O'SuLLivAN — Points on the Curve of Intersection of ttvo Quadrics. 149 

 2°. For the contour 



- V = - i,>i + u, U = (x)i ~ u from (4). Hence, since ,r = pu and 5, = pU, 

 from (2), 



3°. For the contour 



±- ^ -^ 



U = - Wi + e-w 



e-i ^2 62 



2ye, 



t „ e e =7 = - W2 + «, from (4) 



U=w2-u, ^i=pu, .-. from (3), 



iiT 



since Q, (p are each affected by the factor ei , 



., / , mm//w , . ^ iln , , , , 0M ,j.. 



Changing the sign of 11 in (5) and (6), 



e ((..1 + ^0 =^ - ^^ > ^('"1 + '*) = -J^ ' '/'('•" + ^0 = "i^r- ^'^^ 



6 ('1)2 + m) = m . — , (i (w2 + m) = - i — , li (wi + «) = / . — . (8) 



' ^!6 ^ ' (pu ^ ^ (j)U 



Also, since W3 + « = - (co. + ii-i - m), substituting from (8) and (5) 



n / ^ . (bii , . ., Ou , , . tin ... 



((1)3 + U) = - zwc V- , <h (w3 + u) = il ^~ , \L (013 + «) = --—. (9) 



Writing wi + u for u in (7), ws + « for %i in (8), W3 ^■ w for ?« in (9), 

 0(2(Oi + If) = to, ^(2(u, + m) = - ^u, t\) (2aji + ?t) = - -tpu; 

 (2w2 + 74) = - 6m, (j) (2w2 + «) = <j>i(,, -iP (2(1)2 + «■) = - -ipu; (10) 

 (2a)3 + u) = - On, (j) (2(1)3 + m) = - i^?f, j// (2(1)3 + m) = 'A'?'- 



Thus 2(,), is a period of 6, 2(02 a period of (p, 2(U3 a period of \p. 

 The double periods 4(,)i, 4a)2, 4a)3 are periods of all three. 

 As it tends to zero. 6, <p, \p tends to unity. Hence we have 



0(t)i = 0, <pWi = n, \p(jj\ = m ', 



6*0)2 = Mi </)0)2 = 0, \pW2 = I ', (11) 



f^o)3 = M«. ^(1)3 = il, T^Wi = 0. 

 The functions of the half angles will also be required. 



