By D.W. Masser
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43 Finally for all ~0, ~i Ir(~0,v1)I which implies proceeds Theorem r(~0,~1) = O as before. 5 and this The argument completes now the p r o o f of III. It is i n t e r e s t i n g ~12/~ ~ e -K to note that the t r a n s c e n d e n c e in the case of c o m p l e x m u l t i p l i c a t i o n follows of from the i d e n t i t y 2ziC/~12 and the t r a n s c e n d e n c e = (B + 2CT)~I/~I of ni/~i + K (cf. 652). We conclude with Weierstrass e v e n this elliptic appears a corollary referring function with to an a r b i t r a r y algebraic invariants; to be new.
Depending (44) only a contradiction It is e a s y integers on if to see without that loss of g e n e r a l i t y . 2zizs and ¢(zl,z2,z3) p ( 1 0 , 1 z , 1 2 , 1 3 ) ( f ( z l , z 2 , z 3 ) ) A° = Io:o l,=o A~:o it~o (~(~zZ I) )A' (P(~2Z 2) )~e 2~il~z~ where the coefficients p(i0,11,12,13) are y e t to be d e t e r m i n e d . 51 Apart f r o m the c h a n g e auxiliary Hence function f r o m the negative in p a r a m e t e r s , as t h a t appearing calculations integers m~, m~, this of ~2] is the in C o a t e s ' we see same paper that [i~ .
Expansion = -24e -2~ z = putting < = O when = O a consequence Im(E2(z)) that is is Izl ~ i Next s = O or -½. < n so , o(n) treat < y = e -~ cannot a zero each of , (n+2) 3 < 2 ( n + l ) ( n + 2 ) ( n + 3 ) < (200)-i from there is know = monotonicity only this one to must in 1 - have turn. 24 ~ s = 0 or If s = O, o ( n ) e -2~nt considerations solution be at 1 2 ~ (l-y) -4 < # vanish. ~(z) possibility E2(it) we < n2 (n+l) (n+2) ( n + 3 ) y n = [ 1 + g(s,t) Thus and 2~s then Ig(s,t) l < 2~ and 2~ns/sin inequalities e -2~% For sin for t = i.