 



Let t trace the following contour, as illustrated in Figure 37. Beginning at a point 

 t = - R where /? is a large positive real number, let t trace the real axis to the point t = R, 

 except that it goes round above the points a , a„ . . . a along small semicircles. Let t then 

 return to its starting-point along a large semicircle of radius /?. On this contour and everywhere 

 inside it, dz/dt is differentiable. Hence, by the Cauchy integral theorem, / (dz/dt) dt around 

 the contour vanishes. But it can be shown that the contribution of the large semicircle decreases 

 to zero 2iS R -* ao. For, on this semicircle, the absolute value of t equals R and is so large 

 that a , a ... a ^ are relatively negligible and may be omitted. Let 



and write 



t = Re , dt = i R e 



Figure 36 — Semicircular path past a singular point. 



61 



