The Functional Equation is as follows
Does this equation imply that if s is a zero then 1-s is also a zero provided there are no poles that cancel out any zero?
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Yes.$\text{}\text{}$ – Klaus Oct 18 '19 at 09:28
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@Klaus what if s is a non trivial zero. Even then does the statement hold true? Can you provide a proof? – Rohan Asif Oct 18 '19 at 09:51
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Actually only when $s$ is a non-trivial zero. The trivial zeros are caused by the $\sin$. – Klaus Oct 18 '19 at 09:54
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Of course! Thanks for the clarification – Rohan Asif Oct 18 '19 at 09:55
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You need to locate the zeros/poles of $\sin(\pi s/2)$ (easy) and $\Gamma(1-s)$ ($\Gamma(s+1)=s\Gamma(s)$ indicates the poles, for the lack of zeros you'll need to prove $\Gamma(s)\Gamma(1-s)=\pi/\sin(\pi s)$) and to understand how the multiplication/division of meromorphic functions works. – reuns Oct 19 '19 at 02:11