A BICOMPLEX HURWITZ ZETA-FUNCTION 1. Introduction

30 A. Javtokasζ(z 1 + z 2 i 1 , α)e 2 is the only one possible T-holomorphic continuation. Thiscompletes the proof.Finally, the following theorem conrms that the domain T \{1 + O 2 } isthe best possible.Theorem 8. Let ω 0 ∈ 1 + O 2 , thenlim |ζ(ω, α)| = ∞.ω→ω 0(ω /∈1+O 2 )Proof. Let ω = (z 1 −z 2 i 1 )e 1 +(z 1 +z 2 i 1 )e 2 and ω 0 = (z 0 1 −z0 2 i 1)e 1 +(z 0 1 +z 0 2 i 1)e 2 . By hypothesis, ω 0 ∈ 1 + O 2 . Hence, z 0 1 − z0 2 i 1 = 1 or z 0 1 + z0 2 i 1 = 1.Without loss of generality, let us suppose that z 0 1 − z0 2 i 1 = 1. Now, from theidentity (see [4])( |z1 − z 2 i 1 | 2 + |z 1 + z 2 i 1 | 2 ) 1/2|z 1 + z 2 i 2 | =,2it follows that if, for all z 1 + z 2 i 2 ∈ T, ω → ω 0 then z 1 − z 2 i 1 → 1 andz 1 + z 2 i 1 → z 0 1 − z0 2 i 1. Moreover, ζ(s, α) is meromorphically continuable tothe whole complex plane with only a simple pole at s = 1. Therefore,Thenlim |ζ(z 1 − z 2 i 1 , α)| = ∞.z 1 −z 2 i 1 →1limω→ω 0|ζ(ω, α)| = limω→ω 0|ζ(z 1 − z 2 i 1 , α)e 1 + ζ(z 1 + z 2 i 1 , α)e 2 |(ω /∈1+O 2 )(ω /∈1+O 2 )( |ζ(z1 − z 2 i 1 , α)| 2 + |ζ(z 1 + z 2 i 1 , α)| 2= limω→ω 02(ω /∈1+O 2 )= ∞.This completes the proof.) 1/2Acknowledgements. The author is partially supported by Grant fromLithuanian Foundation of Studies and Science.